ch13 dynamics

The design of conveyors for a bottling plant requires knowledge of the forces that act

on them the ability predict the motion of the Education, bottles they transport. Unpublished Work © 2007 by R. C. Hibbeler. To and be published by to Pearson Prentice Hall, Pearson Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

Kinetics of a Particle: Force and Acceleration CHAPTER OBJECTIVES

• To state Newton’s Laws of Motion and Gravitational Attraction and to define mass and weight.

• To analyze the accelerated motion of a particle using the equation of motion with different coordinate systems.

• To investigate central-force motion and apply it to problems in space mechanics.

13.1 Newton’s Laws of Motion Many of the earlier notions about dynamics were dispelled after 1590 when Galileo performed experiments to study the motions of pendulums and falling bodies. The conclusions drawn from these experiments gave some insight as to the effects of forces acting on bodies in motion. The general laws of motion of a body subjected to forces were not known, however, until 1687, when Isaac Newton first presented three basic laws governing the motion of a particle. In a slightly reworded form, Newton’s three laws of motion are First Law: A particle originally at rest, or moving in a straight line with a constant velocity, will remain in this state provided the particle is not subjected to an unbalanced force. Second Law: A particle acted upon by an unbalanced force F experiences an acceleration a that has the same direction as the force and a magnitude that is directly proportional to the force.* Third Law: The mutual forces of action and reaction between two particles are equal, opposite, and collinear. *Stated another way, the unbalanced force acting on the particle is proportional to the time rate of change the particle’s linear momentum. See footnote † on next page. Hall, Pearson Education, Inc., Upper Saddle River,101 Unpublished Work ©of2007 by R. C. Hibbeler. To be published by Pearson Prentice New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

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The first and third laws were used extensively in developing the concepts of statics. Although these laws are also considered in dynamics, it is Newton’s second law of motion that forms the basis for most of this study, since this law relates the accelerated motion of a particle to the forces that act on it. Measurements of force and acceleration can be recorded in a laboratory so that in accordance with the second law, if a known unbalanced force F is applied to a particle, the acceleration a of the particle may be measured. Since the force and acceleration are directly proportional, the constant of proportionality, m, may be determined from the ratio m = F>a.* The positive scalar m is called the mass of the particle. Being constant during any acceleration, m provides a quantitative measure of the resistance of the particle to a change in its velocity. If the mass of the particle is m, Newton’s second law of motion may be written in mathematical form as F = ma This equation, which is referred to as the equation of motion, is one of the most important formulations in mechanics.* As previously stated, its validity is based solely on experimental evidence. In 1905, however, Albert Einstein developed the theory of relativity and placed limitations on the use of Newton’s second law for describing general particle motion. Through experiments it was proven that time is not an absolute quantity as assumed by Newton; as a result, the equation of motion fails to predict the exact behavior of a particle, especially when the particle’s speed approaches the speed of light 10.3 Gm>s2. Developments of the theory of quantum mechanics by Erwin Schrödinger and others indicate further that conclusions drawn from using this equation are also invalid when particles are the size of an atom and move close to one another. For the most part, however, these requirements regarding particle speed and size are not encountered in engineering problems, so their effects will not be considered in this book.

*Recall that the units of force in the SI system and mass in the FPS system are derived from this equation, where N = kg # m>s2 and slug = lb # s2>ft (see Sec. 1.3 of Statics). If, however, the units of force, mass, length, and time were all selected arbitrarily, then it is necessary to write F = kma, where k (a dimensionless constant) would have to be determined experimentally in order to preserve the equality. *Since m is constant, we can also write F = d1mv2>dt, where mv is the particle’s linear momentum.

Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

13.1 NEWTON’S LAWS OF MOTION

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Newton’s Law of Gravitational Attraction. Shortly after formulating his three laws of motion, Newton postulated a law governing the mutual attraction between any two particles. In mathematical form this law can be expressed as F = G

m1m2 r2

(13–1)

where F = force of attraction between the two particles G = universal constant of gravitation; according to experimental evidence G = 66.73110 -122 m3>1kg # s22 m1 , m2 = mass of each of the two particles r = distance between the centers of the two particles Any two particles or bodies have a mutually attractive gravitational force acting between them. In the case of a particle located at or near the surface of the earth, however, the only gravitational force having any sizable magnitude is that between the earth and the particle. This force is termed the “weight” and, for our purpose, it will be the only gravitational force considered.

Mass and Weight. Mass is a property of matter by which we can

compare the response of one body with that of another. As indicated above, this property manifests itself as a gravitational attraction between two bodies and provides a quantitative measure of the resistance of matter to a change in velocity. It is an absolute quantity since the measurement of mass can be made at any location. The weight of a body, however, is not absolute since it is measured in a gravitational field, and hence its magnitude depends on where the measurement is made. From Eq. 13–1, we can develop a general expression for finding the weight W of a particle having a mass m1 = m. Let m2 be the mass of the earth and r the distance between the earth’s center and the particle. Then, if g = Gm2>r2, we have W = mg By comparison with F = ma, we term g the acceleration due to gravity. For most engineering calculations g is measured at a point on the surface of the earth at sea level, and at a latitude of 45°, which is considered the “standard location.” The mass and weight of a body are measured differently in the SI and FPS systems of units, and the method of defining these units should be thoroughly understood.


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m$ W g (slug)

m (kg)

a $ g (ft/s2)

a $ g (m/s2)

W (lb)

W $ mg (N) SI system

FPS system

(a)

(b)

Fig. 13–1

SI System of Units. In the SI system the mass of the body is specified in kilograms, and the weight must be calculated using the equation of motion, F = ma. Hence, if a body has a mass m (kg) and is located at a point where the acceleration due to gravity is g 1m>s22, then the weight is expressed in newtons as W = mg 1N2, Fig. 13–1a. In particular, if the body is located at the “standard location,” the acceleration due to gravity is g = 9.80665 m>s2. For calculations, the value g = 9.81 m>s2 will be used, so that W = mg 1N2

1g = 9.81 m>s22

(13–2)

Therefore, a body of mass 1 kg has a weight of 9.81 N; a 2-kg body weighs 19.62 N; and so on.

FPS System of Units. In the FPS system the weight of the body is

specified in pounds, and the mass must be calculated from F = ma. Hence, if a body has a weight W (lb) and is located at a point where the acceleration due to gravity is g 1ft>s22 then the mass is expressed in slugs as m = W>g (slug), Fig. 13–1b. Since the acceleration of gravity at the standard location is approximately 32.2 ft>s2 1= 9.81 m>s22, the mass of the body measured in slugs is m =

W 1slug2 1g = 32.2 ft>s22 g

(13–3)

Therefore, a body weighing 32.2 lb has a mass of 1 slug; a 64.4-lb body has a mass of 2 slugs; and so on. Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

13.2 THE EQUATION OF MOTION

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13.2 The Equation of Motion When more than one force acts on a particle, the resultant force is determined by a vector summation of all the forces; i.e., FR = ©F. For this more general case, the equation of motion may be written as ©F = ma

(13–4)

To illustrate application of this equation, consider the particle P shown in Fig. 13–2a, which has a mass m and is subjected to the action of two forces, F1 and F2 . We can graphically account for the magnitude and direction of each force acting on the particle by drawing the particle’s free-body diagram, Fig. 13–2b. Since the resultant of these forces produces the vector ma, its magnitude and direction can be represented graphically on the kinetic diagram, shown in Fig. 13–2c.* The equal sign written between the diagrams symbolizes the graphical equivalency between the free-body diagram and the kinetic diagram; i.e., ©F = ma.†* In particular, note that if FR = ©F = 0, then the acceleration is also zero, so that the particle will either remain at rest or move along a straight-line path with constant velocity. Such are the conditions of static equilibrium, Newton’s first law of motion. F2 a P F1 (a) F2 FR $ (F

P

$

ma

F1 Free-body diagram

Kinetic diagram

(b)

(c)

Fig. 13–2 *Recall the free-body diagram considers the particle to be free of its surroundings and shows all the forces acting on the particle. The kinetic diagram pertains to the particle’s motion as caused by the forces. †The equation of motion can also be rewritten in the form ©F - ma = 0. The vector -ma is referred to as the inertia force vector. If it is treated in the same way as a “force vector,” then the state of “equilibrium” created is referred to as dynamic equilibrium. This method for application is often referred to as the D’Alembert principle, named after the French mathematician Jean le Rond d’Alembert. Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

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Inertial Frame of Reference. Whenever the equation of motion

is applied, it is required that measurements of the acceleration be made from a Newtonian or inertial frame of reference. Such a coordinate system does not rotate and is either fixed or translates in a given direction with a constant velocity (zero acceleration). This definition ensures that the particle’s acceleration measured by observers in two different inertial frames of reference will always be the same. For example, consider the particle P moving with an absolute acceleration a P along a straight path as shown in Fig. 13–3. If the observer is fixed in the inertial x, y frame of reference, this acceleration, a P , will be measured by the observer regardless of the direction and magnitude of the velocity vO of the frame of reference. On the other hand, if the observer is fixed in the noninertial x¿, y¿ frame of reference, Fig. 13–3, the observer will not measure the particle’s acceleration as a P . Instead, if the frame is accelerating at a O¿ the particle will appear to have an acceleration of a P>O¿ = a P - a O¿ . Also, if the frame is rotating, as indicated by the curl, then the particle will appear to move along a curved path, in which case it will appear to have other components of acceleration (see Sec. 16.8). In any case, the measured acceleration from this observer cannot be used in Newton’s law of motion to determine the forces acting on the particle. When studying the motions of rockets and satellites, it is justifiable to consider the inertial reference frame as fixed to the stars, whereas dynamics problems concerned with motions on or near the surface of the earth may be solved by using an inertial frame which is assumed fixed to the earth. Even though the earth both rotates about its own axis and revolves about the sun, the accelerations created by these rotations are relatively small and can be neglected in most computations. y¿

y aO¿ x¿ Noninertial frame of reference

O¿ aP P

Path of particle

vO O

Inertial frame of reference

x

Fig. 13–3 Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

13.2 THE EQUATION OF MOTION

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We are all familiar with the sensation one feels when sitting in a car that is subjected to a forward acceleration. Often people think this is caused by a “force” which acts on them and tends to push them back in their seats; however, this is not the case. Instead, this sensation occurs due to their inertia or the resistance of their mass to a change in velocity. Consider the passenger who is strapped to the seat of a rocket sled. Provided the sled is at rest or is moving with constant velocity, then no force is exerted on his back as shown on his free-body diagram.

W

N2

N1 At rest or constant velocity

When the thrust of the rocket engine causes the sled to accelerate, then the seat upon which he is sitting exerts a force F on him which pushes him forward with the sled. In the photo, notice that the inertia of his head resists this change in motion (acceleration), and so his head moves back against the seat and his face, which is nonrigid, tends to distort.

F

W

N2

N1 Acceleration

Upon deceleration the force of the seatbelt F¿ tends to pull his body to a stop, but his head leaves contact with the back of the seat and his face distorts forward, again due to his inertia or tendency to continue to move forward. No force is pulling him forward, although this is the sensation he receives.

F¿

W

N2

N1 Deceleration Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

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13.3 Equation of Motion for a System of Particles

The equation of motion will now be extended to include a system of n particles isolated within an enclosed region in space, as shown in Fig. 13–4a. In particular, there is no restriction in the way the particles are connected, so the following analysis applies equally well to the motion of a solid, liquid, or gas system. At the instant considered, the arbitrary ith particle, having a mass mi , is subjected to a system of internal forces and a resultant external force. The resultant internal force, represented symbolically as fi , is determined from the forces which the other particles exert on the ith particle. Usually these forces are developed by direct contact, although the summation extends over all n particles within the dashed boundary. The resultant external force Fi represents, for example, the effect of gravitational, electrical, magnetic, or contact forces between the ith particle and adjacent bodies or particles not included within the system. The free-body and kinetic diagrams for the ith particle are shown in Fig. 13–4b. Applying the equation of motion yields ©F = ma;

Fi + fi = mia i

When the equation of motion is applied to each of the other particles of the system, similar equations will result. If all these equations are added together vectorially, we obtain ©Fi + ©fi = ©mia i z Fi fi

i

ri G rG Fi fi

y

x

$

Free-body diagram

Inertial coordinate system

mi ai Kinetic diagram

(b)

(a)


13.3 EQUATION OF MOTION FOR A SYSTEM OF PARTICLES

109

The summation of the internal forces, if carried out, will equal zero, since internal forces between particles all occur in equal but opposite collinear pairs. Consequently, only the sum of the external forces will remain, and therefore the equation of motion, written for the system of particles, becomes ©Fi = ©mia i

(13–5)

If rG is a position vector which locates the center of mass G of the particles, Fig. 13–4a, then by definition of the center of mass, mrG = ©miri , where m = ©mi is the total mass of all the particles. Differentiating this equation twice with respect to time, assuming that no mass is entering or leaving the system, yields ma G = ©mia i Substituting this result into Eq. 13–5, we obtain ©F = ma G

(13–6)

Hence, the sum of the external forces acting on the system of particles is equal to the total mass of the particles times the acceleration of its center of mass G. Since in reality all particles must have a finite size to possess mass, Eq. 13–6 justifies application of the equation of motion to a body that is represented as a single particle.

Important Points • The equation of motion is based on experimental evidence and • • • •

is valid only when applied from an inertial frame of reference. The equation of motion states that the unbalanced force on a particle causes it to accelerate. An inertial frame of reference does not rotate, rather it has axes that either translate with constant velocity or are at rest. Mass is a property of matter that provides a quantitative measure of its resistance to a change in velocity. It is an absolute quantity. Weight is a force that is caused by the earth’s gravitation. It is not absolute; rather it depends on the altitude of the mass from the earth’s surface.


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13.4 Equations of Motion: Rectangular Coordinates

When a particle is moving relative to an inertial x, y, z frame of reference, the forces acting on the particle, as well as its acceleration, may be expressed in terms of their i, j, k components, Fig. 13–5. Applying the equation of motion, we have

z

©F = ma Fz P Fx

Fy

z

x

y

Fig. 13–5

For this equation to be satisfied, the respective i, j, k components on the left side must equal the corresponding components on the right side. Consequently, we may write the following three scalar equations: y

x

©Fx i + ©Fy j + ©Fz k = m1ax i + ay j + az k2

©Fx = max ©Fy = may ©Fz = maz

(13–7)

In particular, if the particle is constrained to move only in the x–y plane, then the first two of these equations are used to specify the motion.

Procedure for Analysis The equations of motion are used to solve problems which require a relationship between the forces acting on a particle and the accelerated motion they cause. Free-Body Diagram. • Select the inertial coordinate system. Most often, rectangular or x, y, z coordinates are chosen to analyze problems for which the particle has rectilinear motion. • Once the coordinates are established, draw the particle’s free-body diagram. Drawing this diagram is very important since it provides a graphical representation that accounts for all the forces 1©F2 which act on the particle, and thereby makes it possible to resolve these forces into their x, y, z components. • The direction and sense of the particle’s acceleration a should also be established. If the senses of its components are unknown, for mathematical convenience assume that they are in the same direction as the positive inertial coordinate axes. • The acceleration may be represented as the ma vector on the kinetic diagram.* • Identify the unknowns in the problem. *It is a convention in this text always to use the kinetic diagram as a graphical aid when developing the proofs and theory. The particle’s acceleration or its components will be shown as blue colored vectors near the free-body diagram in the examples. Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

13.4 EQUATIONS OF MOTION: RECTANGULAR COORDINATES

111

Equations of Motion.

• If the forces can be resolved directly from the free-body diagram, apply • •

•

the equations of motion in their scalar component form. If the geometry of the problem appears complicated, which often occurs in three dimensions, Cartesian vector analysis can be used for the solution. Friction. If a moving particle contacts a rough surface, it may be necessary to use the frictional equation, which relates the coefficient of kinetic friction mk to the magnitudes of the frictional and normal forces Ff and N acting at the surfaces of contact, i.e., Ff = mkN. Remember that Ff always acts on the free-body diagram such that it opposes the motion of the particle relative to the surface it contacts. If the particle is on the verge of relative motion then the coefficient of static friction should be used. Spring. If the particle is connected to an elastic spring having negligible mass, the spring force Fs can be related to the deformation of the spring by the equation Fs = ks. Here k is the spring’s stiffness measured as a force per unit length, and s is the stretch or compression defined as the difference between the deformed length l and the undeformed length l0 , i.e., s = l - l0 .

Kinematics.

• If the velocity or position of the particle is to be found, it will be • • • • •

•

necessary to apply the proper kinematic equations once the particle’s acceleration is determined from ©F = ma. If acceleration is a function of time, use a = dv>dt and v = ds>dt which, when integrated, yield the particle’s velocity and position. If acceleration is a function of displacement, integrate a ds = v dv to obtain the velocity as a function of position. If acceleration is constant, use v = v0 + ac t, s = s0 + v0t + 21 ac t2, v2 = v20 + 2ac1s - s02 to determine the velocity or position of the particle. If the problem involves the dependent motion of several particles, use the method outlined in Sec. 12.9 to relate their accelerations. In all cases, make sure the positive inertial coordinate directions used for writing the kinematic equations are the same as those used for writing the equations of motion; otherwise, simultaneous solution of the equations will result in errors. If the solution for an unknown vector component yields a negative scalar, it indicates that the component acts in the direction opposite to that which was assumed.


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EXAMPLE 13.1 The 50-kg crate shown in Fig. 13–6a rests on a horizontal plane for which the coefficient of kinetic friction is mk = 0.3. If the crate is subjected to a 400-N towing force as shown, determine the velocity of the crate in 3 s starting from rest.

P $ 400 N 30'

SOLUTION Using the equations of motion, we can relate the crate’s acceleration to the force causing the motion. The crate’s velocity can then be determined using kinematics.

(a)

Free-Body Diagram. The weight of the crate is W = mg = 50 kg 19.81 m>s22 = 490.5 N. As shown in Fig. 13–6b, the frictional force has a magnitude F = mkNC and acts to the left, since it opposes the motion of the crate. The acceleration a is assumed to act horizontally, in the positive x direction. There are two unknowns, namely NC and a.

y a x 490.5 N 400 N 30'

F $ 0.3 NC NC (b)

Fig. 13–6

Equations of Motion. Using the data shown on the free-body diagram, we have + ©F = ma ; : 400 cos 30° - 0.3NC = 50a (1) x x + c ©Fy = may ; NC - 490.5 + 400 sin 30° = 0 (2) Solving Eq. 2 for NC , substituting the result into Eq. 1, and solving for a yields NC = 290.5 N a = 5.19 m>s2 Kinematics. Note that the acceleration is constant, since the applied force P is constant. Since the initial velocity is zero, the velocity of the crate in 3 s is + 2 1: v = v0 + act = 0 + 5.19132 Ans. = 15.6 m>s : 490.5 N 400 N 30'

$

50a

F $ 0.3 NC NC

(c)

NOTE: We can also use the alternative procedure of drawing the crate’s free-body and kinetic diagrams, Fig. 13–6c, prior to applying the equations of motion. Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

113


EXAMPLE 13.2 A 10-kg projectile is fired vertically upward from the ground, with an initial velocity of 50 m>s, Fig. 13–7a. Determine the maximum height to which it will travel if (a) atmospheric resistance is neglected; and (b) atmospheric resistance is measured as FD = 10.01v22 N, where v is the speed at any instant, measured in m>s.

z

SOLUTION In both cases the known force on the projectile can be related to its acceleration using the equation of motion. Kinematics can then be used to relate the projectile’s acceleration to its position. Part (a) Free-Body Diagram. As shown in Fig. 13–7b, the projectile’s weight is W = mg = 1019.812 = 98.1 N. We will assume the unknown acceleration a acts upward in the positive z direction.

(a)

Equation of Motion. + c ©Fz = maz ;

z

2

-98.1 = 10a,

a = -9.81 m>s

The result indicates that the projectile, like every object having free-flight motion near the earth’s surface, is subjected to a constant downward acceleration of 9.81 m>s2.

a 98.1 N

Kinematics. Initially, z0 = 0 and v0 = 50 m>s, and at the maximum height z = h, v = 0. Since the acceleration is constant, then

(b)

v2 = v20 + 2ac1z - z02

1+ c 2

0 = 15022 + 21-9.8121h - 02 h = 127 m

Ans. 2

Part (b) Free-Body Diagram. Since the force FD = 10.01v 2 N tends to retard the upward motion of the projectile, it acts downward as shown on the free-body diagram, Fig. 13–7c. Equation of Motion. + c ©Fz = maz ;

a 98.1 N

-0.01v2 - 98.1 = 10a,

a = -0.001v2 - 9.81

Kinematics. Here the acceleration is not constant since FD depends on the velocity. Since a = f1v2, we can relate a to position using 1+ c 2 a dz = v dv;

z FD

1-0.001v2 - 9.812 dz = v dv

(c)

Fig. 13–7

Separating the variables and integrating, realizing that initially z0 = 0, v0 = 50 m>s (positive upward), and at z = h, v = 0, we have h

0

0 v dv = -500 ln1v2 + 98102 ` 2 L0 L50 0.001v + 9.81 50 h = 114 m

dz = -

Ans.

NOTE: The answer indicates a lower elevation than that obtained in part (a) due to atmospheric resistance. Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

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EXAMPLE 13.3

A

B

The baggage truck A shown in the photo has a weight of 900 lb and tows a 550-lb cart B and a 325-lb cart C. For a short time the driving frictional force developed at the wheels of the truck is FA = 140t2 lb, where t is in seconds. If the truck starts from rest, determine its speed in 2 seconds. Also, what is the horizontal force acting on the coupling between the truck and cart B at this instant? Neglect the size of the truck and carts.

C

900 lb

550 lb

325 lb

FA NA

NC

NB (a)

SOLUTION Free-Body Diagram. As shown in Fig. 13–8a, it is the frictional driving force that gives both the truck and carts an acceleration. Here we have considered all three vehicles. Equation of Motion. Only motion in the horizontal direction has to be considered. 900 + 550 + 325 + ©F = ma ; ; 40t = a ba x x 32.2 a = 0.7256t Kinematics. Since the acceleration is a function of time, the velocity of the truck is obtained using a = dv>dt with the initial condition that v0 = 0 at t = 0. We have v

L0 900 lb

T FA NA (b)

Fig. 13–8

2

dv =

L0

0.7256t dt;

2

v = 0.3628t2 ` = 1.45 ft>s 0

Ans.

Free-Body Diagram. In order to determine the force between the truck and cart B, we can consider a free-body diagram of the truck so that we can “expose” the coupling force T as external to the free-body diagram, Fig. 13–8b. Equation of Motion. When t = 2 s, then 900 + ©F = ma ; ; 40122 - T = a b[0.7256122] x x 32.2 T = 39.4 lb Ans. NOTE: Try and obtain this same result by considering a free-body diagram of carts B and C.


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EXAMPLE 13.4 A smooth 2-kg collar C, shown in Fig. 13–9a, is attached to a spring having a stiffness k = 3 N>m and an unstretched length of 0.75 m. If the collar is released from rest at A, determine its acceleration and the normal force of the rod on the collar at the instant y = 1 m. SOLUTION Free-Body Diagram. The free-body diagram of the collar when it is located at the arbitrary position y is shown in Fig. 13–9b. Note that the weight is W = 219.812 = 19.62 N. Furthermore, the collar is assumed to be accelerating so that “a” acts downward in the positive y direction. There are four unknowns, namely, NC , Fs , a, and u.

-NC + Fs cos u = 0

(1)

+ T ©Fy = may ;

19.62 - Fs sin u = 2a

(2)

From Eq. 2 it is seen that the acceleration depends on the magnitude and direction of the spring force. Solution for NC and a is possible once Fs and u are known. The magnitude of the spring force is a function of the stretch s of the spring; i.e., Fs = ks. Here the unstretched length is AB = 0.75 m, Fig. 13–9a; therefore, Since k = 3 N>m, then

s = CB - AB = 4y2 + 10.7522 - 0.75.

Fs = ks = 3 A 4y2 + 10.7522 - 0.75 B

u

y

B

k $ 3 N/m

C

Equations of Motion. + ©F = ma ; : x x

0.75 m

A

(a)

a

x

19.62 N

y (b)

Fs

u NC

Fig. 13–9

(3)

From Fig. 13–9a, the angle u is related to y by trigonometry. tan u =

y 0.75

(4)

Substituting y = 1 m into Eqs. 3 and 4 yields Fs = 1.50 N and u = 53.1°. Substituting these results into Eqs. 1 and 2, we obtain NC = 0.900 N

Ans. 2

a = 9.21 m>s T

Ans.

NOTE: This is not a case of constant acceleration, since the spring force

changes both its magnitude and direction as the collar moves downward.


116

C H A P T E R 13

KINETICS

OF A


AND


EXAMPLE 13.5

Datum sA C

sB

A B (a)

The 100-kg block A shown in Fig. 13–10a is released from rest. If the masses of the pulleys and the cord are neglected, determine the speed of the 20-kg block B in 2 s. SOLUTION Free-Body Diagrams. Since the mass of the pulleys is neglected, then for pulley C, ma = 0 and we can apply ©Fy = 0 as shown in Fig. 13–10b. The free-body diagrams for blocks A and B are shown in Fig. 13–10c and d, respectively. One can see that for A to remain static requires T = 490.5 N, whereas for B to remain static requires T = 196.2 N. Hence A will move down while B moves up. Here we will assume both blocks accelerate downward, in the direction of +sA and +sB . The three unknowns are T, aA , and aB . Equations of Motion. Block A (Fig. 13–10c):

T T

+ T ©Fy = may ;

981 - 2T = 100aA

(1)

196.2 - T = 20aB

(2)

Block B (Fig. 13–10d): + T ©Fy = may ; 2T (b)

2T

aA sA 981 N (c) T

aB

sB

196.2 N (d)

Fig. 13–10

Kinematics. The necessary third equation is obtained by relating aA to aB using a dependent motion analysis. Using the technique developed in Sec. 12.9, the coordinates sA and sB measure the positions of A and B from the fixed datum, Fig. 13–10a. It is seen that 2sA + sB = l where l is constant and represents the total vertical length of cord. Differentiating this expression twice with respect to time yields (3) 2aA = -aB Notice that in writing Eqs. 1 to 3, the positive direction was always assumed downward. It is very important to be consistent in this assumption since we are seeking a simultaneous solution of equations. The solution yields T = 327.0 N aA = 3.27 m>s2 aB = -6.54 m>s2 Hence when block A accelerates downward, block B accelerates upward. Since aB is constant, the velocity of block B in 2 s is thus 1+ T2

v = v0 + aBt = 0 + 1-6.542122 = -13.1 m>s The negative sign indicates that block B is moving upward.

Ans.


PROBLEMS

117

PROBLEMS 13–1. Determine the gravitational attraction between two spheres which are just touching each other. Each sphere has a mass of 10 kg and a radius of 200 mm. 13–2. By using an inclined plane to retard the motion of a falling object, and thus make the observations more accurate, Galileo was able to determine experimentally that the distance through which an object moves in free fall is proportional to the square of the time for travel. Show that this is the case, i.e., s r t2, by determining the time tB , tC , and tD needed for a block of mass m to slide from rest at A to points B, C, and D, respectively. Neglect the effects of friction.

sD $

9m

sC $

80 km/h

4m 2m sB $ A

B

Prob. 13–4

C D

*13–4. A crate having a mass of 60 kg falls horizontally off the back of a truck which is traveling at 80 km>h. Determine the coefficient of kinetic friction between the road and the crate if the crate slides 45 m on the ground with no tumbling along the road before coming to rest. Assume that the initial speed of the crate along the road is 80 km>h.

20'

Probs. 13–1/2 13–3. The 300-kg bar B, originally at rest, is being towed over a series of small rollers. Determine the force in the cable when t = 5 s, if the motor M is drawing in the cable for a short time at a rate of v = 10.4t22 m>s, where t is in seconds 10 … t … 6 s2. How far does the bar move in 5 s? Neglect the mass of the cable, pulley, and the rollers.

13–5. The crane lifts the 700-kg bin with an initial acceleration of 3 m>s2. Determine the force in each of the supporting cables due to this motion.

M

B 5

A

v

5

4 4 3

3

C

B

Prob. 13–3

Prob. 13–5


118

C H A P T E R 13

KINETICS

OF A


13–6. The baggage truck A has a mass of 800 kg and is used to pull the two cars, each with mass 300 kg. If the tractive force F on the truck is F = 480 N, determine the initial acceleration of the truck. What is the acceleration of the truck if the coupling at C suddenly fails? The car wheels are free to roll. Neglect the mass of the wheels.

AND


*13–8. The 200-kg crate is suspended from the cable of a crane. Determine the force in the cable when t = 2 s if the crate is moving upward with (a) a constant velocity of 2 m>s, and (b) a speed of v = 10.2t2 + 22 m>s, where t is in seconds.

A C

B F

v

Prob. 13–6 Prob. 13–8 13–7. The 500-kg fuel assembly for a nuclear reactor is being lifted out from the core of the nuclear reactor using the pulley system shown. It is hoisted upward with a constant acceleration such that s = 0 and v = 0 when t = 0, and s = 2.5 m when t = 1.5 s. Determine the tension in the cable at A during the motion.

13–9. The elevator E has a mass of 500 kg, and the counterweight at A has a mass of 150 kg. If the motor supplies a constant force of 5 kN on the cable at B, determine the speed of the elevator when t = 3 s, starting from rest. Neglect the mass of the pulleys and cable. 13–10. The elevator E has a mass of 500 kg and the counterweight at A has a mass of 150 kg. If the elevator attains a speed of 10 m>s after it rises 40 m, determine the constant force developed in the cable at B. Neglect the mass of the pulleys and cable.

A

s

E B A

Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, Prob. 13–7 This publication is protected by Copyright and written permission Probs. should 13–9/10 New Jersey. All rights reserved. be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

PROBLEMS 13–11. The water-park ride consists of an 800-lb sled which slides from rest down the incline and then into the pool. If the frictional resistance on the incline is Fr = 30 lb, and in the pool for a short distance Fr = 80 lb, determine how fast the sled is traveling when s = 5 ft.

119

13–14. Each of the two blocks has a mass m. The coefficient of kinetic friction at all surfaces of contact is m. If a horizontal force P moves the bottom block, determine the acceleration of the bottom block in each case.

B P

A (a) B

100 ft

P

A (b)

Prob. 13–14 Prob. 13–11

100 ft

s

*13–12. A car of mass m is traveling at a slow velocity v0 . If it is subjected to the drag resistance of the wind, which is proportional to its velocity, i.e., FD = kv, determine the distance and the time the car will travel before its velocity becomes 0.5v0 . Assume no other frictional forces act on the car.

13–15. The driver attempts to tow the crate using a rope that has a tensile strength of 200 lb. If the crate is originally at rest and has a weight of 500 lb, determine the greatest acceleration it can have if the coefficient of static friction between the crate and the road is ms = 0.4, and the coefficient of kinetic friction is mk = 0.3.

30'

v

FD

Prob. 13–12 13–13. Determine the normal force the 10-kg crate A exerts on the smooth cart if the cart is given an acceleration of a = 2 m>s2 down the plane. Also, what is the acceleration of the crate? Set u = 30°.

Prob. 13–15 *13–16. The 3.5-Mg engine is suspended from a 500-kg spreader beam and hoisted by a crane which gives it an acceleration of 4 m>s2 when it has a velocity of 2 m>s. Determine the force in chains AC and AD during the lift.

C 60'

A

60'

A

B D

E

a

u Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, Prob. 13–13 Prob. 13–16 New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

120

C H A P T E R 13

KINETICS

OF A


13–17. The bullet of mass m is given a velocity due to gas pressure caused by the burning of powder within the chamber of the gun. Assuming this pressure creates a force of F = F0 sin1pt>t02 on the bullet, determine the velocity of the bullet at any instant it is in the barrel. What is the bullet’s maximum velocity? Also, determine the position of the bullet in the barrel as a function of time.

AND


13–19. A 40-lb suitcase slides from rest 20 ft down the smooth ramp. Determine the point where it strikes the ground at C. How long does it take to go from A to C? *13–20. Solve Prob. 13–19 if the suitcase has an initial velocity down the ramp of vA = 10 ft>s and the coefficient of kinetic friction along AB is mk = 0.2. A

20 ft F 30'

F0

4 ft t0

t

C R

Prob. 13–17

13–18. The 400-lb cylinder at A is hoisted using the motor and the pulley system shown. If the speed of point B on the cable is increased at a constant rate from zero to vB = 10 ft>s in t = 5 s, determine the tension in the cable at B to cause the motion.

B

Probs. 13–19/20

13–21. The winding drum D is drawing in the cable at an accelerated rate of 5 m>s2. Determine the cable tension if the suspended crate has a mass of 800 kg.

B

D

vA

A

Prob. 13–18

Prob. 13–21


PROBLEMS 13–22. At a given instant the 5-lb weight A is moving downward with a speed of 4 ft>s. Determine its speed 2 s later. Block B has a weight of 6 lb, and the coefficient of kinetic friction between it and the horizontal plane is mk = 0.3. Neglect the mass of the pulleys and cord.

B

121

*13–24. At a given instant the 10-lb block A is moving downward with a speed of 6 ft>s. Determine its speed 2 s later. Block B has a weight of 4 lb, and the coefficient of kinetic friction between it and the horizontal plane is mk = 0.2. Neglect the mass of the pulleys and cord.

B

A

A

Prob. 13–22

Prob. 13–24

13–23. A force F = 15 lb is applied to the cord. Determine how high the 30-lb block A rises in 2 s starting from rest. Neglect the weight of the pulleys and cord.

13–25. A freight elevator, including its load, has a mass of 500 kg. It is prevented from rotating due to the track and wheels mounted along its sides. If the motor M develops a constant tension T = 1.50 kN in its attached cable, determine the velocity of the elevator when it has moved upward 3 m starting from rest. Neglect the mass of the pulleys and cables.

B

C

M

F

A

Prob. 13–23

Prob. 13–25


122

C H A P T E R 13

KINETICS

OF A


13–26. At the instant shown the 100-lb block A is moving down the plane at 5 ft>s while being attached to the 50-lb block B. If the coefficient of kinetic friction is mk = 0.2, determine the acceleration of A and the distance A slides before it stops. Neglect the mass of the pulleys and cables.

A

5 4

AND


*13–28. The 400-kg mine car is hoisted up the incline using the cable and motor M. For a short time, the force in the cable is F = 13200t22 N, where t is in seconds. If the car has an initial velocity v1 = 2 m>s when t = 0, determine its velocity when t = 2 s. 13–29. The 400-kg mine car is hoisted up the incline using the cable and motor M. For a short time, the force in the cable is F = 13200t22 N, where t is in seconds. If the car has an initial velocity v1 = 2 m>s at s = 0 and t = 0, determine the distance it moves up the plane when t = 2 s.

3

M C

v1 $ 2 m/s

D

B 17

8

15

Prob. 13–26

Probs. 13–28/29

13–27. The safe S has a weight of 200 lb and is supported by the rope and pulley arrangement shown. If the end of the rope is given to a boy B of weight 90 lb, determine his acceleration if in the confusion he doesn’t let go of the rope. Neglect the mass of the pulleys and rope.

"13–30. The tanker has a weight of 80011062 lb and is traveling forward at v0 = 3 ft>s in still water when the engines are shut off. If the drag resistance of the water is proportional to the speed of the tanker at any instant and can be approximated by FD = 140011032v2 lb, where v is in ft>s, determine the time needed for the tanker’s speed to become 1.5 ft>s. Given the initial velocity of v0 = 3 ft>s through what distance must the tanker travel before it stops?

S v B

Prob. 13–27

FD

Prob. 13–30


PROBLEMS 13–31. The spring mechanism is used as a shock absorber for railroad cars. Determine the maximum compression of spring HI if the fixed bumper R of a 5-Mg railroad car, rolling freely at 2 m>s, strikes the plate P. Bar AB slides along the guide paths CE and DF. The ends of all springs are attached to their respective members and are originally unstretched.

123

13–34. The boy having a weight of 80 lb hangs uniformly from the bar. Determine the force in each of his arms in t = 2 s if the bar is moving upward with (a) a constant velocity of 3 ft>s, and (b) a speed of v = 14t22 ft>s, where t is in seconds.

k $ 80 kN/m B

F

D I

H P

R

C E

A

k¿ $ 160 kN/m

Prob. 13–34 2 m/s

k $ 80 kN/m

Prob. 13–31 *13–32. The 2-kg collar C is free to slide along the smooth shaft AB. Determine the acceleration of collar C if (a) the shaft is fixed from moving, (b) collar A, which is fixed to shaft AB, moves downward at constant velocity along the vertical rod, and (c) collar A is subjected to a downward acceleration of 2 m>s2. In all cases, the collar moves in the plane.

13–35. The 10-kg block A rests on the 50-kg plate B in the position shown. Neglecting the mass of the rope and pulley, and using the coefficients of kinetic friction indicated, determine the time needed for block A to slide 0.5 m on the plate when the system is released from rest.

13–33. The 2-kg collar C is free to slide along the smooth shaft AB. Determine the acceleration of collar C if collar A is subjected to an upward acceleration of 4 m>s2. B

45'

0.5 m

C mAB $ 0.2

A

C

A B

mBC $ 0.1

30'

Probs. 13–32/33

Prob. 13–35


124

C H A P T E R 13

KINETICS

OF A


*13–36. Determine the acceleration of block A when the system is released. The coefficient of kinetic friction and the weight of each block are indicated. Neglect the mass of the pulleys and cord.

AND


13–38. An electron of mass m is discharged with an initial horizontal velocity of v0 . If it is subjected to two fields of force for which Fx = F0 and Fy = 0.3F0 , where F0 is constant, determine the equation of the path, and the speed of the electron at any time t.

y

++++++++++++++ ++++++++++++++

A

v0 20 lb B

80 lb

mk $ 0.2

60'

Prob. 13–36

13–37. The conveyor belt is moving at 4 m>s. If the coefficient of static friction between the conveyor and the 10-kg package B is ms = 0.2, determine the shortest time the belt can stop so that the package does not slide on the belt.

x

Prob. 13–38

13–39. The conveyor belt delivers each 12-kg crate to the ramp at A such that the crate’s speed is vA = 2.5 m>s, directed down along the ramp. If the coefficient of kinetic friction between each crate and the ramp is mk = 0.3, determine the speed at which each crate slides off the ramp at B. Assume that no tipping occurs. Take u = 30°. vA $ 2.5 m/s

B

A 3m

u

Prob. 13–37

B

Prob. 13–39


125

PROBLEMS *13–40. A parachutist having a mass m opens his parachute from an at-rest position at a very high altitude. If the atmospheric drag resistance is FD = kv2, where k is a constant, determine his velocity when he has fallen for a time t. What is his velocity when he lands on the ground? This velocity is referred to as the terminal velocity, which is found by letting the time of fall t : q .

13–43. Blocks A and B each have a mass m. Determine the largest horizontal force P which can be applied to B so that A will not slip up B.The coefficient of static friction between A and B is ms . Neglect any friction between B and C.

A

P

u B C

FD

Probs. 13–42/43 v

*13–44. Each of the three plates has a mass of 10 kg. If the coefficients of static and kinetic friction at each surface of contact are ms = 0.3 and mk = 0.2, respectively, determine the acceleration of each plate when the three horizontal forces are applied.

Prob. 13–40 18 N

13–41. Block B rests on a smooth surface. If the coefficients of static and kinetic friction between A and B are ms = 0.4 and mk = 0.3, respectively, determine the acceleration of each block if someone pushes horizontally on block A with a force of (a) F = 6 lb, and (b) F = 50 lb.

D 100 N

C 15 N

B A

Prob. 13–44

F

A 20 lb B 30 lb

13–45. Crate B has a mass m and is released from rest when it is on top of cart A, which has a mass 3m. Determine the tension in cord CD needed to hold the cart from moving while B is sliding down A. Neglect friction. B

Prob. 13–41

13–42. Blocks A and B each have a mass m. Determine the largest horizontal force P which can be applied to B so that A will not move relative to B. All surfaces are smooth.

D

C

u

A

Prob. 13–45


126

C H A P T E R 13

KINETICS

OF A


"13–46. The tractor is used to lift the 150-kg load B with the 24-m-long rope, boom, and pulley system. If the tractor is traveling to the right at a constant speed of 4 m>s, determine the tension in the rope when sA = 5 m. When sA = 0, sB = 0. 13–47. The tractor is used to lift the 150-kg load B with the 24-m-long rope, boom, and pulley system. If the tractor is traveling to the right with an acceleration of 3 m>s2 and has a velocity of 4 m>s at the instant sA = 5 m, determine the tension in the rope at this instant. When sA = 0, sB = 0.

AND


13–49. Block A has a mass mA and is attached to a spring having a stiffness k and unstretched length l0 . If another block B, having a mass mB , is pressed against A so that the spring deforms a distance d, determine the distance both blocks slide on the smooth surface before they begin to separate. What is their velocity at this instant? 13–50. Block A has a mass mA and is attached to a spring having a stiffness k and unstretched length l0 . If another block B, having a mass mB , is pressed against A so that the spring deforms a distance d, show that for separation to occur it is necessary that d 7 2mkg1mA + mB2>k, where mk is the coefficient of kinetic friction between the blocks and the ground. Also, what is the distance the blocks slide on the surface before they separate?

12 m

sB

k B

A

B

A sA

Probs. 13–49/50

Probs. 13–46/47 *13–48. Block B has a mass m and is hoisted using the cord and pulley system shown. Determine the magnitude of force F as a function of the block’s vertical position y so that when F is applied the block rises with a constant acceleration aB . Neglect the mass of the cord and pulleys.

13–51. The block A has a mass mA and rests on the pan B, which has a mass mB . Both are supported by a spring having a stiffness k that is attached to the bottom of the pan and to the ground. Determine the distance d the pan should be pushed down from the equilibrium position and then released from rest so that separation of the block will take place from the surface of the pan at the instant the spring becomes unstretched.

2m

1m

A B B

y d

k

A

Prob. 13–48

Prob. 13–51


13.5 EQUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES

127

13.5 Equations of Motion:

Normal and Tangential Coordinates

When a particle moves over a curved path which is known, the equation of motion for the particle may be written in the tangential, normal, and binormal directions. We have ©F = ma ©Ftut + ©Fnun + ©Fbub = ma t + ma n Here ©Ft , ©Fn , ©Fb represent the sums of all the force components acting on the particle in the tangential, normal, and binormal directions, respectively, Fig. 13–11. Note that there is no motion of the particle in the binormal direction, since the particle is constrained to move along the path. The above equation is satisfied provided ©Ft = mat ©Fn = man ©Fb = 0

(13–8)

Recall that at 1= dv>dt2 represents the time rate of change in the magnitude of velocity. Consequently, if ©Ft acts in the direction of motion, the particle’s speed will increase, whereas if it acts in the opposite direction, the particle will slow down. Likewise, an 1= v2>r2 represents the time rate of change in the velocity’s direction. Since this vector always acts in the positive n direction, i.e., toward the path’s center of curvature, then ©Fn , which causes a n , also acts in this direction. For example, when the particle is constrained to travel in a circular path with a constant speed, there is a normal force exerted on the particle by the constraint in order to change the direction of the particle’s velocity 1a n2. Since this force is always directed toward the center of the path, it is often referred to as the centripetal force.

The centrifuge is used to subject a passenger to very large normal accelerations caused by high rotations. Realize that these accelerations are caused by the unbalanced normal force exerted on the passenger by the seat of the centrifuge.

b (Fbub

O

t n

(Fnun

(Ftut P



128

C H A P T E R 13

KINETICS

OF A


AND


Procedure for Analysis When a problem involves the motion of a particle along a known curved path, normal and tangential coordinates should be considered for the analysis since the acceleration components can be readily formulated. The method for applying the equations of motion, which relate the forces to the acceleration, has been outlined in the procedure given in Sec. 13.4. Specifically, for t, n, b coordinates it may be stated as follows: Free-Body Diagram.

• Establish the inertial t, n, b coordinate system at the particle and draw the particle’s free-body diagram.

• The particle’s normal acceleration an always acts in the positive • •

n direction. If the tangential acceleration a t is unknown, assume it acts in the positive t direction. Identify the unknowns in the problem.

Equations of Motion. • Apply the equations of motion, Eqs. 13–8. Kinematics.

• Formulate the tangential and normal components of •

acceleration; i.e., at = dv>dt or at = v dv>ds and an = v2>r. If the path is defined as y = f1x2, the radius of curvature at the point where the particle is located can be obtained from r = [1 + 1dy>dx22]3>2> ƒ d2y>dx2 ƒ .


129


EXAMPLE 13.6 Determine the banking angle u for the race track so that the wheels of the racing cars shown in Fig. 13–12a will not have to depend upon friction to prevent any car from sliding up or down the track. Assume the cars have negligible size, a mass m, and travel around the curve of radius r with a speed v.

u

(a) b

SOLUTION Before looking at the following solution, give some thought as to why it should be solved using t, n, b coordinates. Free-Body Diagram. As shown in Fig. 13–12b, and as stated in the problem, no frictional force acts on the car. Here NC represents the resultant of the ground on all four wheels. Since an can be calculated, the unknowns are NC and u. Equations of Motion. Using the n, b axes shown, + ©F = ma ; : n n + c ©Fb = 0;

an

NC

n

u W $ mg

v2 NC sin u = m r

(1)

NC cos u - mg = 0

(2)

(b)

Fig. 13–12

Eliminating NC and m from these equations by dividing Eq. 1 by Eq. 2, we obtain v2 tan u = gr v2 u = tan-1 a b Ans. gr NOTE: The result is independent of the mass of the car. Also, a force summation in the tangential direction is of no consequence to the solution. If it were considered, then at = dv>dt = 0, since the car moves with constant speed. A further analysis of this problem is discussed in Prob. 21–48. Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

130

C H A P T E R 13

KINETICS

OF A


AND


EXAMPLE 13.7 The 3-kg disk D is attached to the end of a cord as shown in Fig. 13–13a. The other end of the cord is attached to a ball-and-socket joint located at the center of a platform. If the platform is rotating rapidly, and the disk is placed on it and released from rest as shown, determine the time it takes for the disk to reach a speed great enough to break the cord. The maximum tension the cord can sustain is 100 N, and the coefficient of kinetic friction between the disk and the platform is mk = 0.1. Motion of platform

D 1m

(a)

SOLUTION Free-Body Diagram. The frictional force has a magnitude F = mkND = 0.1ND and a sense of direction that opposes the relative motion of the disk with respect to the platform. It is this force that gives the disk a tangential component of acceleration causing v to increase, thereby causing T to increase until it reaches 100 N. The weight of the disk is W = 319.812 = 29.43 N. Since an can be related to v, the unknowns are ND , at , and v.

b

29.43 N F $ 0.1 ND t

at

Equations of Motion.

T

ND (b)

Fig. 13–13

©Fn = man ; an

n

©Ft = mat ; ©Fb = 0;

v2 b 1 0.1ND = 3at T = 3a

ND - 29.43 = 0

(1) (2) (3)

Setting T = 100 N, Eq. 1 can be solved for the critical speed vcr of the disk needed to break the cord. Solving all the equations, we obtain ND = 29.43 N at = 0.981 m>s2 vcr = 5.77 m>s Kinematics. Since at is constant, the time needed to break the cord is vcr = v0 + att 5.77 = 0 + 10.9812t t = 5.89 s

Ans.


131


EXAMPLE 13.8 Design of the ski jump shown in the photo requires knowing the type of forces that will be exerted on the skier and his approximate trajectory. If in this case the jump can be approximated by the parabola shown in Fig. 13–14a, determine the normal force on the 150-lb skier the instant he arrives at the end of the jump, point A, where his velocity is 65 ft>s. Also, what is his acceleration at this point? SOLUTION Why consider using n, t coordinates to solve this problem?

y y$

Free-Body Diagram. The free-body diagram for the skier when he is at A is shown in Fig. 13–14b. Since the path is curved, there are two components of acceleration, a n and a t . Since an can be calculated, the unknowns are at and NA .

x 200 ft

A

Equations of Motion. + c ©Fn = man ;

(a)

2

NA - 150 =

+ ©F = ma ; ; t t

0 =

150 1652 a b r 32.2

(1)

150 a 32.2 t

(2)

[1 + 1dy>dx22]3>2 ƒ d2y>dx2 ƒ

`

x=0

=

n an

The radius of curvature r for the path must be determined at point 1 1 1 A(0, -200 ft). Here y = 200 x2 - 200, dy>dx = 100 x, d2y>dx2 = 100 , so that at x = 0, r =

1 x2 # 200 200

[1 + 1022]3>2 1 ƒ 100 ƒ

t

at

= 100 ft

Substituting into Eq. 1 and solving for NA , we have NA = 347 lb

150 lb

(b)

NA

Ans. Fig. 13–14

Kinematics. From Eq. 2, at = 0 Thus, an =

16522 v2 = = 42.2 ft>s2 r 100

aA = an = 42.2 ft>s2 c

Ans.

NOTE: Apply the equation of motion in the y direction and show that when the skier is in mid air the acceleration is 32.2 ft>s2. Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

132

C H A P T E R 13

KINETICS

OF A



AND

EXAMPLE 13.9 v0 $ 1 m/s ds $ 0.5 du

SOLUTION Free-Body Diagram. The free-body diagram for a package, when it is located at the general position u, is shown in Fig. 13–15b.The package must have a tangential acceleration a t , since its speed is always increasing as it slides downward.The weight is W = 219.812 = 19.62 N. Specify the three unknowns. Equations of Motion. v2 +b©Fn = man; -NB + 19.62 cos u = 2 (1) 0.5 +R©Ft = mat; 19.62 sin u = 2at (2)

u du r $ 0.5 m

(a)

NB an

at

u 19.62 N

t

n (b)

Fig. 13–15

Packages, each having a mass of 2 kg, are delivered from a conveyor to a smooth circular ramp with a velocity of v0 = 1 m>s as shown in Fig. 13–15a. If the effective radius of the ramp is 0.5 m, determine the angle u = umax at which each package begins to leave the surface.

At the instant u = umax , the package leaves the surface of the ramp so that NB = 0. Therefore, there are three unknowns, v, at , and u. Kinematics. The third equation for the solution is obtained by noting that the magnitude of tangential acceleration at may be related to the speed of the package v and the angle u. Since at ds = v dv and ds = r du = 0.5 du, Fig. 13–15a, we have v dv at = (3) 0.5 du To solve, substitute Eq. 3 into Eq. 2 and separate the variables. This gives v dv = 4.905 sin u du Integrate both sides, realizing that when u = 0°, v0 = 1 m>s. v

L1

v dv = 4.905

u

L0°

sin u du

u v2 v ` = -4.905 cos u ` ; v2 = 9.8111 - cos u2 + 1 2 1 0°

Substituting into Eq. 1 with NB = 0 and solving for cos umax yields 2 19.62 cos umax = [9.8111 - cos umax2 + 1] 0.5 43.24 cos umax = 58.86 umax = 42.7° Ans. NOTE: The speed of the package is increasing because its tangential acceleration is increasing with u, Eq. 2. Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

133

PROBLEMS

PROBLEMS *13–52. Determine the mass of the sun, knowing that the distance from the earth to the sun is 149.611062 km. Hint: Use Eq. 13–1 to represent the force of gravity acting on the earth. 13–53. The 1.40-Mg helicopter is traveling at a constant speed of 40 m>s along the horizontal curved path while banking at u = 30°. Determine the force acting normal to the blade, i.e., in the y¿ direction, and the radius of curvature of the path.

13–57. The 600-kg wrecking ball is suspended from the crane by a cable having a negligible mass. If the ball has a speed v = 8 m>s at the instant it is at its lowest point, u = 0°, determine the tension in the cable at this instant. Also, determine the angle u to which the ball swings before it stops.

13–54. The 1.40-Mg helicopter is traveling at a constant speed of 33 m>s along the horizontal curved path having a radius of curvature of r = 300 m. Determine the force the blade exerts on the frame and the bank angle u. y y¿

u

u

12 m

r

Probs. 13–53/54 13–55. The plane is traveling at a constant speed of 800 ft/s along the curve y = 20110-62x2 + 5000, where x and y are in feet. If the pilot has a weight of 180 lb, determine the normal and tangential components of the force the seat exerts on the pilot when the plane is at its lowest point. *13–56. The jet plane is traveling at a constant speed of 1000 ft>s along the curve y = 20110-62x2 + 5000, where x and y are in feet. If the pilot has a weight of 180 lb, determine the normal and tangential components of the force the seat exerts on the pilot when y = 10 000 ft.

Prob. 13–57

13–58. Prove that if the block is released from rest at point B of a smooth path of arbitrary shape, the speed it attains when it reaches point A is equal to the speed it attains when it falls freely through a distance h; i.e., v = 22gh.

y B

y $ 20(10#6)x2 " 5000

h A

x

Probs. 13–55/56

Prob. 13–58


134

C H A P T E R 13

KINETICS

OF A


13–59. The sled and rider have a total mass of 80 kg and start from rest at A(10 m, 0). If the sled descends the smooth slope, which may be approximated by a parabola, determine the normal force that the ground exerts on the sled at the instant it arrives at point B. Neglect the size of the sled and rider. Hint: Use the result of Prob. 13–58. *13–60. The sled and rider have a total mass of 80 kg and start from rest at A(10 m, 0). If the sled descends the smooth slope, which may be approximated by a parabola, determine the normal force that the ground exerts on the sled at the instant it arrives at point C. Neglect the size of the sled and rider. Hint: Use the result of Prob. 13–58.

AND


13–63. If the crest of the hill has a radius of curvature r = 200 ft, determine the maximum constant speed at which the car can travel over it without leaving the surface of the road. Neglect the size of the car in the calculation. The car has a weight of 3500 lb.

v

r $ 200 ft

y

y$

1 20

x2 # 5 A

5m

C 2m

x

Prob. 13–63

B 10 m

Probs. 13–59/60

13–61. At the instant u = 60°, the boy’s center of mass G has a downward speed vG = 15 ft>s. Determine the rate of increase in his speed and the tension in each of the two supporting cords of the swing at this instant. The boy has a weight of 60 lb. Neglect his size and the mass of the seat and cords.

*13–64. The airplane, traveling at a constant speed of 50 m>s, is executing a horizontal turn. If the plane is banked at u = 15°, when the pilot experiences only a normal force on the seat of the plane, determine the radius of curvature r of the turn. Also, what is the normal force of the seat on the pilot if he has a mass of 70 kg.

13–62. At the instant u = 60°, the boy’s center of mass G is momentarily at rest. Determine his speed and the tension in each of the two supporting cords of the swing when u = 90°. The boy has a weight of 60 lb. Neglect his size and the mass of the seat and cords. u

10 ft

u

r

G

Unpublished Work © 2007 by R. C. Hibbeler. To be13–61/62 published by Pearson Prentice Hall, PearsonProb. Education, Probs. 13–64Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

135

PROBLEMS 13–65. The 150-lb man lies against the cushion for which the coefficient of static friction is ms = 0.5. Determine the resultant normal and frictional forces the cushion exerts on him if, due to rotation about the z axis, he has a constant speed v = 20 ft>s. Neglect the size of the man.Take u = 60°.

*13–68. The 200-kg snowmobile with passenger is traveling down the hill at a constant speed of 6 m>s. Determine the resultant normal force and the resultant frictional force exerted on the tracks at the instant it reaches point A. Neglect the size of the snowmobile.

13–66. The 150-lb man lies against the cushion for which the coefficient of static friction is ms = 0.5. If he rotates about the z axis with a constant speed v = 30 ft>s, determine the smallest angle u of the cushion at which he will begin to slip off.

13–69. The 200-kg snowmobile with passenger is traveling down the hill such that when it is at point A, it is traveling at 4 m>s and increasing its speed at 2 m>s2. Determine the resultant normal force and the resultant frictional force exerted on the tracks at this instant. Neglect the size of the snowmobile.

z

y

x 8 ft G

5m A

u

10 m

y $ #5(10#3)x3

Probs. 13–68/69

Probs. 13–65/66 13–67. Determine the constant speed of the passengers on the amusement-park ride if it is observed that the supporting cables are directed at u = 30° from the vertical. Each chair including its passenger has a mass of 80 kg. Also, what are the components of force in the n, t, and b directions which the chair exerts on a 50-kg passenger during the motion?

13–70. A collar having a mass of 0.75 kg and negligible size slides over the surface of a horizontal circular rod for which the coefficient of kinetic friction is mk = 0.3. If the collar is given a speed of 4 m>s and then released at u = 0°, determine how far, s, it slides on the rod before coming to rest. z

4m

b u

6m

100 mm

t

y

u

n

s

x

v

Unpublished Work © 2007 by R. C.13–67 Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, Prob. Prob. 13–70 New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

136

C H A P T E R 13

KINETICS

OF A


13–71. The roller coaster car and passenger have a total weight of 600 lb and starting from rest at A travel down the track that has the shape shown. Determine the normal force of the tracks on the car when the car is at point B, where it has a velocity of 15 ft>s. Neglect friction and the size of the car and passenger.

AND


13–73. The 5-kg pendulum bob B is released from rest when u = 0°. Determine the initial tension in the cord and also at the instant the bob reaches point D, u = 45°. Neglect the size of the bob.

y A B

A

u $ 45'

p y $ 40 cos ( –– 40 x)

40 ft B

2m x D

40 ft

Prob. 13–73 20 ft

Prob. 13–71 *13–72. The smooth block B, having a mass of 0.2 kg, is attached to the vertex A of the right circular cone using a light cord. The cone is rotating at a constant angular rate about the z axis such that the block attains a speed of 0.5 m>s. At this speed, determine the tension in the cord and the reaction which the cone exerts on the block. Neglect the size of the block.

13–74. A ball having a mass of 2 kg and negligible size moves within a smooth vertical circular slot. If it is released from rest when u = 10°, determine the force of the slot on the ball when the ball arrives at points A and B.

10'

z

u A

200 mm

r $ 0.8 m A

B

400 mm

300 mm

10' B

Prob. 13–72

Prob. 13–74


137

PROBLEMS 13–75. The rotational speed of the disk is controlled by a 30-g smooth contact arm AB which is spring-mounted on the disk. When the disk is at rest, the center of mass G of the arm is located 150 mm from the center O, and the preset compression in the spring is 20 mm. If the initial gap between B and the contact at C is 10 mm, determine the (controlling) speed vG of the arm’s mass center, G, which will close the gap. The disk rotates in the horizontal plane. The spring has a stiffness of k = 50 N>m, and its ends are attached to the contact arm at D and to the disk at E.

13–77. The 35-kg box has a speed of 2 m>s when it is at A on the smooth ramp. If the surface is in the shape of a parabola, determine the normal force on the box at the instant x = 3 m. Also, what is the rate of increase in its speed at this instant?

y

2 m/s

O

k $ 50 N/m

DG

1 y $ 4 # –– x2 9

BC

A 150 mm

A

x E 10 mm

Prob. 13–77 Prob. 13–75 *13–76. The 2-kg spool S fits loosely on the inclined rod for which the coefficient of static friction is ms = 0.2. If the spool is located 0.25 m from A, determine the minimum constant speed the spool can have so that it does not slip down the rod. z 5

13–78. The man has a mass of 80 kg and sits 3 m from the center of the rotating platform. Due to the rotation his # speed is increased from rest by v = 0.4 m>s2. If the coefficient of static friction between his clothes and the platform is ms = 0.3, determine the time required to cause him to slip.

3

4

S 0.25 m

3m 10 m

A

Prob. 13–76

Prob. 13–78


138

C H A P T E R 13

KINETICS

OF A


13–79. The collar A, having a mass of 0.75 kg, is attached to a spring having a stiffness of k = 200 N>m. When rod BC rotates about the vertical axis, the collar slides outward along the smooth rod DE. If the spring is unstretched when s = 0, determine the constant speed of the collar in order that s = 100 mm. Also, what is the normal force of the rod on the collar? Neglect the size of the collar.

AND


13–81. If the bicycle and rider have a total weight of 180 lb, determine the resultant normal force acting on the bicycle when it is at point A while it is freely coasting at vA = 6 ft>s. Also, compute the increase in the bicyclist’s speed at this point. Neglect the resistance due to the wind and the size of the bicycle and rider. y

C

k $ 200 N/m D

A

vA $ 6 ft/s

5 ft 20 ft

p y $ 20 cos — x 20

A E s

x B

Prob. 13–79

Prob. 13–81

*13–80. The block has a weight of 2 lb and it is free to move along the smooth slot in the rotating disk. The spring has a stiffness of 2.5 lb>ft and an unstretched length of 1.25 ft. Determine the force of the spring on the block and the tangential component of force which the slot exerts on the side of the block, when the block is at rest with respect to the disk and is traveling with a constant speed of 12 ft>s.

13–82. The 5-lb packages ride on the surface of the conveyor belt. If the belt starts from rest and increases to a constant speed of 12 ft>s in 2 s, determine the maximum angle u so that none of the packages slip on the inclined surface AB of the belt. The coefficient of static friction between the belt and a package is ms = 0.3. At what angle f do the packages first begin to slip off the surface of the belt after the belt is moving at its constant speed of 12 ft>s? Neglect the size of the packages.

B 6 in.

f

k $ 2.5 lb/ft

A

Prob. 13–80

u

Prob. 13–82


13.6 EQUATIONS OF MOTION: CYLINDRICAL COORDINATES

13.6 Equations of Motion:

139

(Fzuz

Cylindrical Coordinates

(Fuuu

When all the forces acting on a particle are resolved into cylindrical components, i.e., along the unit-vector directions ur , uu , uz , Fig. 13–16, the equation of motion may be expressed as ©F = ma ©Frur + ©Fuuu + ©Fzuz = marur + mauuu + mazuz To satisfy this equation, the respective ur , uu , uz components on the left side must equal the corresponding components on the right side. Consequently, we may write the following three scalar equations of motion: ©Fr = mar

P

(Frur

z

O u

r


©Fu = mau ©Fz = maz

(13–9)

Fig. 13–16

If the particle is constrained to move only in the r–u plane, then only the first two of Eqs. 13–9 are used to specify the motion.

Tangential and Normal Forces. The most straightforward type of problem involving cylindrical coordinates requires the determination of the resultant force components ©Fr , ©Fu , ©Fz which cause a particle to move with a known acceleration. If, however, the particle’s accelerated motion is not completely specified at the given instant, then some information regarding the directions or magnitudes of the forces acting on the particle must be known or computed in order to solve Eqs. 13–9. For example, the force P causes the particle in Fig. 13–17a to move along a path r = f1u2. The normal force N which the path exerts on the particle is always perpendicular to the tangent of the path, whereas the frictional force F always acts along the tangent in the opposite direction of motion. The directions of N and F can be specified relative to the radial coordinate by using the angle c (psi), Fig. 13–17b, which is defined between the extended radial line and the tangent to the curve. r $ f (u)

r $ f (u)

Tangent

Tangent

c N

r O

r

F

u P

O (a)

Fig. 13–17

u (b)

As the car of weight W descends the spiral track, the resultant normal force which the track exerts on the car can be represented by its three cylindrical components. -Nr creates a radial acceleration, -ar , Nu creates a transverse acceleration au , and the difference, W - Nz creates an azimuthal acceleration -az .


140

C H A P T E R 13

KINETICS

Tangent dr

du

c

rc

c $ 75'

2a

(13–10)

Procedure for Analysis

Tangent

u $ 30'

r dr>du

If c is calculated as a positive quantity, it is measured from the extended radial line to the tangent in a counterclockwise sense or in the positive direction of u. If it is negative, it is measured in the opposite direction to positive u. For example, consider the cardioid r = a11 + cos u2, shown in Fig. 13–18. Because dr>du = -a sin u, then when u = 30°, tan c = a11 + cos 30°2>1-a sin 30°2 = -3.732, or c = -75°, measured clockwise, as shown in the figure.

Fig. 13–17 (cont.)

r


tan c =

(c)

O

AND

ds u

O


This angle can be obtained by noting that when the particle is displaced a distance ds along the path, Fig. 13–17c, the component of displacement in the radial direction is dr and the component of displacement in the transverse direction is r du. Since these two components are mutually perpendicular, the angle c can be determined from tan c = r du>dr, or

r $ f (u)

r du

OF A

Cylindrical or polar coordinates are a suitable choice for the analysis of a problem for which data regarding the angular motion of the radial line r are given, or in cases where the path can be conveniently expressed in terms of these coordinates. Once these coordinates have been established, the equations of motion can be applied in order to relate the forces acting on the particle to its acceleration components. The method for doing this has been outlined in the procedure for analysis given in Sec. 13.4.The following is a summary of this procedure. Free-Body Diagram.

• Establish the r, u, z inertial coordinate system and draw the Fig. 13–18

• •

particle’s free-body diagram. Assume that a r , a u , a z act in the positive directions of r, u, z if they are unknown. Identify all the unknowns in the problem.

Equations of Motion. • Apply the equations of motion, Eqs. 13–9. Kinematics. • Use the methods# of$ Sec. 12.8 to determine r and the time # $ $ derivatives r, r, u, u, #z, and then evaluate the acceleration $ $ ## $ components ar = r - ru2, au = ru + 2ru, az = z. • If any of the acceleration components is computed as a negative quantity, it indicates that it acts in its negative coordinate direction. • When taking the time derivatives of r = f1u2, it is very important to use the chain rule of calculus. Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

141


EXAMPLE 13.10 The 2-lb block in Fig. 13–19a moves on a smooth horizontal track, such that its path is specified in polar coordinates by the parametric equations r = 110t22 ft and u = 10.5t2 rad, where t is in seconds. Determine the magnitude of the tangential force F causing the motion at the instant t = 1 s. SOLUTION

F

Free-Body Diagram. As shown on the block’s free-body diagram, Fig. 13–19b,the normal force of the track on the block,N,and the tangential force F are located at an angle c from the r and u axes. This angle can be obtained from Eq. 13–10. To do so, we must first express the path as r = f1u2 by eliminating the parameter t between r and u. This yields r = 40u2. Also, when t = 1 s, u = 0.511 s2 = 0.5 rad. Thus, r 40u2 = = 0.25 ` dr>du 4012u2 u = 0.5 rad c = 14.04°

r u (a)

tan c =

Because c is a positive quantity, it is measured counterclockwise from the r axis to the tangent (the same direction as u) as shown in Fig. 13–19b. There are presently four unknowns: F, N, ar and au . Equations of Motion. 2 + T ©Fr = mar ; F cos 14.04° - N sin 14.04° = a (1) 32.2 r 2 c + ©Fu = mau ; F sin 14.04° + N cos 14.04° = a (2) 32.2 u Kinematics. Since the motion is specified, the coordinates and the required time derivatives can be calculated and evaluated at t = 1 s. r = 10t2 ` # r = 20t `

t=1 s

t=1 s

= 10 ft

u = 0.5t `

t=1 s

u 14.04' N

Tangent au

F

r c = 14.04'

ar

u (b)

Fig. 13–19

= 0.5 rad

# = 20 ft>s u = 0.5 rad>s

$ $ r = 20 ft>s2 u = 0 # $ ar = r - ru2 = 20 - 1010.522 = 17.5 ft>s2 $ ## au = ru + 2ru = 10102 + 2120210.52 = 20 ft>s2 Substituting into Eqs. 1 and 2 and solving, we get F = 1.36 lb Ans. N = 0.942 lb NOTE: The tangential axis is in the direction of F, and the normal axis is in the direction of N. Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

142

C H A P T E R 13

KINETICS

OF A


AND


EXAMPLE 13.11 The smooth 2-kg cylinder C in Fig. 13–20a has a peg P through its center which passes through the slot# in arm OA. If the arm rotates in the vertical plane at a constant rate u = 0.5 rad>s, determine the force that the arm exerts on the peg at the instant u = 60°.

u 0.4 m

O · u $ 0.5 rad/s

r C

SOLUTION Why is it a good idea to use polar coordinates to solve this problem? Free-Body Diagram. The free-body diagram for the cylinder is shown in Fig. 13–20b. The force on the peg, FP , acts perpendicular to the slot in the arm. As usual, a r and a u are assumed to act in the directions of positive r and u, respectively. Identify the four unknowns. Equations of Motion. Using the data in Fig. 13–20b, we have +R©Fr = mar ; +b©Fu = mau ;

P

(1) (2)

Kinematics. From Fig. 13–20a, r can be related to u by the equation

A

r =

(a)

FP

u

u NC

r (b)

Fig. 13–20

0.4 = 0.4 csc u sin u

Since d1csc u2 = -1csc u cot u2 du and d1cot u2 = -1csc2 u2 du, then r and the necessary time derivatives become # u = 0.5 r = 0.4 csc u $ # # u = 0 r = -0.41csc u cot u2u

19.62 N

ar

19.62 sin u - NC sin u = 2ar 19.62 cos u + FP - NC cos u = 2au

= -0.2 csc u cot u # # $ r = -0.21-csc u cot u21u2 cot u - 0.2 csc u1-csc2 u2u

au

= 0.1 csc u1cot2 u + csc2 u2 u

Evaluating these formulas at u = 60°, we get # u = 0.5 r = 0.462 $ # u = 0 r = -0.133 $ r = 0.192 # $ ar = r - ru2 = 0.192 - 0.46210.522 = 0.0770 $ ## au = ru + 2ru = 0 + 21-0.133210.52 = -0.133 Substituting these results into Eqs. 1 and 2 with u = 60° and solving yields NC = 19.4 N

FP = -0.356 N

Ans.

The negative sign indicates that FP acts opposite to the direction shown in Fig. 13–20b. Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

143


EXAMPLE 13.12 A can C, having a mass of 0.5 kg, moves along a grooved horizontal slot shown in Fig. 13–21a.The slot is in the form of a spiral, which is defined by the equation r = 10.1u2 m, where u is in radians. If the arm OA is # rotating at a constant rate u = 4 rad>s in the horizontal plane, determine the force it exerts on the can at the instant u = p rad. Neglect friction and the size of the can. SOLUTION Free-Body Diagram. The driving force FC acts perpendicular to the arm OA, whereas the normal force of the wall of the slot on the can, NC , acts perpendicular to the tangent to the curve at u = p rad, Fig. 13–21b. As usual, ar and a u are assumed to act in the positive directions of r and u, respectively. Since the path is specified, the angle c which the extended radial line r makes with the tangent, Fig. 13–21c, can be determined from Eq. 13–10. We have r = 0.1u, so that dr>du = 0.1, and therefore r 0.1u tan c = = = u dr>du 0.1

u

r $ 0.1 u r

A

O

· u $ 4 rad/s

C (a)

FC ar r

f NC f au

Tangent

-1

When u = p, c = tan p = 72.3°, so that f = 90° - c = 17.7°, as shown in Fig. 13–21c. Identify the four unknowns in Fig. 13–21b. Equations of Motion. Fig. 13–21b, we have + ©F = ma ; ; r r + T ©Fu = mau ;

u

Using f = 17.7° and the data shown in

NC cos 17.7° = 0.5ar FC - NC sin 17.7° = 0.5au

Kinematics. The time derivatives of r and u are # u = 4 rad>s r = 0.1u $ # # u = 0 r = 0.1u = 0.1142 = 0.4 m>s $ $ r = 0.1u = 0 At the instant u = p rad, # $ ar = r - ru2 = 0 - 0.11p21422 = -5.03 m>s2 $ ## au = ru + 2ru = 0 + 210.42142 = 3.20 m>s2

(b)

(1) (2) r $ 0.1 u

u$p

r c f Tangent

u (c)

Fig. 13–21

Substituting these results into Eqs. 1 and 2 and solving yields NC = -2.64 N FC = 0.800 N

Ans.

What does the negative sign for NC indicate? Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

144

C H A P T E R 13

KINETICS

OF A


AND


PROBLEMS 13–83. A particle, having a mass of 1.5 kg, moves along a path defined by the equations r = 14 + 3t2 m, u = 1t2 + 22 rad, and z = 16 - t32 m, where t is in seconds. Determine the r, u, and z components of force which the path exerts on the particle when t = 2 s.

*13–84. The path of motion of a 5-lb particle in the horizontal plane is described in terms of polar coordinates as r = 12t + 12 ft and u = 10.5t2 - t2 rad, where t is in seconds. Determine the magnitude of the unbalanced force acting on the particle when t = 2 s.

13–87. The 4-kg spool slides along the rotating rod. At the instant shown, the angular rate of rotation of the rod is # $ u = 6 rad>s, which is increasing at u = 2 rad>s2. At this same instant, the spool is moving outward along the rod at 3 m>s, which is increasing at 1 m>s2 when r = 0.5 m. Determine the radial frictional force and the normal force of the rod on the spool at this instant.

r $ 0.5 m ••

u $ 2 rad/s2 O

13–85. The spring-held follower AB has a weight of 0.75 lb and moves back and forth as its end rolls on the contoured surface of the cam, where r = 0.2 ft and z = 10.1 sin u2 ft. If the cam is rotating at a constant rate of 6 rad>s, determine the force at the end A of the follower when u = 90°. In this position the spring is compressed 0.4 ft. Neglect friction at the bearing C.

•

u $ 6 rad/s

vs $ 3 m/s as $ 1 m/s2

Prob. 13–87 13–86. The spring-held follower AB has a weight of 0.75 lb and moves back and forth as its end rolls on the contoured surface of the cam, where r = 0.2 ft and z = 10.1 sin 2u2 ft. If the cam is rotating at a constant rate of 6 rad>s, determine the maximum and minimum force the follower exerts on the cam if the spring is compressed 0.2 ft when u = 45°.

*13–88. The boy of mass 40 kg is sliding down the spiral slide at a constant speed such that his position, measured from the top of the chute, has components r = 1.5 m, u = 10.7t2 rad, and z = 1-0.5t2 m, where t is in seconds. Determine the components of force Fr , Fu , and Fz which the slide exerts on him at the instant t = 2 s. Neglect the size of the boy. z

u z $ 0.1 sin 2u z · u $ 6 rad/s

0.2 ft

r $ 1.5 m

C B

A

k $ 12 lb/ft z

Probs. 13–85/86

Prob. 13–88


145

PROBLEMS 13–89. The girl has a mass of 50 kg. She is seated on the horse of the merry-go-round which undergoes constant # rotational motion u = 1.5 rad>s. If the path of the horse is defined by r = 4 m, z = 10.5 sin u2 m, determine the maximum and minimum force Fz the horse exerts on her during the motion.

13–91. The particle has a mass of 0.5 kg and is confined to move along the smooth horizontal slot due to the rotation of the arm OA. Determine the force of the rod on the particle and the normal force of the slot on the particle when u =# 30°. The rod is rotating with a constant angular velocity u = 2 rad>s. Assume the particle contacts only one side of the slot at any instant. *13–92. Solve $Problem 13–91 if# the arm has an angular acceleration of u = 3 rad>s2 and u = 2 rad>s at this instant. Assume the particle contacts only one side of the slot at any instant.

z

A

u r

· u $ 2 rad/s

r 0.5 m

u

O

Prob. 13–89

13–90. The 0.5-lb particle is guided along the circular path using the slotted arm guide. If the arm has $an angular velocity # u = 4 rad>s and an angular acceleration u = 8 rad>s2 at the instant u = 30°, determine the force of the guide on the particle. Motion occurs in the horizontal plane.

Probs. 13–91/92

13–93. A smooth can C, having a mass of 3 kg, is lifted from a feed at A to a ramp at B by a rotating rod. If the rod # maintains a constant angular velocity of u = 0.5 rad>s, determine the force which the rod exerts on the can at the instant u = 30°. Neglect the effects of friction in the calculation and the size of the can so that r = 11.2 cos u2 m. The ramp from A to B is circular, having a radius of 600 mm. B

r u 0.5 ft

· u $ 0.5 rad/s u

0.5 ft

Prob. 13–90

C r 600 mm

A

600 mm

Prob. 13–93


146

C H A P T E R 13

KINETICS

OF A


13–94. The 2-lb collar slides along the smooth horizontal spiral rod, r = 12u2 ft, where u is in radians. If its angular rate # of rotation is constant and equals u = 4 rad>s, determine the tangential force P needed to cause the motion and the normal force that the rod exerts on the spool at the instant u = 90°. 13–95.

Solve Prob. 13–94 if the spiral rod is vertical.

AND


13–97. The smooth particle has a mass of 80 g. It is attached to an elastic cord extending from O to P and due to the slotted arm guide moves along the horizontal circular path r = 10.8 sin u2 m. If the cord has a stiffness k = 30 N>m and an unstretched length of 0.25 m, determine the force of the guide on the particle # when u = 60°. The guide has a constant angular velocity u = 5 rad>s. $ # 13–98. Solve Prob. 13–97 if u = 2 rad>s2 when u = 5 rad>s and u = 60°.

P P r $ 2u r r · u $ 4 rad/s

· u $ 5 rad/s

0.4 m

u O

u

Probs. 13–94/95

O

Probs. 13–97/98 *13–96. The forked rod is used to move the smooth 2-lb particle around the horizontal path in the shape of a limaçon, r = 12 + cos u2 ft. If u = 10.5t22 rad, where t is in seconds, determine the force which the rod exerts on the particle at the instant t = 1 s. The fork and path contact the particle on only one side.

13–99. Determine the normal and frictional driving forces that the partial spiral track# exerts on the 200-kg $ motorcycle at the instant u = 53 p rad, u = 0.4 rad>s, and u = 0.8 rad>s2. Neglect the size of the motorcycle.

r $ (5u) m 2 ft

r

u r

u · u

3 ft

Prob. 13–96

Prob. 13–99


PROBLEMS *13–100. Using a forked rod, a smooth cylinder C having a mass of 0.5 kg is forced to move along the vertical slotted path r = 10.5u2 m, where u is in radians. If the angular position of the arm is u = 10.5t22 rad, where t is in seconds, determine the force of the rod on the cylinder and the normal force of the slot on the cylinder at the instant t = 2 s. The cylinder is in contact with only one edge of the rod and slot at any instant.

147

13–102. The smooth surface of the vertical cam is defined in part by the curve r = 10.2 cos u + 0.32 m. If the forked rod is # rotating with a constant angular velocity of u = 4 rad>s, determine the force the cam and the rod exert on the 2-kg roller when u = 30°. The attached spring has a stiffness k = 30 N>m and an unstretched length of 0.1 m.

r C u

u

r $ 0.5 u

Prob. 13–100

Prob. 13–102

13–101. The ball has a mass of 2 kg and a negligible size. It is originally traveling around the horizontal circular path of radius r0 = 0.5 m such that the angular rate of rotation is # u0 = 1 rad>s. If the attached cord ABC is drawn down through the hole at a constant speed of 0.2 m>s, determine the tension the cord exerts on the ball at the instant r = 0.25 m. Also, compute the angular velocity of the ball at this instant. Neglect the effects of friction between the ball and horizontal plane. Hint: First show that$ the equation of motion # in the u ## direction yields au = ru + 2ru = 11>r21d1r2u2>dt2 = 0. # When integrated, r2u = c, where the constant c is determined from the problem data.

13–103. The collar has a mass of 2 kg and travels along the smooth horizontal rod defined by the equiangular spiral r = 1eu2 m, where u is in radians. Determine the tangential force F and the normal force N acting on the collar when u = 90°, if the force F maintains a constant # angular motion u = 2 rad>s.

A B

u

F

r r0

· u0 r

0.2 m/s

r $ eu

u

C F

Prob. 13–101

Prob. 13–103


148

C H A P T E R 13

KINETICS

OF A


*13–104. The smooth surface of the vertical cam is defined in part by the curve r = 10.2 cos u + 0.32 m. The forked rod is $ rotating with an angular acceleration of u = 2 rad>s2, and # when u = 45° the angular velocity is u = 6 rad>s. Determine the force the cam and the rod exert on the 2-kg roller at this instant. The attached spring has a stiffness k = 100 N>m and an unstretched length of 0.1 m.

AND


13–106. Using air pressure, the 0.5-kg ball is forced to move through the tube lying in the horizontal plane and having the shape of a logarithmic spiral. If the tangential force exerted on the ball due to the air is 6 N, determine the rate of increase in the ball’s speed at the instant u = p>2. What direction does it act in?

r

r $ 0.2e0.1u

u

r F$6N

u

Prob. 13–106 Prob. 13–104 13–105. The pilot of an airplane executes a vertical loop which in part follows the path of a “four-leaved rose,” r = 1-600 cos 2u2 ft, where u is in radians. If his speed at A is a constant vP = 80 ft>s, determine the vertical reaction the seat of the plane exerts on the pilot when the plane is at A. He weighs 130 lb. Hint: To determine the time derivatives necessary to compute the acceleration components ar and au , take the first and second time derivatives of r = 40011 + cos u2. Then, for further information, use # Eq. 12–26 to determine u. Also, take the time derivative of $ # Eq. 12–26, noting that vC = 0, to determine u. 80 ft/s

13–107. Solve Prob. 13–106 if the tube lies in a vertical plane.

r $ 0.2e0.1u

A r r $#600 cos 2u

r

u

F$6N

u

Prob. 13–105

Prob. 13–107


149

PROBLEMS # *13–108. The arm is rotating at a rate of u = 5 rad>s when $ u = 2 rad>s2 and u = 90°. Determine the normal force it must exert on the 0.5-kg particle if the particle is confined to move along the slotted path defined by the horizontal hyperbolic spiral ru = 0.2 m.

13–110. The tube # rotates in the horizontal plane at a constant rate of u = 4 rad>s. If a 0.2-kg ball B starts at the # origin O with an initial radial velocity of r = 1.5 m>s and moves outward through the tube, determine the radial and transverse components of the ball’s velocity at the instant it leaves the outer end at C, r = 0.5 m. Hint: Show that the equation of motion in the r direction is $ r - 16r = 0. The solution is of the form r = Ae -4t + Be4t. Evaluate the integration constants A and B, and determine the time t when r = 0.5 m. Proceed to obtain vr and vu . z

0.2 r $— u

· u $ 4 rad/s O

r u· $ 5 rad/s, u·· $ 2 rad/s2

u

r

0.5 m

u $ 90'

y

B

x

C

Prob. 13–108 Prob. 13–110 13–109. The collar, which has a weight of 3 lb, slides along the smooth rod lying in the horizontal plane and having the shape of a parabola r = 4>11 - cos u2, where u is in radians and r is #in feet. If the collar’s angular rate is constant and equals u = 4 rad>s, determine the tangential retarding force P needed to cause the motion and the normal force that the collar exerts on the rod at the instant u = 90°.

13–111. A 0.2-kg spool slides down along a smooth rod. If # the rod has a constant angular rate of rotation u = 2 rad>s in the vertical plane, show that the equations of $ motion for the spool are r - 4r - 9.81 sin u = 0 and # 0.8r + Ns - 1.962 cos u = 0, where Ns is the magnitude of the normal force of the rod on the spool. Using the methods of differential equations, it can be shown that the solution of the first of these equations is # r = C1e -2t + C2e2t - 19.81>82 sin 2t. If r, r, and u are zero when t = 0, evaluate the constants C1 and C2 to determine r at the instant u = p>4 rad.

P

r u

Prob. 13–109

u u $ 2 rad/s

r

Prob. 13–111


150

C H A P T E R 13

KINETICS

OF A



AND

*13.7 Central-Force Motion and Space Mechanics P F du

r u

1 dA $ — r 2du 2

O (a)

If a particle is moving only under the influence of a force having a line of action which is always directed toward a fixed point, the motion is called central-force motion. This type of motion is commonly caused by electrostatic and gravitational forces. In order to determine the motion, we will consider the particle P shown in Fig. 13–22a, which has a mass m and is acted upon only by the central force F. The free-body diagram for the particle is shown in Fig. 13–22b. Using polar coordinates (r, u), the equations of motion, Eqs. 13–9, become -F = mc

d2r du 2 ra b d dt dt2

0 = mar

d2u dr du + 2 b dt dt dt2

(13–11)

The second of these equations may be written in the form 1 d 2 du c ar bd = 0 r dt dt so that integrating yields

u r u F (b)

Fig. 13–22

r2

du = h dt

(13–12)

Here h is a constant of integration. From Fig. 13–22a notice that the shaded area described by the radius r, as r moves through an angle du, is dA = 12 r2 du. If the areal velocity is defined as dA 1 du h = r2 = dt 2 dt 2

(13–13)

then, it is seen that the areal velocity for a particle subjected to centralforce motion is constant. In other words, the particle will sweep out equal segments of area per unit of time as it travels along the path. To obtain the path of motion, r = f1u2, the independent variable t must be eliminated from Eqs. 13–11. Using the chain rule of calculus and Eq. 13–12, the time derivatives of Eqs. 13–11 may be replaced by dr dr du h dr = = 2 dt du dt r du 2 d h dr d h dr du d h dr h dr = a b = a b = c a 2 bd dt r2 du du r2 du dt du r du r2 dt2 Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

151

13.7 CENTRAL-FORCE MOTION AND SPACE MECHANICS

Substituting a new dependent variable (xi) j = 1>r into the second equation, we have d2j d2r = -h2j2 2 2 dt du Also, the square of Eq. 13–12 becomes a

du 2 b = h2j4 dt

Substituting these last two equations into the first of Eqs. 13–11 yields -h2j2 or

d2j du2

d2j du2

- h2j3 = -

+ j =

F mh2j2

This satellite is subjected to a central force and its orbital motion can be closely predicted using the equations developed in this section.

F m (13–14)

This differential equation defines the path over which the particle travels when it is subjected to the central force F. * For application, the force of gravitational attraction will be considered. Some common examples of central-force systems which depend on gravitation include the motion of the moon and artificial satellites about the earth, and the motion of the planets about the sun. As a typical problem in space mechanics, consider the trajectory of a space satellite or space vehicle launched into free-flight orbit with an initial velocity v0 , Fig. 13–23. It will be assumed that this velocity is initially parallel to the tangent at the surface of the earth, as shown in the figure.†*Just after the satellite is released into free flight, the only force acting on it is the gravitational force of the earth. (Gravitational attractions involving other bodies such as the moon or sun will be neglected, since for orbits close to the earth their effect is small in comparison with the earth’s gravitation.) According to Newton’s law of gravitation, force F will always act between the mass centers of the earth and the satellite, Fig. 13–23. From Eq. 13–1, this force of attraction has a magnitude of F = G

Free-flight trajectory v0 Satellite F

F

r $ r0

Power-flight trajectory

Launching

Fig. 13–23

Mem r2

where Me and m represent the mass of the earth and the satellite, respectively, G is the gravitational constant, and r is the distance between *In the derivation, F is considered positive when it is directed toward point O. If F is oppositely directed, the right side of Eq. 13–14 should be negative. †The case where v0 acts at some initial angle u to the tangent is best described using the conservation of angular momentum (see Prob. 15–103). Unpublished Work © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc., Upper Saddle River, New Jersey. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department.

152

C H A P T E R 13

KINETICS

OF A


AND


the mass centers. To obtain the orbital path, we set j = 1>r in the foregoing equation and substitute the result into Eq. 13–14. We obtain d2j 2

du

+ j =

GMe h2

(13–15)

This second-order ordinary differential equation has constant coefficients and is nonhomogeneous. The solution is represented as the sum of the complementary and particular solutions. The complementary solution is obtained when the term on the right is equal to zero. It is jc = C cos1u - f2 where C and f are constants of integration. The particular solution is jp =

GMe h2

Thus, the complete solution to Eq. 13–15 is j = jc + jp GMe 1 = = C cos1u - f2 + r h2

D directrix P

A

u#f r

x

F

f u

x¿

p D

Fig. 13–24

focus

(13–16)

The validity of this result may be checked by substitution into Eq. 13–15. Equation 13–16 represents the free-flight trajectory of the satellite. It is the equation of a conic section expressed in terms of polar coordinates. As shown in Fig. 13–24, a conic section is defined as the locus of point P, which moves in a plane in such a way that the ratio of its distance from a fixed point F to its distance from a fixed line is constant. The fixed point is called the focus, and the fixed line DD is called the directrix. The constant ratio is called the eccentricity of the conic section and is denoted by e. Thus, FP e = PA which may be written in the form FP = r = e1PA2 = e[p - r cos1u - f2] or

1 1 1 = cos1u - f2 + r p ep

Comparing this equation with Eq. 13–16, it is seen that the eccentricity of the conic section for the trajectory is e =

Ch2 GMe

(13–17)



153

and the fixed distance from the focus to the directrix is p =

1 C

(13–18)

Provided the polar angle u is measured from the x axis (an axis of symmetry since it is perpendicular to the directrix), the angle f is zero, Fig. 13–24, and therefore Eq. 13–16 reduces to GMe 1 = C cos u + r h2

(13–19)

The constants h and C are determined from the data obtained for the position and velocity of the satellite at the end of the power-flight trajectory. For example, if the initial height or distance to the space vehicle is r0 (measured from the center of the earth) and its initial speed is v0 at the beginning of its free flight, Fig. 13–25, then the constant h may be obtained from Eq. 13–12. When u = f = 0°, the velocity v0 has no radial component; therefore, from Eq. 12–25, v0 = r01du>dt2, so that h = r20 or

du dt

h = r0v0

(13–20)

To determine C, use Eq. 13–19 with u = 0°, r = r0 , and substitute Eq. 13–20 for h: C =

GMe 1 a1 b r0 r0v20

(13–21)

The equation for the free-flight trajectory therefore becomes GMe GMe 1 1 b cos u + 2 2 = a1 2 r r0 r0v0 r0v0

(13–22)

The type of path taken by the satellite is determined from the value of the eccentricity of the conic section as given by Eq. 13–17. If e = 0

free-flight trajectory is a circle

e = 1

free-flight trajectory is a parabola

e 6 1 e 7 1

free-flight trajectory is an ellipse free-flight trajectory is a hyperbola

(13–23)


154

C H A P T E R 13

KINETICS

OF A


AND


Each of these trajectories is shown in Fig. 13–25. From the curves it is seen that when the satellite follows a parabolic path, it is “on the border” of never returning to its initial starting point. The initial launch velocity, v0 , required for the satellite to follow a parabolic path is called the escape velocity. The speed, ve , can be determined by using the second of Eqs. 13–23 with Eqs. 13–17, 13–20, and 13–21. It is left as an exercise to show that ve =

2GMe A r0

(13–24)

The speed vc required to launch a satellite into a circular orbit can be found using the first of Eqs. 13–23. Since e is related to h and C, Eq. 13–17, C must be zero to satisfy this equation (from Eq. 13–20, h cannot be zero); and therefore, using Eq. 13–21, we have vc =

GMe A r0

(13–25)

Provided r0 represents a minimum height for launching, in which frictional resistance from the atmosphere is neglected, speeds at launch which are less than vc will cause the satellite to reenter the earth’s atmosphere and either burn up or crash, Fig. 13–25. Hyperbolic trajectory e&1 Parabolic trajectory e$1 Elliptical trajectory e%1 v0

Crash trajectory

Circular trajectory e$0

v 0 % vc r0



155

b rp O

ra

b

a

a

Fig. 13–26

All the trajectories attained by planets and most satellites are elliptical, Fig. 13–26. For a satellite’s orbit about the earth, the minimum distance from the orbit to the center of the earth O (which is located at one of the foci of the ellipse) is rp and can be found using Eq. 13–22 with u = 0°. Therefore; rp = r0

(13–26)

This minimum distance is called the perigee of the orbit. The apogee or maximum distance ra can be found using Eq. 13–22 with u = 180°.* Thus, ra =

r0 12GMe>r0v202 - 1

(13–27)

With reference to Fig. 13–26, the semimajor axis a of the ellipse is a =

rp + ra

(13–28) 2 Using analytical geometry, it can be shown that the minor axis b is determined from the equation b = 2rpra

(13–29)

*Actually, the terminology perigee and apogee pertains only to orbits about the earth. If any other heavenly body is located at the focus of an elliptical orbit, the minimum and maximum distances are referred to respectively as the periapsis and apoapsis of the orbit.


156

C H A P T E R 13

KINETICS

OF A


AND


Furthermore, by direct integration, the area of an ellipse is

A = pab =

p 1r + ra22rpra 2 p

(13–30)

The areal velocity has been defined by Eq. 13–13, dA>dt = h>2. Integrating yields A = hT>2, where T is the period of time required to make one orbital revolution. From Eq. 13–30, the period is

T =

p 1r + ra22rpra h p

(13–31)

In addition to predicting the orbital trajectory of earth satellites, the theory developed in this section is valid, to a surprisingly close approximation, at predicting the actual motion of the planets traveling around the sun. In this case the mass of the sun, Ms , should be substituted for Me when the appropriate formulas are used. The fact that the planets do indeed follow elliptic orbits about the sun was discovered by the German astronomer Johannes Kepler in the early seventeenth century. His discovery was made before Newton had developed the laws of motion and the law of gravitation, and so at the time it provided important proof as to the validity of these laws. Kepler’s laws, developed after 20 years of planetary observation, are summarized as follows: 1. Every planet moves in its orbit such that the line joining it to the sun sweeps over equal areas in equal intervals of time, whatever the line’s length. 2. The orbit of every planet is an ellipse with the sun placed at one of its foci. 3. The square of the period of any planet is directly proportional to the cube of the minor axis of its orbit. A mathematical statement of the first and second laws is given by Eqs. 13–13 and 13–22, respectively. The third law can be shown from Eq. 13–31 using Eqs. 13–19, 13–28, and 13–29.



157

EXAMPLE 13.13 A satellite is launched 600 km from the surface of the earth, with an initial velocity of 30 Mm>h acting parallel to the tangent at the surface of the earth, Fig. 13–27. Assuming that the radius of the earth is 6378 km and that its mass is 5.976110242 kg, determine (a) the eccentricity of the orbital path, and (b) the velocity of the satellite at apogee. SOLUTION Part (a). The eccentricity of the orbit is obtained using Eq. 13–17.The constants h and C are first determined from Eqs. 13–20 and 13–21. Since rp = r0 = 6378 km + 600 km = 6.97811062 m v0 = 30 Mm>h = 8333.3 m>s

v0 $ 30 Mm/h

ra

rp

then h = rpv0 = 6.9781106218333.32 = 58.1511092 m2>s

O

A vA

GMe 1 C = a1 b rp rpv20 600 km

66.73110 -122[5.976110242] 1 f = 25.4110 -92 m-1 = e 1 6.97811062 6.9781106218333.322 Hence,

2.54110 -82[58.1511092]2 Ch2 e = = = 0.215 6 1 GMe 66.73110 -122[5.976110242]

Fig. 13–27

Ans.

From Eq. 13–23, observe that the orbit is an ellipse. Part (b). If the satellite were launched at the apogee A shown in Fig. 13–27, with a velocity vA , the same orbit would be maintained provided that h = rpv0 = ravA = 58.1511092 m2>s Using Eq. 13–27, we have rp 6.97811062 ra = = = 10.80411062 2GMe 2[66.73110 -122][5.976110242] -1 -1 rpv20 6.9781106218333.322 Thus, vA =

58.1511092 10.80411062

= 5382.2 m>s = 19.4 Mm>h

Ans.

NOTE: The farther the satellite is from the earth, the slower it moves,

which is to be expected since h is constant.


158

C H A P T E R 13

KINETICS

OF A


AND


PROBLEMS In the following problems, except where otherwise indicated, assume that the radius of the earth is 6378 km, the earth’s mass is 5.976110242 kg, the mass of the sun is 1.99110302 kg, and the gravitational constant is G = 66.73110-122 m3>1kg # s22. *13–112. The rocket is in circular orbit about the earth at an altitude of h = 4 Mm. Determine the minimum increment in speed it must have in order to escape the earth’s gravitational field.

13–115. The rocket is traveling in free flight along an elliptical trajectory A¿A. The planet has a mass 0.60 times that of the earth’s. If the rocket has an apoapsis and periapsis as shown in the figure, determine the speed of the rocket when it is at point A.

r $ 3.20 Mm A

B

O

6.40 Mm

A¿

16 Mm

Prob. 13–115

h $ 4 Mm

*13–116. An elliptical path of a satellite has an eccentricity e = 0.130. If it has a speed of 15 Mm>h when it is at perigee, P, determine its speed when it arrives at apogee, A. Also, how far is it from the earth’s surface when it is at A? Prob. 13–112 13–113. Prove Kepler’s third law of motion. Hint: Use Eqs. 13–19, 13–28, 13–29, and 13–31. 13–114. A satellite is to be placed into an elliptical orbit about the earth such that at the perigee of its orbit it has an altitude of 800 km, and at apogee its altitude is 2400 km. Determine its required launch velocity tangent to the earth’s surface at perigee and the period of its orbit.

A

P

Prob. 13–116 2400 km

Prob. 13–114

800 km

13–117. A communications satellite is to be placed into an equatorial circular orbit around the earth so that it always remains directly over a point on the earth’s surface. If this requires the period to be 24 hours (approximately), determine the radius of the orbit and the satellite’s velocity.


PROBLEMS 13–118. The rocket is traveling in free flight along an elliptical trajectory A¿A. The planet has no atmosphere, and its mass is 0.60 times that of the earth’s. If the rocket has the apogee and perigee shown, determine the rocket’s velocity when it is at point A. Take G = 34.4110 -921lb # ft22>slug 2, Me = 409110212 slug, 1 mi = 5280 ft.

A

B

O r $ 2000 mi

13–121. The rocket is traveling in free flight along an elliptical trajectory A¿A. The planet has no atmosphere, and its mass is 0.70 times that of the earth’s. If the rocket has an apoapsis and periapsis as shown in the figure, determine the speed of the rocket when it is at point A.

A¿

r $ 3 Mm A

4000 mi

O

A¿

6 Mm

13–119. If the rocket is to land on the surface of the planet, determine the required free-flight speed it must have at A¿ so that the landing occurs at B. How long does it take for the rocket to land, in going from A¿ to B? The planet has no atmosphere, and its mass is 0.6 times that of the earth’s. Take G = 34.4110 -921lb # ft22>slug 2, Me = 409110212 slug, 1 mi = 5280 ft.

B

B

10 000 mi

Prob. 13–118

A

159

O r $ 2000 mi

9 Mm

Prob. 13–121

13–122. If the rocket in Prob. 13–121 is to land on the surface of the planet, determine the required free-flight speed it must have at A¿ so that it strikes the planet at B. How long does it take for the rocket to land, going from A¿ to B along an elliptical path?

A¿ r $ 3 Mm A

4000 mi

B

O

A¿

10 000 mi

Prob. 13–119 *13–120. The speed of a satellite launched into a circular orbit about the earth is given by Eq. 13–25. Determine the speed of a satellite launched parallel to the surface of the earth so that it travels in a circular orbit 800 km from the earth’s surface.

6 Mm

9 Mm

Prob. 13–122


160

C H A P T E R 13

KINETICS

OF A


13–123. A satellite S travels in a circular orbit around the earth. A rocket is located at the apogee of its elliptical orbit for which e = 0.58. Determine the sudden change in speed that must occur at A so that the rocket can enter the satellite’s orbit while in free flight along the blue elliptical trajectory. When it arrives at B, determine the sudden adjustment in speed that must be given to the rocket in order to maintain the circular orbit.

AND


13–126. The rocket is traveling in a free-flight elliptical orbit about the earth such that e = 0.76 and its perigee is 9 Mm as shown. Determine its speed when it is at point B. Also determine the sudden decrease in speed the rocket must experience at A in order to travel in a circular orbit about the earth.

A 120 Mm

B

B

A

S

10 Mm

9 Mm

Prob. 13–126

Prob. 13–123 *13–124. An asteroid is in an elliptical orbit about the sun such that its perihelion distance is 9.3011092 km. If the eccentricity of the orbit is e = 0.073, determine the aphelion distance of the orbit. 13–125. A satellite is in an elliptical orbit around the earth such that e = 0.156. If its perigee is 5 Mm, determine its velocity at this point and also the distance OB when it is at point B, located 135° away as shown.

13–127. A rocket is in free-flight elliptical orbit around the planet Venus. Knowing that the periapsis and apoapsis of the orbit are 8 Mm and 26 Mm, respectively, determine (a) the speed of the rocket at point A¿, (b) the required speed it must attain at A just after braking so that it undergoes an 8-Mm free-flight circular orbit around Venus, and (c) the periods of both the circular and elliptical orbits. The mass of Venus is 0.816 times the mass of the earth.

B

135'

P

5 Mm

O

O

A¿

A

8 Mm

18 Mm

Prob. 13–125

Prob. 13–127



161

Design Project 13–1D. Design of a Ramp Catapult The block B has a mass of 20 kg and is to be catapulted from the table. Design a catapulting mechanism that can be attached to the table and to the container of the block, using cables and pulleys. Neglect the mass of the container, assume the operator can exert a constant tension of 120 N on a single cable during operation, and that the maximum movement of his arm is 0.5 m. The coefficient of kinetic friction between the table and container is mk = 0.2. Submit a drawing of your design, and calculate the maximum range R to where the block will strike the ground. Compare your value with that of others in the class. B

0.75 m R

Fig. 13–1D


162

C H A P T E R 13

KINETICS

OF A


AND


Chapter Review Kinetics Kinetics is the study of the relationship between forces and the acceleration they cause. This relationship is based on Newton’s second law of motion, expressed mathematically as ©F = ma. (See page 102.) The mass m is the proportionality constant between the resultant force ©F acting on the particle and the acceleration a caused by this resultant. The mass represents the quantity of matter contained within the particle. It measures the resistance to a change in its motion. Before applying the equation of motion, it is important to first draw the particle’s free-body diagram in order to account for all of the forces that act on the particle. Graphically, this diagram is equal to the kinetic diagram, which shows the result of the forces, that is, the ma vector. (See page 105.)

F2 FR = (F

P

$

ma

F1 Free-body diagram

Kinetic diagram

Inertial Coordinate Systems When applying the equation of motion, it is important to make measurements of the acceleration from an inertial coordinate system. This system has axes that do not rotate but are either fixed or translate with a constant velocity. Various types of inertial coordinate systems can be used to apply ©F = ma in component form. (See page 106.) Rectangular x, y, z axes are used to describe rectilinear motion along each of the axes. Normal and tangential n, t axes are often used when the path is known. Recall that a n is always directed in the +n direction. It indicates the change in the velocity direction. Also recall that a t is tangent to the path. It indicates the change in the velocity magnitude.

a P

Path of particle

vO O

Inertial frame of reference

©Fx = max , ©Fy = may , ©Fz = maz

©Ft = mat , ©Fn = man , ©Fb = 0 at = dv>dt or a t = v dv>ds, and a n = v2>r e =

[1 + 1dy>dx22]3>2 ƒ d2y>dx2 ƒ


CHAPTER REVIEW

Cylindrical coordinates are useful when angular motion of the radial coordinate r is specified or when the path can conveniently be described with these coordinates. For some problems, the direction of the forces on the free-body diagram will require finding the angle c between the extended radial coordinate and the tangent to the curve. (See pages 106, 123, and 135.)

# $ $ # # $ ©Fr = m1r - ru 22, ©Fu = m1ru + 2r u2, ©Fz = mz tan c =

Central-Force Motion When a single force acts upon a particle, such as during the free-flight trajectory of a satellite in a gravitational field, then the motion is referred to as central-force motion. The orbit depends upon the eccentricity e; and as a result, the trajectory can either be circular, parabolic, elliptical, or hyperbolic. (See page 147.)

163

r dr>du

Free flight trajectory v0 Satellite F

F

r $ r0

Power flight trajectory

Launching


ch13 dynamics

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