BDA 2013 Dynamics Chapter 3 Kinetics of Particles: Work & Energy Method En. Saifulnizan Bin Jamian Faculty of Mechanical Engineering and Manufacturing Kole Kolejj Univ Univer ersi siti ti Tekn Teknol olog ogii Tun Tun Huss Hussei ein n Onn 86400 86400 Parit Parit Raja, Raja, Johor Johor [C16-101-11
[email protected] [email protected]]] [ 07-453 6200 EXT 1379 ] © Copyright 2005
Contents Sample Problem 13.6
Introduction Work of a Force Principle of Work & Energy Applications of the Principle of Work & Energy Power and Efficiency Sample Problem 13.1 Sample Problem 13.2 Sample Problem 13.3 Sample Problem 13.4 Sample Problem 13.5 Potential Energy Conservative Forces Conservation of Energy Motion Under a Conservative Central Force
Sample Problem 13.7 Sample Problem 13.9
Introduction • Previously, problems dealing with the motion of particles were r r solved through the fundamental equation of motion, F = ma. Current chapter introduces two additional methods of analysis. • Method of work work and energy energy:: directly relates force, mass, mass, velocity and displacement. displacement. • Method of impulse and momentum: momentum: directly relates force, mass, velocity, and time.
Work of a Force r
• Differential vector d r is the particle displacement displacement . • Work of the force is r
r
dU = F • d r
= F ds cosα = F xdx+ F ydy + F z dz • Work is a scala scalarr quantity, quantity, i.e., i.e., it has magni magnitude tude and and sign but not direction. • Dimensions of wo work are length × force. Units are )(1 m) 1 J ( joule ) = (1 N )( 1ft ⋅ lb = 1.356 J
Work of a Force • Work of a force during a finite displacement, A2 r
r
U 1→2 = ∫ F • d r A1 s2
s2
s1
s1
= ∫ ( F cos α )ds = ∫ F t ds A2
= ∫ ( F x dx + F y dy + F z dz ) A1
• Work is represented by the area under the curve of F t plotted against s.
Work of a Force • Work of a constant force in rectilinear motion, U 1→2 = ( F cos α ) ∆x • Work of the force of gravity, dU = F x dx + F y dy + F z dz
= −W dy y2
U 1→2 = − ∫ W dy y1
= −W ( y 2 − y1 ) = −W ∆y • Work of the weight is equal to product of weight W and vertical displacement ∆ y. • Work of the weight is positive when ∆ y < 0, i.e., when the weight moves down.
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
U 1→2 = − ∫ kx dx = 1 kx12 − 1 kx22 2 2 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, U 1→2 = − 12 ( F 1 + F 2 ) ∆x
Work of a Force Work of a gravitational force (assume particle M occupies fixed position O while particle m follows path shown), Mm dU = − Fdr = −G dr r 2 r 2
Mm Mm Mm U 1→2 = − ∫ G dr = G G − r 2 r 1 r 2 r 1
Work of a Force Forces which do not do work (ds = 0 or cos α = 0): • reaction at frictionless pin supporting rotating body, • reaction at frictionless surface when body in contact moves along surface, • reaction at a roller moving along its track, and • weight of a body when its center of gravity moves horizontally.
Particle Kinetic Energy: Principle of Work & Energy
r
• Consider a particle of mass m acted upon by force F dv F t = 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 ∫ F t ds = m ∫ v dv = 12 mv2 − 12 mv1
U 1→2 = T 2 − T 1
T = 12 mv 2 = kinetic energy r
• The work of the force F is 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
Applications of the Principle of Work and Energy • Wish to determine velocity of pendulum bob at A2. Consider work & kinetic energy. r
• Force P acts normal to path and does no work. T 1 + U 1→2 = T 2 0 + ml =
1 2
mv22
v2 = 2 gl
• 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.
Applications of the Principle of Work and Energy • Principle of work and energy cannot be applied to directly determine the acceleration of the pendulum bob. • Calculating the tension in the cord requires supplementing the method of work and energy with an application of Newton’s second law. • As the bob passes through A2 ,
∑ F n = m an P − mg = man = m
v2 =
2 gl
P = mg + m
v22
l 2 gl l
= 3mg
Power and Efficiency • Power = rate at which work is done. r
=
r
dU F • d r dt
r
=
r
dt
= F • v
• Dimensions of power are work/time or force*velocity. Units for power are J m 1 W (watt) = 1 = 1 N ⋅ s s
• η = efficiency
=
output wor k
=
power output
input work power input
Example 1: SP13.1 SOLUTION: • Evaluate the change in kinetic energy. • Determine the distance required for the work to equal the kinetic energy change.
An automobile weighing 19.62 kN is driven down a 5o incline at a speed of 100 km/h when the brakes are applied causing a constant total breaking force of 7 kN. Determine the distance traveled by the automobile as it comes to a stop.
Example 1: SP13.1 SOLUTION: • Evaluate the change in kinetic energy.
v1 = 100
km 1000 m 1 h = 27.78 m s h 1 km 3600 s
T 1 = 12 mv12 = 12 (2000 kg)(27.78 m/s 2 ) = 771.73 kJ 2
v2 = 0
T 2 = 0
• Determine the distance required for the work to equal the kinetic energy change. U 1→2 = (− 7 kN ) x + (19.62 kN)(sin 5°) x
= −(5.29 kN ) x T 1 + U 1→2 = T 2 771.73 kJ − (5.29 kN ) x = 0
x = 145.9 m
Example 2: SP13.2 SOLUTION: • Apply the principle of work and energy separately to blocks A and B. • When the two relations are combined, the work of the cable forces cancel. Solve for the velocity. 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 µ k = 0.25 and that the pulley is weightless and frictionless.
Example 2: SP13.2 SOLUTION: • Apply the principle of work and energy separately to blocks A and B. 2 W A = (200 kg ) 9.81 m s = 1962 N
F A = µ k N A = µ k W A = 0.25(1962 N ) = 490 N T 1 + U 1→2 = T 2 : 0 + F C (2 m ) − F A (2 m ) = 12 m A v
2
2 F C (2 m ) − (490 N )(2 m ) = 12 (200 kg )v
W B = (300 kg ) 9.81 m s 2 = 2940 N T 1 + U 1→2 = T 2 : 2 0 − F c (2 m ) + W B (2 m ) = 12 m B v
− F c (2 m ) + (2940 N )(2 m ) = 12 (300 kg )v 2
Example 2: SP13.2 • When the two relations are combined, the work of the cable forces cancel. Solve for the velocity. F C (2 m ) − (490 N )(2 m ) = 1 (200 kg )v 2 2
− F c (2 m ) + (2940 N )(2 m ) = 12 (300 kg )v 2 (2940 N )(2 m ) − (490 N )(2 m ) = 12 (200 kg + 300 kg )v 2 2 4900 J = 12 (500 kg )v
v = 4.43 m s
Example 3 : SP13.3 SOLUTION: • Apply the principle of work and energy between the initial position and the point at which the spring is fully compressed and the velocity is zero. The only unknown in the relation is the friction coefficient.
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 • Apply the principle of work and energy and is held by cables so that it is initially for the rebound of the package. The compressed 120 mm. The package has a only unknown in the relation is the velocity of 2.5 m/s in the position shown velocity at the final position. 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.
Example 3 : SP13.3 SOLUTION: • Apply principle of work and energy between initial position and the point at which spring is fully compressed. 2
T 1 = 1 mv12 = 1 (60 kg)(2.5 m s ) = 187.5 J 2 2
T 2 = 0
(U 1→2 ) f = − µ k W x
= − µ k (60 kg )(9.81m s 2 )(0.640 m ) = −(377 J ) µ k P min = kx0 = (20 kN m )(0.120 m ) = 2400 N P max = k ( x0 + ∆ x ) = (20 kN m )(0.160 m ) = 3200 N
(U 1→2 )e = − 12 ( P min + P max )∆x
= − 12 (2400 N + 3200 N )(0.040 m ) = −112.0 J U 1→2 = (U 1→2 ) f + (U 1→2 )e = − (377 J ) µ k − 112 J T 1 + U 1→2 = T 2 : 187.5 J - (377 J ) µ k − 112 J = 0
µ k = 0.20
Example 3 : SP13.3 • Apply the principle of work and energy for the rebound of the package. T 2 = 0
T 3= 12 mv32 = 12 (60kg )v32
U 2→3 = (U 2→3 ) f + (U 2→3 )e = − (377 J ) µ k + 112 J
= +36.5 J T 2 + U 2→3 = T 3 : 0 + 36.5 J = 12 (60 kg )v32 v3 = 1.103 m s
Example 4: SP13.4 SOLUTION: • Apply principle of work and energy to determine velocity at point 2. • Apply Newton’s second law to find normal force by the track at point 2. A 1000 kg 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.
• Apply principle of work and energy to determine velocity at point 3. • Apply Newton’s second law to find minimum radius of curvature at point 3 such that a positive normal force is exerted by the track.
Example 4: SP13.4 SOLUTION:
• Apply principle of work and energy to determine velocity at point 2.
T 1 = 0
T 2 = 12 mv22 =
1 W 2 v2 2 g
U 1→2 = +W (12 m) T 1 + U 1→2 = T 2 :
0 + mg (12 m) =
v22 = 24 g = 24(9.81)
1 mv22 2
v2 = 15.3 m s
• Apply Newton’s second law to find normal force by the track at point 2.
+ ↑ ∑ F n = m an :
− mg + N = m an = m
v22
=m
ρ 2
2(12 m)g 6m N = 49.1kN
N = 5 mg
Example 4: SP13.4
• Apply principle of work and energy to determine velocity at point 3 .
T 1 + U 1→3 = T 3
0 + mg (12 m − 4.5 m) =
v = 15 g = 15(9.81) 2 3
1 2
mv32
v3 = 12.1 m s
• Apply Newton’s second law to find minimum radius of curvature at point 3 such that a positive normal force is exerted by the track.
+ ↓ ∑ F n = m an : mg = m an 2
=m
v3
ρ 3
=m
2(15 m) g ρ 3
ρ 3 = 15 m
Example 5: SP13.5 SOLUTION: Force exerted by the motor cable has same direction as the dumbwaiter velocity. Power delivered by motor is equal to Fv D , v D = 2.5 m/s.
The dumbwaiter D and its load have a combined weight of 300 kg, while the counterweight C weighs 400 kg.
• In the first case, bodies are in uniform motion. Determine force exerted by motor cable from conditions for static equilibrium.
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 2.5 m/s and an acceleration of 0.75 m/s2, both directed upwards.
• In the second case, both bodies are accelerating. Apply Newton’s second law to each body to determine the required motor cable force.
Example 5: SP13.5 • In the first case, bodies are in uniform motion. Determine force exerted by motor cable from conditions for static equilibrium. Free-body C:
+ ↑ ∑ F y = 02:T − (400) (9.81) N = 0
T = 19.62 N
Free-body D:
+ ↑ ∑ F y = 0 : F + T − (300) (9.81) N = 0 F = (300) (9.81) N − T = (300) (9.81) N − 19.62 N = 9.81 N Power = FvD = (9.81 N) (2.5 m/s) = 2453 J s Power = (2453 J s)
1 hp 746 J s
= 3.3 hp
Example 5: SP13.5 • In the second case, both bodies are accelerating. Apply Newton’s second law to each body to determine the required motor cable force. a D = 0.75 m s ↑ 2
aC = − 12 a D = 0.375 m s ↓ 2
Free-body C:
+ ↓ ∑ F y = mC aC : (400) (9.81) − 2T = 400(0.375) T = 18.87 N Free-body D:
+ ↑ ∑ F y = m D a D : F + T − (300) (9.81) = 300 (0.75) F + 1887 − (300) (9.81) = 225 F = 1281 N Power = FvD = (1281 N) (2.5 m/s) = 3203 J/s Power = (3203 J s)
1 hp 746 J s
= 4.3 hp
Potential Energy
r
• Work of the force of gravity W , U 1→2 = W y1 − W y 2 • Work is independent of path followed; depends only on the initial and final values of Wy. V g = Wy
= potential energy of the body with respect to force of gravity. U 1→2 = V g − V g 1
2
• Choice of datum from which the elevation y is measured is arbitrary. • Units of work and potential energy are the same: V g = Wy = N ⋅ m = J
Potential Energy • Previous expression for potential energy of a body with respect to gravity is only valid when the weight of the body can be assumed constant. • For a space vehicle, the variation of the force of gravity with distance from the center of the earth should be considered. • Work of a gravitational force, U 1→2 =
GMm r 2
−
GMm r 1
• Potential energy V g when the variation in the force of gravity can not be neglected, V g = −
GMm r
=−
WR
2
r
Potential Energy • Work of the force exerted by a spring depends only on the initial and final deflections of the spring, 2 2 U 1→2 = 12 kx1 − 12 kx2
• The potential energy of the body with respect to the elastic force, V e = 12 kx 2 U 1→2 = (V e )1 − (V e )2 • Note that the preceding expression for V e is valid only if the deflection of the spring is measured from its undeformed position.
Conservative Forces
• Concept of potential energy can be applied if the work of the force is independent of the path followed by its point of application. U 1→2 = V ( x1 , y1 , z 1 ) − V ( x2 , y 2 , z 2 ) Such forces are described as conservative forces. • For any conservative force applied on a closed path, r
r
∫ F • d r = 0 • Elementary work corresponding to displacement between two neighboring points, dU = V ( , y , z ) − V ( + dx, y + dy, z + dz )
= −dV ( x, y, z ) ∂V ∂V ∂V dx + dy + dz ∂ y ∂ z ∂ x
F x dx + F y dy + F z dz = − r
∂V ∂V ∂V + + = −gradV ∂ ∂ ∂ z x y
F = −
Conservation of Energy • Work of a conservative force, U 1→2 = V 1 − V 2
• Concept of work and energy, U 1→2 = T 2 − T 1
• Follows that T 1 + V 1 = T 2 + V 2 E = T + V = constant T 1 = 0 V 1 = W l T 1 + V 1 = W l T 2 = 12 mv22 = T 2 + V 2 = W l
1 W 2 g
• When a particle moves under the action of conservative forces, the total mechanical energy is constant.
(2 g l ) = W l V 2 = 0 • 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 constant.
Motion Under a Conservative Central Force • When a particle moves under a conservative central force, both the principle of conservation of angular momentum r 0 mv0 sin φ 0 = rmv sin φ and the principle of conservation of energy T 0 + V 0 = T + V 1 mv 2 0 2
−
GMm r 0
= 12 mv 2 −
GMm r
may be applied. • Given r , the equations may be solved for v and ϕ. • At minimum and maximum r , ϕ = 90o. Given the launch conditions, the equations may be solved for r min , r max , vmin, and vmax.
Example 6 : SP13.6 SOLUTION: • Apply the principle of conservation of energy between positions 1 and 2. • The elastic and gravitational potential energies at 1 and 2 are evaluated from the given information. The initial kinetic energy is zero. A 9 kg collar slides without friction along a vertical rod as shown. The spring attached to the collar has an undeflected length of 100 mm and a constant of 540 N/m. If the collar is released from rest at position 1, determine its velocity after it has moved 150 mm to position 2.
• Solve for the kinetic energy and velocity at 2.
Example 6 : SP13.6 SOLUTION: • Apply the principle of conservation of energy between positions 1 and 2.
= 12 kx12 = 12 (540 N m)(0.1 m)2 = 2.7 J V 1 = V e + V g = 2.7 J
Position 1: V e
T 1 = 0 2 2 1 1 Position 2: V e = 2 kx2 = 2 (540 N m)(0.15 m ) = 6.1 J V g = Wy = (99.81 N )(− 0.15 m) = −13.3 J V 2 = V e + V g = (6.1 J) − (13.35) = −7.2 J T 2 = 12 mv2 = 2
1 2
9v2 = 4.5 v2 2
2
Conservation of Energy:
T 1 + V 1 = T 2 + V 2 0 + 2.7 J = 4.5v22 − 7.2 J v2 = 1.48 m s ↓
Example 7 : SP13.7 SOLUTION: • Since the pellet must remain in contact with the loop, the force exerted on the pellet must be greater than or equal to zero. Setting the force exerted by the loop to zero, solve for the minimum velocity at D.
The 200 g 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.
• Apply the principle of conservation of energy between points A and D. Solve for the spring deflection required to produce the required velocity and kinetic energy at D.
Example 7 : SP13.7 SOLUTION: • Setting the force exerted by the loop to zero, solve for the minimum velocity at D.
+ ↓ ∑ F n = man : W = man
mg = m v D2 r
v D2 = rg = (0.6 m)(9.81m s 2 ) = 5.89 m 2 s 2 • Apply the principle of conservation of energy between points A and D.
V 1 = V e + V g = 12 kx 2 + 0 = 12 (540 N m) x 2 = 270 x 2 T 1 = 0 V 2 = V e + V g = 0 + Wy = (0.29.81) N(1.2 m) = 2.35 J T 2 = 12 mv D = 2
1 2
(0.2) (5.89) = 0.589 J
T 1 + V 1 = T 2 + V 2
0 + 270 x 2 = 0.589 J + 2.35 J
x = 0.104 m = 104 mm
Example 8 : SP13.9 SOLUTION: • For motion under a conservative central force, the principles of conservation of energy and conservation of angular momentum may be applied simultaneously.
A satellite is launched in a direction parallel to the surface of the earth with a velocity of 36900 km/h from an altitude of 500 km. Determine (a) the maximum altitude reached by the satellite, and (b) the maximum allowable error in the direction of launching if the satellite is to come no closer than 200 km to the surface of the earth
• Apply the principles to the points of minimum and maximum altitude to determine the maximum altitude. • Apply the principles to the orbit insertion point and the point of minimum altitude to determine maximum allowable orbit insertion angle error.
Example 8 : SP13.9 • Apply the principles of conservation of energy and conservation of angular momentum to the points of minimum and maximum altitude to determine the maximum altitude. Conservation of energy: 1 mv 2 0 2
T A + V A = T A′ + V A′
GMm
−
= 12 mv12 −
r 0
GMm r 1
Conservation of angular momentum: r r 0 mv0 = r 1mv1 v1 = v0 0 r 1 Combining, 2 1 v 2 1 − r 0 2 0 2 r 1
=
GM r 0 1 − r 0 r 1
1+
r 0 r 1
=
2GM r 0v02
r 0 = 6370 km + 500 km = 6870 km v0 = 36900 km h = 10.25 × 106 m s
(
)(
)2
GM = gR 2 = 9.81m s 2 6.37 × 106 m = 398 × 1012 m3 s 2 r 1 = 60.4 × 106 m = 60400 km
Example 8 : SP13.9 • Apply the principles to the orbit insertion point and the point of minimum altitude to determine maximum allowable orbit insertion angle error. Conservation of energy: T 0 + V 0 = T A + V A
1 mv 2 0 2
−
GMm r 0
2 = 12 mvmax −
Conservation of angular momentum: r 0 mv0 sin φ 0 = r min mvmax
vmax = v0 sin φ 0
GMm r min
r 0 r min
Combining and solving for sin ϕ 0, sin φ 0 = 0.9801 ϕ 0 = 90° ± 11.5°
allowable error = ±11.5°
Example from Hibbeler 10th Edition • 14.7 Design consideration for the bumper B on the 5-Mg train car require use of a nonlinear spring having the load-deflection characteristics shown in the graph. Select the proper value of k so that the maximum deflection of the spring is limited to 0.2 m when the car, traveling at 4 m/s, strikes the rigid stop. Neglect the mass of the car wheels. • Answer
Example from Hibbeler 10th Edition • 14.18 Determine the heigh h to the top of the incline D to which the 200-kg roller coaster car will reach, if it is launched at B with a speed just sufficient for it to round the top of the loop at C without leaving the track. The radius of curvature at C is 25 m. • Answer
Example from Hibbeler 10th Edition • 14.74 The collar has a weight of 8 lb. If it is released from rest at a height of h= 2 ft from the top of the uncompressed spring, determine the speed of the collar after it falls and compresses the spring 0.3 ft. • Answer
Example from Hibbeler 10th Edition • 14.75 The 2-kg collar is attached to a spring that has an unstretched length of 3 m. If the collar is drawn to point B and released from rest, determine its speed when it arrives at point A. • Answer
Example from Hibbeler 10th Edition • 14.77 The 5-lb collar is released from rest at A and travels along the smooth guide. Determine its speed when its center reaches point C and the normal force it exerts on the rod at this point. The spring has an unstretched length of 12 in., and point C is located just before the end of the curved portion of the rod. • Answer
Other Examples : Hibbeler 10th edition • 14.3, 29, 78a, b, 83 , 78 MathCad Soln
