RVCC
ENGR 133-51 - Engineering Mechanics II: Dynamics
Kinetics of Particles: Energy and Momentum Methods Chapter
13
Systems of Particles Chapter
14 Vector Mechanics for Engineers: Dynamics F. Beer & Al
Introduction • Previously, problems dealing with the motion of particles were solved through the fundamental equation of motion, F ma. Current chapter introduces two additional methods of analysis.
• Method of work and energy: directly relates force, mass, velocity and displacement (F, m, v, d). • Method of impulse and momentum: directly relates force, mass, velocity, and time (F, m, v, t).
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Work of a Force • Differential vector dr is the particle displacement. • Work of the force is
dU F dr F ds cos ..............................
• Work is a scalar quantity, i.e., it has magnitude and sign but not direction. It can be >0, <0 (F opposite to displacement), zero (F perpendicular to displacement). • Dimensions of work are length force. Units are 1 J joule 1 N 1 m 1ft lb 1.356 J
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Work of a Force • Object in a curvilinear motion • Work of a force during a finite displacement, A2 U12 F dr A1 s2
s2
s1
s1
F cos ds Ft ds or
A2
F dx F dy F dz x
y
z
A1
The work done by the external total force F to move the particle from A1 to A2, is equal to the integral of the tangential component of that force with respect to the distance along the path. • Work is represented by the area under the curve of Ft plotted against s. 13/14- 4
Work of a Force • Work of a constant force in rectilinear motion, A2
U12 F cos dx A1
U12 F cos x • Work of the force of gravity, from previous slide: W = -W j
dU Fx dx Fy dy Fz dz Wdy y2
U12 Fx dx Fy dy Fz dz Wdy y1
W ( y2 y1 ) Wy • Work of the weight is equal to product of weight W and vertical displacement y. • Work of the weight is positive when the body moves down. The work is independent of the path and is equal to weight times vertical displacement 13/14- 5
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Work of a Force • Magnitude of the force exerted by a spring is proportional to deflection, F kx k spring constant N/m or lb/in. • Work of the force exerted by spring, dU F dx kx dx x2
U12 kx dx ........................ x1
• Work of the force exerted by spring is positive when x2 < x1, i.e., when the spring is returning to its undeformed position. • Work of the force exerted by the spring is equal to negative of area under curve of F plotted against x, U12 12 F1 F2 x F = kx is actually a static relationship which is true only when elements of the spring have no acceleration. 13/14- 7
Particle Kinetic Energy: Principle of Work & Energy • Consider a particle of mass m acted upon by force F dv Ft mat m dt dv ds dv m mv ds dt ds F t ds mv dv
• Integrating from A1 to A2 , s2
v2
s1
v1
2 2 1 1 F ds m v dv mv mv 2 1 2 2 t
U12 T2 T1
T 12 mv2 kinetic energy (definition of) The work of the force Fis equal to the change in kinetic energy of the particle. • Units of work and kinetic energy are the same: 2 m m T 12 mv 2 kg kg 2 m N m J s s Kinetic energy is always positive 13/14- 8
Work of a Force Work of a gravitational force (assume particle M occupies fixed position O while particle m follows path shown), The work of F while m moves from A to A’ Mm dU Fdr G 2 dr r r2
Mm Mm Mm U12 G 2 dr G G r2 r1 r r1
Mm since F G 2 r
3 m where G 66.73 1012 kg s 2 MG MG m and W m 2 mg g 2 9.81 2 R R s
Mm WR 2 mgR2 F G 2 2 2 r r r 13/14- 9
Applications of the Principle of Work and Energy • Wish to determine velocity of pendulum bob at A2. In A1 no speed. Consider work & kinetic energy. Draw FBD in a general point A
Consider the two forces and the work they do from A1 to A2. What is the work done by P? None! Because……………
• Velocity found without determining expression for acceleration and integrating. • All quantities are scalars and can be added directly. • Forces which do no work are eliminated from the problem.
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Applications of the Principle of Work and Energy • On the other side, principle of work and energy cannot be applied to directly determine the acceleration of the pendulum bob.
v2 2 gl
• Moreover, calculating the tension in the cord (that does no work) requires supplementing the method of work and energy with an application of Newton’s second law. • Suppose you want to know P when it passes in A2:………….
where v2 2 gl
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Sample Problem 13.1
An automobile weighing 4000 lb is driven down a 5o incline at a speed of 60 mi/h when the brakes are applied causing a constant total breaking force of 1500 lb. Determine the distance traveled by the automobile as it comes to a stop.
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Problem 1
Calculate the velocity v of the 50-kg crate when it reaches the bottom of the chute at B if it is given an initial velocity of 4 m/s down the chute at A. The coefficient of kinetic friction is 0.30.
Hints: 1. Draw FBD 2. Evaluate work. 3. Apply work-energy principle.
FBD
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Sample Problem 13.2 SOLUTION: • Apply the principle of work and energy separately to blocks A and B.
Two blocks are joined by an inextensible cable as shown. If the system is released from rest, determine the velocity of block A after it has moved 2 m. Assume that the coefficient of friction between block A and the plane is mk = 0.25 and that the pulley is weightless and frictionless.
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Problem 4
Hints: 1. Apply work-energy principle to the two crates independently
Forces that do work are the ones parallel to the path 13/14- 15
Sample Problem 13.3
A spring is used to stop a 60 kg package which is sliding on a horizontal surface. The spring has a constant k = 20 kN/m and is held by cables so that it is initially compressed 120 mm. The package has a velocity of 2.5 m/s in the position shown and the maximum deflection of the spring is 40 mm. Determine (a) the coefficient of kinetic friction between the package and surface and (b) the velocity of the package as it passes again through the position shown. 13/14- 16
Problem 5
P is a force that is pushing the crate upward. It is not the force of the spring.
Here is already stretched of 0.5 m…! 13/14- 17
Homework:
Solving Problems on Your Own Page ~769
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Sample Problem 13.4
A 2000 lb car starts from rest at point 1 and moves without friction down the track shown. Determine: a) the force exerted by the track on the car at point 2, and b) the minimum safe value of the radius of curvature at point 3.
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Problem 6
Packages having a mass of 2 kg are delivered from a conveyor to a smooth circular ramp with a velocity of vO = 1m/s. If the radius of the ramp is 0.5 m, determine the angle θ = θmax at which each package begins to leave the surface. FBD
Hints: 1. Draw FBD. 2. Work of weight is… 3. Use work-energy principle and Newton’s second law.
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Problem 7 same problem Packages having a mass of 2 kg are delivered from a conveyor to a smooth circular ramp with a velocity of vO = 1m/s. If the radius of the ramp is 0.5 m, determine the angle θ = θmax at which each package begins to leave the surface.
Let’s solve the same problem using instead only the equations of motion
Hints: 1. Draw FBD. 2. Apply motion for tangential and normal coordinates. 3.
You then need another equation
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When and how to use the Principle of Work and Energy
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When and how to use the Principle of Work and Energy
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Power and Efficiency • Power rate at which work is done.
dU F dr Power F v dt dt
• Dimensions of power are work/time or force*velocity. Units for power are J m ft lb 1 W (watt) 1 1 N or 1 hp 550 746 W s s s
efficiency of a machine (constant is the rate of U) output work input work power output power input
........... because of friction forces
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Problem 8 The power winch A hoists the 800 lb log up the 30o incline at a constant speed of 4 ft/sec. If the power output of the winch is 6 hp, compute the coefficient of kinetic friction μk between the log and the incline.
Hints: 1. FBD 2. Use power equation to find T 3. Apply Newton
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Problem 8.b Continuing on the previous problem, if the power is suddenly increased to 8 hp, what is the corresponding instantaneous acceleration a of the log? [still consider the same temporary v = 4 ft/s; need to use Newton]
Applying Newton:
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Sample Problem 13.5
The dumbwaiter D and its load have a combined weight of 600 lb, while the counterweight C weighs 800 lb. Determine the power delivered by the electric motor M when the dumbwaiter (a) is moving up at a constant speed of 8 ft/s and (b) has an instantaneous velocity of 8 ft/s and an acceleration of 2.5 ft/s2, both directed upwards. 13/14- 27
If velocity is not constant! If a problem states that the power is constant (or you need to find the average power) and time and v are your variables (no info on x), you can solve as: Separate the variables
Equation where x does not appear
If you need x instead of t:
Equation where t does not appear 13/14- 28
If velocity is not constant but acceleration is If velocity is not constant but acceleration is
P mav
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Problem 9 Hints: 1. Apply kinematics to find acceleration (constant) 2. Consider previous slide reasoning to find P
Eq. of Power
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How to work with Power
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Homework:
Solving Problems on Your Own Page~ 769
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Conservative Forces Work done by a force to move particle from point 1 to point 2 is independent of the path followed
Force is conservative
Weight force
Conservative, since it depends only on the vertical displacement, not on the path
Spring force
Conservative, since it depends only on the compression or extension of the spring
Friction force
Non-Conservative, since it depends on the path (the longer, the more dissipation)
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Work-Energy Principle Highlighting Conservative Forces • We have seen:
U12 T2 T1
where U12 represents the work done by all forces acting on the particle
U12conservative U12nonconservative T2 T1
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Gravitational Potential Energy Potential energy is a measure of the amount of work a conservative force will do
A
Weight force
A rock that falls on a pile of brick from a point A at +y high, will brake the pile. In A the brick has a gravitational potential energy that is associated to its weight, since capable of doing (positive) work At –y the gravitational potential energy is negative since the weight does a negative work when the particle is moved back at y=0 (datum) where Vg = 0 13/14- 35
Potential Energy • If gravitational force is the only force acting on the body, the work of the force of gravity can be written as : U12 W y1 W y2 • Work is independent of path followed; it depends only on the initial and final values of the function Wy.
V g Wy potential energy of the body with respect to force of gravity.
U12 Vg 1 Vg 2 Vg mgy mgh • The choice of datum from which the elevation y is measured is arbitrary. • Units of work and potential energy are the same: Vg Wy N m J 13/14- 36
Elastic Potential Energy Potential energy is a measure of the amount of work a conservative force will do Spring force
The spring force has always the capacity or potential of doing (positive) work on a particle when it is under a deformed position. The work is done when returning to the unstretched position
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Potential Energy • Work of the force exerted by a spring depends only on the initial and final deflections of the spring, U12 12 kx12 12 kx22
• The potential energy of the body with respect to the elastic force, Ve 12 kx 2
U12 Ve 1 Ve 2
• Note that the preceding expression for Ve is valid only if the deflection of the spring is measured from its undeformed position. 13/14- 38
Conservation of Energy • Work of a conservative force,
U12conservative V1 V2
(1)
• Concept of work and energy,
U12 T2 T1
(2)
• If only conservative forces are acting on the body, it follows that
T1 V1 T2 V2 T1 0 V1 W T1 V1 W
Recall v2
2 gl
E T V constant • When a particle moves under the action of conservative forces, the total mechanical energy is constant. In A1 the energy is entirely potential In A2 the energy is entirely kinetic 13/14- 39
Summarizing • Work of a conservative force,
U12conservative V1 V2
(1)
• Concept of work and energy,
U12 conservative nonconservative T2 T1
(2)
•Notice that: we started this chapter on equation (2). •The work was inclusive of all forces we called “external and internal”. •We notice that some of these forces are conservative (gravity and elastic); we now indicate them with V1 and V2 •This allow us to separate conservative forces and non-conservative forces (an external pushing force P)
•If we only have conservative forces, it means equation (2) becomes:
U12 V1 V2
T1 V1 T2 V2
If also non-conservative forces are used, then we should write a general equation
T1 V1 U12 nonconservative T2 V2
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Conservation of Energy • Friction forces are not conservative. Total mechanical energy of a system involving friction decreases. • Mechanical energy is dissipated by friction into thermal energy. • Total energy is always constant, but other forms of energy could take place: a generator converts mechanical energy into electrical, a gasoline engine converts chemical energy into mechanical, a nuclear reactor converts mass into thermal energy. 13/14- 41
Conservation of Energy If we consider only conservative forces, the equation: T1 + V1 = T2 + V2
is referred as the conservation of (mechanical) energy It states that during motion the kinetic energy must be transformed into potential energy and vice-versa. That is, if the body increases its kinetic energy, its potential energy will decrease
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Total energy of the ball at the initial position 1
E T1 V1 0 Wh Wh 2
At distance h/2
v 2 v02 2ag ( y y0 ) 2 g
3
h 2
v gh The total energy E at point 2 is:
The total energy E at point 3 is:
Note:Vg = Wy where y is positive upward from the datum and negative downward from the datum 13/14- 43
When and how to use the Principle of Conservation of Energy
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When and how to use the Principle of Conservation of Energy
Summarizing: we started with the principle of work & energy : T1 + U1→2 = T2 If the forces acting on the body are conservative forces (spring and/or gravitational) then the equation becomes the Conservation of energy: T1 + V1 = T2 + V2
If also forces non-conservative act on the body then the equation becomes: T1 + V1 + U1→2 non-cons. = T2 + V2 13/14- 45
Sample Problem 13.6
A 20 lb collar slides without friction along a vertical rod as shown. The spring attached to the collar has an undeflected length of 4 in. and a constant of 3 lb/in.
If the collar is released from rest at position 1, determine its velocity after it has moved 6 in. to position 2. 13/14- 46
Problem 10
The 10 kg slider moves with negligible friction up the inclined guide. The attached spring has a stiffness of 60 N/m and is stretched 0.6 m in position A, where the slider is released from rest. The 250 N force is constant and the pulley offers negligible resistance to the motion of the cord. Calculate the velocity vc of the slider as it passes point C.
Hints: 1. In the energy principle, consider that you also have forces that may do work (i.e.T and friction, but friction is not considered here). 2. T is constant but is its path constant….? 3. What about the reactions of the guides on the slider, do they do work? 4. Choose datum at A to compare with slides.
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Sample Problem 13.7
The 0.5 lb pellet is pushed against the spring and released from rest at A. Neglecting friction, determine the smallest deflection of the spring for which the pellet will travel around the loop and remain in contact with the loop at all times. 13/14- 48
Problem 11 The gantry structure in the photo is used to test the response of an airplane during a crash. The plane, having a mass of 8 Mg, is hoisted back until θ = 60o and then the pull-back cable AC is released when the plane is at rest. Determine the speed of the plane just before it crashes into the ground at θ = 15o. Also, what is the maximum tension developed in the supporting cable during the motion? Neglect the effect of lift caused by the wings during the motion and the size of the airplane.
Hints: 1. Apply the conservation of energy. 2. What is the work of the force of the cable? 3. How are you going to evaluate it? 4. Choose datum at the top of the gantry.
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Linear Momentum of a Particle Useful in many cases involving velocity and needed for using the principle of impulse and momentum. • Replacing the acceleration by the derivative of the velocity yields
dv F m dt d dL m v L dt dt L linear momentum of the particle
Linear Momentum Conservation Principle: If the resultant force on a particle is zero, the linear momentum of the particle remains constant in both magnitude and direction. 13/14- 50
Problem 12 1 mi = 5280 ft
When evaluating velocity, you need to calculate it as distance covered divided by time.
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And for a System of Particles?
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Sample Problem 14.2 (linear momentum for a system of particles)
A 20-lb projectile is moving with a velocity of 100 ft/s when it explodes into 5 and 15-lb fragments. Immediately after the explosion, the fragments travel in the directions qA = 45o and qB = 30o. Determine the velocity of each fragment. 13/14- 53
Problem 13 (linear momentum with no external forces) Car A (4000lb) and car B (3700 lb) are at rest on a flatcar also at rest. Car A and B then accelerate and quickly reach constant speeds relative to the flatcar of 7.65 ft/s and 7.50 ft/s, respectively, before decelerating to a stop at the opposite end of the flatcar. Knowing that the speed of the flatcar is 1.02 ft/s to the right when the cars are moving at constant speeds, determine the weight of the flatcar. Neglect friction and rolling resistance.
Hints: 1. All initial velocities are zero 2. Are there any horizontal external forces?
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What happens if there is an external force acting on the system?
Linear momentum will not be conserved! That is, the final momentum of the system will be different from the initial one because of the action of the external force
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Principle of Impulse and Momentum This method is typically used in problems involving Force, Time and Velocity. • From Newton’s second law, d F mv mv linear momentum dt Fdt d mv t2
F dt m v m v 2 1
t1 t2
Fdt Imp 12 impulse of the force F
t1
mv1 Imp 12 mv2 13/14- 56
Principle of Impulse and Momentum mv1 Imp12 mv2 The final momentum of the particle can be obtained by adding vectorially its initial momentum and the impulse of the force during the time interval. • Dimensions of the (linear) impulse of a force are force*time. • Units for the impulse of a force are
N s kg m s 2 s kg m s Notice that, since Δt will always be small, F has to be high in order for the Impulse to be comparable to the linear momentum. 13/14- 57
Principle of Impulse and Momentum If more forces act on a particle
mv1 Imp 12 mv2 If no external forces act on a particle
mv1 mv2
Total momentum is conserved
This occurs either if the resultant of the external forces is zero, or (as seen in next slides) when Δt is very short and all the external forces are non-impulsive (and will be neglected).
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Impulsive Motion • Force acting on a particle during a very short time interval that is large enough to cause a significant change in momentum is called an impulsive force. • When impulsive forces act on a particle, mv1 F t mv2 • When a baseball is struck by a bat, contact occurs over a short time interval but force is large enough to change sense of ball motion. • Nonimpulsive forces are forces for which Ft is small and therefore, may be neglected (weight of a body, force of a spring only if the time interval is small!). 13/14- 59
When and how to use the Principle of Impulse and Momentum
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When and how to use the Principle of Impulse and Momentum
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Sample Problem 13.10
An automobile weighing 4000 lb is driven down a 5o incline at a speed of 60 mi/h when the brakes are applied, causing a constant total braking force of 1500 lb. Determine the time required for the automobile to come to a stop. 13/14- 62
Sample Problem 13.11
A 4 oz baseball is pitched with a velocity of 80 ft/s. After the ball is hit by the bat, it has a velocity of 120 ft/s in the direction shown. If the bat and ball are in contact for 0.015 s, determine the average impulsive force exerted on the ball during the impact.
13/14- 63
Problem 12 Not the typical impulse/momentum problem! The 100 kg stone is originally at rest on the smooth horizontal surface. If a towing force of 200 N, acting at an angle of 45o is applied to the stone for 10 seconds, determine the final velocity and the normal force which the surface exerts on the stone during the time interval. Hints: • the problem involves F, v, t: what principle do we use to solve it….? • ………………………….! • Draw FBD. • Are forces constant or do we need an integration?
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Problem 13
Blocks A and B have a mass of 3 kg and 5kg, respectively. If the system is released from rest, determine the velocity of block B in 6 seconds. Neglect the mass of the pulleys and the cord. Hints: •Apply principle of Impulse and Momentum to A and B independently •Draw FBD before starting any calculation •Apply equation of dependent motion
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Conservation of Linear Momentum for a System of Particles For a system of particles:
mv1 F dt mv2
or
mv1 F t mv2
Equation 1
Where the integral (or the sum) involves only impulsive external* forces (the impulsive internal forces cancel each other (action-reaction)). If the sum of the external impulses is zero, then:
mv1 mv2
Equation 2
The total momentum of the particles is conserved This principle is often applied when particles moving freely collide or interact Note that the momentum is conserved but in general their total energy is not conserved In problems with system of particles, often Equation 2 is applied and allows the evaluation of a velocity; then Equation 1 is applied to only one particle to find the impulsive force(s). 13/14- 66
Sample Problem 13.12
A 10 kg package drops from a chute into a 24 kg cart with a velocity of 3 m/s. Knowing that the cart is initially at rest and can roll freely, determine (a) the final velocity of the cart, (b) the impulse exerted by the cart on the package, and (c) the fraction of the initial energy lost in the impact.
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Problem 15
An 800 kg rigid pile P is driven into the ground using a 300 kg hammer H. The hammer falls from rest at a height yo = 0.5 m and strikes the top of the pile. Determine the impulse which the hammer imparts on the pile if the pile is surrounded entirely by loose sand so that after striking, the hammer does not rebound off the pile.
Hints: • To apply the impulse –momentum, you need velocities before and after impact. Note that the hammer is in free fall, therefore conservation of energy… • Note that weight of hammer, of pile and reaction of sand are all non-impulsive… • Since there is no rebound, the two objects have the same final speed. •Consider three (3) different moments: 0 before the hammer falls; 1 when the hammer hits the pile, 2 when the two move together.
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Recalling the example on the textbook y x
t=2
t=1
t=0
WH RI -RI
=
+
v2
WP Rs The hammer is not released yet.
Right at the impact before they move: WH, Wp and Rs are all negligible.
They move together
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Homework:
Solving Problems on Your Own Page ~811
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Problem 16
Hints: 1. Are there any horizontal external forces…?
13- 7373 13/14-
Problem 17
Hints: 1. At the beginning the boat is at rest
a) Women dives first:
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Homework:
Solving Problems on Your Own Page ~880
13/14- 75
Impact • Impact: Collision between two bodies which occurs during a small time interval and during which the bodies exert large forces on each other. • Line of Impact: Common normal to the surfaces in contact during impact. Direct Central Impact
• Central Impact: Impact for which the mass centers of the two bodies lie on the line of impact; otherwise, it is an eccentric impact.. • Direct Impact: Impact for which the velocities of the two bodies are directed along the line of impact. • Oblique Impact: Impact for which one or both of the bodies move along a line other than the line of impact.
Oblique Central Impact 13/14- 76
Direct Central Impact • Bodies moving in the same straight line, vA > vB . • Upon impact the bodies undergo a period of deformation, at the end of which, they are in contact and moving at a common velocity u. • A period of restitution follows during which the bodies either regain their original shape or remain permanently deformed (the restitution impulse pushes the particles apart). • We wish to determine the final velocities of the two bodies. Considering a system with both particles, no external forces act. Then the total momentum of the two body system is preserved: mAvA mB vB mAvA mB vB • A second relation between the final velocities is required. Apply the principle to each particle. 13/14- 77
Direct Central Impact Particle A deforms under the action of B
• Period of deformation: m Av A Pdt m Au
During the period of restitution (when A regain its shape) consider again the action of B on A:
• Period of restitution: It is possible to define a coefficient of restitution:
It depends on: materials, shape, velocity and size of particles
m Au Rdt m AvA
e coefficient of restitution Rdt u vA Pdt vA u 0 e 1 13/14- 78
Direct Central Impact
vB u e u vB
• A similar analysis of particle B yields
vB vA ev A vB
• Combining the relations leads to the desired second relation between the final velocities. • Perfectly plastic impact, e = 0: vB vA v
? .................................
• Perfectly elastic impact, e = 1: Total energy and total momentum conserved.
? .......................
Note that only in the case of perfectly elastic impact the total energy of the two particles as well as their total momentum is conserved. The principle of work and energy cannot be used for impact problems since it is not known how the internal forces of deformation and restitution vary. However, knowing the particle’s velocities before and after the collision, the energy loss during collision can be calculated as U T T 12
2
1
If perfectly elastic impact If plastic impact 13/14- 79
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Problem 16
Hints: •Consider four (4) different moments: 0 before the ball is released after stretching the cord; 1 when the ball hits the ceiling, 2 when the ball bounce back after the collision and finally 3 when is will be all stretched. • To apply the impulse –momentum, you need velocity before the impact, therefore conservation of energy… • Apply impulse-momentum (impact ballceiling)… • Apply equation of conservation of energy to the ball just after the collision to determine stretch. •Weight of ball is non-impulsive 13/14- 81
Oblique Central Impact • Final velocities are unknown in magnitude and direction. Four equations are required (choosing n and t axes).
Consider particles perfectly smooth and frictionless (i.e. no external impulse)
Impulse is only due to internal forces along n axis • No tangential impulse component; tangential component of momentum for each particle is conserved. • Normal component of total momentum of the two particles is conserved. • Normal components of relative velocities before and after impact are related by the coefficient of restitution.
v A t vA t
vB t vB t
m A v A n mB v B n m A vA n mB vB n
vB n vA n evA n vB n 13/14- 82
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Sample Problem 13.15
The magnitude and direction of the velocities of two identical frictionless balls before they strike each other are as shown. Assuming e = 0.9, determine the magnitude and direction of the velocity of each ball after the impact.
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Problem 17
Two smooth disks A and B, having a mass of 1 kg and 2 kg respectively, collide with the velocities shown. If the coefficient of restitution for the disks is e = 0.75, determine the velocity of each disk after the collision. Hints: • What kind of impact? • Assume the four unknown components of v after impact all in the positive directions (if come out negative means opposite directions)
Components of initial velocities (along x and y or n and t):
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Sample Problem 13.14
A ball is thrown against a frictionless, vertical wall. Immediately before the ball strikes the wall, its velocity has a magnitude v and forms angle of 30o with the horizontal. Knowing that e = 0.90, determine the magnitude and direction of the velocity of the ball as it rebounds from the wall. 13/14- 86
Sample Problem 13.17
A 30 kg block is dropped from a height of 2 m onto the the 10 kg pan of a spring scale. Assuming the impact to be perfectly plastic, determine the maximum deflection of the pan. The constant of the spring is k = 20 kN/m.
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Homework:
Solving Problems on Your Own Page ~832
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