Dynamics - Chapter 13 [Beer7]

Seventh Edition

CHAPTER

1 3

VECTOR MECHANICS FOR ENGINEERS:

DYNAMICS Ferdinand P. Beer E. Russell Johnston, Jr. Lecture Notes: J. Walt Oler Texas Tech University

Kinetics of Particles: Energy and Momentum Methods

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Contents Introduction Sample Problem 13.6 Work of a Force Sample Problem 13.7 Principle of Work & Energy Sample Problem 13.9 Applications of the Principle of Work & Principle Energy of Impulse and Momentum Power and Efficiency Impulsive Motion Sample Problem 13.1 Sample Problem 13.10 Sample Problem 13.2 Sample Problem 13.11 Sample Problem 13.3 Sample Problem 13.12 Sample Problem 13.4 Impact Sample Problem 13.5 Direct Central Impact Potential Energy Oblique Central Impact Conservative Forces Problems Involving Energy and Momentum Conservation of Energy Sample Problem 13.14 Motion Under a Conservative Central Force Sample Problem 13.15 Sample Problems 13.16 Sample Problem !3.17 © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Vector Mechanics for Engineers: Dynamics 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. • Method of impulse and momentum: directly relates force, mass, velocity, and time.

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Vector Mechanics for Engineers: Dynamics Work of a Force  • Differential vector dr is the particle displacement. • Work of the force is   dU = F • dr = F ds cos α = Fx dx + Fy dy + Fz dz • Work is a scalar quantity, i.e., it has magnitude and sign but not direction. • Dimensions of work are length × force. Units are 1 J ( joule ) = (1 N )(1 m ) 1ft ⋅ lb = 1.356 J

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Work of a Force • Work of a force during a finite displacement, U1→2 = = =

A2 

 F • d r ∫

A1 s2

s2

s1

s1

∫ ( F cos α ) ds = ∫ Ft ds

A2

∫ ( Fx dx + Fy dy + Fz dz )

A1

• Work is represented by the area under the curve of Ft plotted against s.

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Work of a Force • Work of a constant force in rectilinear motion,

U1→2 = ( F cos α ) ∆x • Work of the force of gravity, dU = Fx dx + Fy dy + Fz dz = −W dy y2

U1→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. © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Vector Mechanics for Engineers: Dynamics 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

U1→2 = − ∫ kx dx = 12 kx12 − 12 kx22 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 1 x, U 1→2 = − 2 ( F1 + F2 ) ∆x © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Vector Mechanics for Engineers: Dynamics 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 2 dr r r2

Mm

r1

r2

U1→2 = − ∫ G

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dr = G

Mm Mm −G r2 r1

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Vector Mechanics for Engineers: Dynamics 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.

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Vector Mechanics for Engineers: Dynamics 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

∫ Ft ds = m ∫ v dv = 12 mv2 − 12 mv1

T = 12 mv 2 = kinetic energy  • The work of the force F is equal to the change in kinetic energy of the particle. U1→2 = T2 − T1

• Units of work and kinetic energy are the same: 2 m  m   2 T = 12 mv = kg  =  kg 2 m = N ⋅ m = J s  s  © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Vector Mechanics for Engineers: Dynamics Applications of the Principle of Work and Energy • Wish to determine velocity of pendulum bob at A2. Consider work & kinetic energy.  • Force P acts normal to path and does no work. T1 + U1→2 = T2 0 + Wl =

1W 2 v2 2 g

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. © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Vector Mechanics for Engineers: Dynamics 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 ,

∑ Fn = m an

v2 = 2 gl

W v22 P −W = g l W 2 gl P =W + = 3W g l

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Vector Mechanics for Engineers: Dynamics Power and Efficiency • Power = rate at which work is done.   dU F • dr = = dt 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

or 1 hp = 550

ft ⋅ lb = 746 W s

• η = efficiency output work = input work power output = power input © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Vector Mechanics for Engineers: Dynamics Sample Problem 13.1 SOLUTION: • Evaluate the change in kinetic energy. • Determine the distance required for the work to equal the kinetic energy change.

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. © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Vector Mechanics for Engineers: Dynamics Sample Problem 13.1 SOLUTION: • Evaluate the change in kinetic energy.  mi  5280 ft  h  v1 =  60    = 88 ft s h  mi  3600 s   T1 = 12 mv12 = 12 ( 4000 32.2 )( 88) 2 = 481000 ft ⋅ lb v2 = 0

T2 = 0

• Determine the distance required for the work to equal the kinetic energy change. U1→2 = ( − 1500 lb ) x + ( 4000 lb )( sin 5°) x = −(1151 lb ) x

T1 + U1→2 = T2

481000 ft ⋅ lb − (1151 lb ) x = 0 x = 418 ft

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 13.2 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 µk = 0.25 and that the pulley is weightless and frictionless.

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 13.2 SOLUTION: • Apply the principle of work and energy separately to blocks A and B.

(

)

W A = ( 200 kg ) 9.81 m s 2 = 1962 N

FA = µ k N A = µ k W A = 0.25(1962 N ) = 490 N

T1 + U1→2 = T2 : 0 + FC ( 2 m ) − FA ( 2 m ) = 12 m Av 2 FC ( 2 m ) − ( 490 N )( 2 m ) = 12 ( 200 kg ) v 2

(

)

WB = ( 300 kg ) 9.81 m s 2 = 2940 N T1 + U1→2 = T2 :

0 − Fc ( 2 m ) + WB ( 2 m ) = 12 m B v 2 − Fc ( 2 m ) + ( 2940 N )( 2 m ) = 12 ( 300 kg ) v 2 © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 13.2 • When the two relations are combined, the work of the cable forces cancel. Solve for the velocity. FC ( 2 m ) − ( 490 N )( 2 m ) = 12 ( 200 kg ) v 2 − Fc ( 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 4900 J = 12 ( 500 kg ) v 2 v = 4.43 m s

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 13.3 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 and is held by cables so that it is initially energy for the rebound of the package. compressed 120 mm. The package has a The 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. © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 13.3 SOLUTION: • Apply principle of work and energy between initial position and the point at which spring is fully compressed. T1 = 12 mv12 = 12 ( 60 kg )( 2.5 m s ) 2 = 187.5 J

(U1→2 ) f

= − µ kW x

(

T2 = 0

)

= − µ k ( 60 kg ) 9.81m s 2 ( 0.640 m ) = −( 377 J ) µ k Pmin = kx0 = ( 20 kN m )( 0.120 m ) = 2400 N

Pmax = k ( x0 + ∆x ) = ( 20 kN m )( 0.160 m ) = 3200 N

(U1→2 ) e = − 12 ( Pmin + Pmax ) ∆x = − 12 ( 2400 N + 3200 N )( 0.040 m ) = −112.0 J U1→ 2 = ( U1→ 2 ) f + ( U1→ 2 ) e = −( 377 J ) µ k − 112 J T1 + U1→ 2 = T2 :

187.5 J - ( 377 J ) µ k − 112 J = 0

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µ k = 0.20 13 - 20

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Vector Mechanics for Engineers: Dynamics Sample Problem 13.3 • Apply the principle of work and energy for the rebound of the package. T 3= 12 mv32 = 12 ( 60kg ) v32

T2 = 0

U 2→3 = ( U 2→3 ) f + ( U 2→3 ) e = −( 377 J ) µ k + 112 J = +36.5 J T2 + U 2→3 = T3 : 0 + 36.5 J = 12 ( 60 kg ) v32 v3 = 1.103 m s

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 13.4 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 2000 lb car starts from rest at point 1 • Apply principle of work and energy to and moves without friction down the determine velocity at point 3. track shown. • Apply Newton’s second law to find Determine: minimum radius of curvature at point 3 such that a positive normal force is b) the force exerted by the track on exerted by the track. the car at point 2, and c) the minimum safe value of the radius of curvature at point 3.

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 13.4 SOLUTION: • Apply principle of work and energy to determine velocity at point 2. T2 = 12 mv22 =

T1 = 0

U1→ 2 = +W ( 40 ft ) T1 + U1→ 2 = T2 :

1W 2 v2 2g

0 + W ( 40 ft ) =

(

v22 = 2( 40 ft ) g = 2( 40 ft ) 32.2 ft s 2

)

1W 2 v2 2g v2 = 50.8 ft s

• Apply Newton’s second law to find normal force by the track at point 2. + ↑ ∑ Fn = m an : W v22 W 2( 40 ft ) g − W + N = m an = = g ρ 2 g 20 ft N = 5W © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

N = 10000 lb 13 - 23

Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 13.4 • Apply principle of work and energy to determine velocity at point 3. T1 + U1→3 = T3

0 + W ( 25 ft ) =

v32 = 2( 25 ft ) g = 2( 25 ft )( 32.2 ft s )

1W 2 v3 2g v3 = 40.1ft 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. + ↓ ∑ Fn = m an : W = m an W v32 W 2( 25 ft ) g = = g ρ3 g ρ3

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ρ3 = 50 ft

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 13.5 SOLUTION: Force exerted by the motor cable has same direction as the dumbwaiter velocity. Power delivered by motor is equal to FvD, vD = 8 ft/s. 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. © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

• In the first case, bodies are in uniform motion. Determine force exerted by motor cable from conditions for static equilibrium. • In the second case, both bodies are accelerating. Apply Newton’s second law to each body to determine the required motor cable force. 13 - 25

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Vector Mechanics for Engineers: Dynamics Sample Problem 13.5 • In the first case, bodies are in uniform motion. Determine force exerted by motor cable from conditions for static equilibrium. Free-body C: + ↑ ∑ Fy = 0 :

2T − 800 lb = 0

T = 400 lb

Free-body D: + ↑ ∑ Fy = 0 : F + T − 600 lb = 0 F = 600 lb − T = 600 lb − 400 lb = 200 lb Power = Fv D = ( 200 lb )( 8 ft s ) = 1600 ft ⋅ lb s Power = (1600 ft ⋅ lb s ) © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

1 hp = 2.91 hp 550 ft ⋅ lb s 13 - 26

Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 13.5 • 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 = 2.5 ft s 2 ↑

aC = − 12 a D = 1.25 ft s 2 ↓

Free-body C: + ↓ ∑ Fy = mC aC : 800 − 2T =

800 (1.25) 32.2

T = 384.5 lb

Free-body D: 600 ( 2.5) 32.2 F + 384.5 − 600 = 46.6

+ ↑ ∑ Fy = m D a D : F + T − 600 =

F = 262.1 lb

Power = Fv D = ( 262.1 lb )( 8 ft s ) = 2097 ft ⋅ lb s Power = ( 2097 ft ⋅ lb s ) © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

1 hp = 3.81 hp 550 ft ⋅ lb s 13 - 27

Seventh Edition

Vector Mechanics for Engineers: Dynamics Potential Energy  • Work of the force of gravity W , U1→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.

( )1 − (Vg ) 2

U1→2 = V g

• 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 © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Vector Mechanics for Engineers: Dynamics 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, GMm GMm U1→2 = − r2 r1 • Potential energy Vg when the variation in the force of gravity can not be neglected, GMm WR 2 Vg = − =− r r © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Vector Mechanics for Engineers: Dynamics Potential Energy • Work of the force exerted by a spring depends only on the initial and final deflections of the spring, U1→2 = 12 kx12 − 12 kx22 • The potential energy of the body with respect to the elastic force, Ve = 12 kx 2

U1→2 = ( 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.

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Vector Mechanics for Engineers: Dynamics 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. U1→2 = V ( x1 , y1 , z1 ) − V ( x2 , y 2 , z 2 ) Such forces are described as conservative forces. • For any conservative force applied on a closed path,   ∫ F • dr = 0 • Elementary work corresponding to displacement between two neighboring points, dU = V ( x, y, z ) − V ( x + dx, y + dy, z + dz ) = − dV ( x, y, z )  ∂V ∂V ∂V  Fx dx + F y dy + Fz dz = − dx + dy + dz  ∂y ∂z   ∂x   ∂V ∂V ∂V  F = − + +  = −grad V  ∂x ∂y ∂z  © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Vector Mechanics for Engineers: Dynamics Conservation of Energy • Work of a conservative force, U1→ 2 = V1 − V2

• Concept of work and energy, U1→ 2 = T2 − T1

• Follows that T1 + V1 = T2 + V2 E = T + V = constant T1 = 0 V1 = W T1 + V1 = W T2 = 12 mv22 = T2 + V2 = W

1W ( 2 g) = W V2 = 0 2g

• When a particle moves under the action of conservative forces, the total mechanical energy is constant. • 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.

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Vector Mechanics for Engineers: Dynamics Motion Under a Conservative Central Force • When a particle moves under a conservative central force, both the principle of conservation of angular momentum r0 mv0 sin φ 0 = rmv sin φ and the principle of conservation of energy T0 + V0 = T + V 1 mv 2 0 2

−

GMm 1 2 GMm = 2 mv − r0 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 rmin, rmax, vmin, and vmax. © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Vector Mechanics for Engineers: Dynamics Sample Problem 13.6 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 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.

• Solve for the kinetic energy and velocity at 2.

If the collar is released from rest at position 1, determine its velocity after it has moved 6 in. to position 2. © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Vector Mechanics for Engineers: Dynamics Sample Problem 13.6 SOLUTION: • Apply the principle of conservation of energy between positions 1 and 2. 2 Position 1: Ve = 12 kx12 = 12 ( 3 lb in.)( 8 in. − 4 in.) = 24 in. ⋅ lb

V1 = Ve + Vg = 24 in. ⋅ lb + 0 = 2 ft ⋅ lb T1 = 0

2 2 Position 2: Ve = 12 kx2 = 12 ( 3 lb in.)(10 in. − 4 in.) = 54 in. ⋅ lb

Vg = Wy = ( 20 lb )( − 6 in.) = −120 in. ⋅ lb

V2 = Ve + Vg = 54 − 120 = −66 in. ⋅ lb = −5.5 ft ⋅ lb T2 = 12 mv22 =

1 20 2 v2 = 0.311v22 2 32.2

Conservation of Energy: T1 + V1 = T2 + V2 0 + 2 ft ⋅ lb = 0.311v22 − 5.5 ft ⋅ lb

v2 = 4.91ft s ↓ © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Vector Mechanics for Engineers: Dynamics Sample Problem 13.7 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 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.

• 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.

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Vector Mechanics for Engineers: Dynamics Sample Problem 13.7 SOLUTION: • Setting the force exerted by the loop to zero, solve for the minimum velocity at D. 2 + ↓ ∑ Fn = man : W = man mg = m vD r 2 vD = rg = ( 2 ft )( 32.2 ft s ) = 64.4 ft 2 s 2

• Apply the principle of conservation of energy between points A and D. V1 = Ve + Vg = 12 kx 2 + 0 = 12 ( 36 lb ft ) x 2 = 18 x 2 T1 = 0 V2 = Ve + Vg = 0 + Wy = ( 0.5 lb )( 4 ft ) = 2 ft ⋅ lb 2 T2 = 12 mvD =

(

)

1 0.5 lb 2 2 64 . 4 ft s = 0.5 ft ⋅ lb 2 32.2 ft s 2

T1 + V1 = T2 + V2 0 + 18 x 2 = 0.5 + 2 © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

x = 0.3727 ft = 4.47 in. 13 - 37

Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 13.9 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.

• 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 Determine (a) the maximum altitude insertion angle error. 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 © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Vector Mechanics for Engineers: Dynamics Sample Problem 13.9 • 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: TA + VA = TA′ + VA′

1 mv 2 0 2

−

GMm 1 2 GMm = 2 mv1 − r0 r1

Conservation of angular momentum: r r0 mv0 = r1mv1 v1 = v0 0 r1 Combining, 2  r0 2GM 1 v 2 1 − r0  = GM 1 − r0  1 + =   2 0 2 r0  r1  r1 r0v02  r1  r0 = 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 r1 = 60.4 × 106 m = 60400 km © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Vector Mechanics for Engineers: Dynamics Sample Problem 13.9 • 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: GMm 1 mv 2 − GMm = 1 mv 2 T0 + V0 = TA + V A − 0 max 2 2 r0 rmin Conservation of angular momentum: r r0 mv0 sin φ0 = rmin mvmax vmax = v0 sin φ0 0 rmin Combining and solving for sin ϕ0, sin φ0 = 0.9801

ϕ 0 = 90° ± 11.5°

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

allowable error = ±11.5°

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Vector Mechanics for Engineers: Dynamics Principle of Impulse and Momentum • 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

• Dimensions of the impulse of a force are force*time. • Units for the impulse of a force are

(

2

)

N ⋅ s = kg ⋅ m s ⋅ s = kg ⋅ m s

t2

  ∫ Fdt = Imp1→2 = impulse of the force F

t1

  mv1 + Imp1→2 = 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.

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Vector Mechanics for Engineers: Dynamics 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  F∆t is small and therefore, may be neglected.

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Vector Mechanics for Engineers: Dynamics Sample Problem 13.10 SOLUTION: • Apply the principle of impulse and momentum. The impulse is equal to the product of the constant forces and the time interval. 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.

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Vector Mechanics for Engineers: Dynamics Sample Problem 13.10 SOLUTION: • Apply the principle of impulse and momentum.   mv1 + ∑ Imp1→2 = mv2 Taking components parallel to the incline, mv1 + ( W sin 5°) t − Ft = 0  4000   ( 88 ft s ) + ( 4000 sin 5°) t − 1500t = 0  32.2  t = 9.49 s

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 13.11 SOLUTION: • Apply the principle of impulse and momentum in terms of horizontal and vertical component equations.

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.

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 13.11 SOLUTION: • Apply the principle of impulse and momentum in terms of horizontal and vertical component equations.   mv1 + Imp1→ 2 = mv2 x component equation: − mv1 + Fx ∆t = mv2 cos 40° 4 16 ( 80) + Fx ( 0.15) = 4 16 (120 cos 40°) 32.2 32.2 Fx = 89 lb

−

y component equation: y

0 + Fy ∆t = mv2 sin 40° x

4 16 (120 cos 40°) 32.2 Fy = 39.9 lb    F = ( 89 lb ) i + ( 39.9 lb ) j , F = 97.5 lb Fy ( 0.15) =

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 13.12 SOLUTION:

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.

• Apply the principle of impulse and momentum to the package-cart system to determine the final velocity. • Apply the same principle to the package alone to determine the impulse exerted on it from the change in its momentum.

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 13.12 SOLUTION: • Apply the principle of impulse and momentum to the package-cart system to determine the final velocity. y x

(

)

  m p v1 + ∑ Imp1→ 2 = m p + mc v2 x components:

(

)

m p v1 cos 30° + 0 = m p + mc v2

(10 kg )( 3 m/s ) cos 30° = (10 kg + 25 kg ) v2 v2 = 0.742 m/s

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13 - 48

Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 13.12 • Apply the same principle to the package alone to determine the impulse exerted on it from the change in its momentum. y x

  m p v1 + ∑ Imp1→ 2 = m p v2 x components:

m p v1 cos 30° + Fx ∆t = m p v2

(10 kg )( 3 m/s ) cos 30° + Fx ∆t = (10 kg ) v2 y components:

Fx ∆t = −18.56 N ⋅ s

− m p v1 sin 30° + Fy ∆t = 0

− (10 kg )( 3 m/s ) sin 30° + Fy ∆t = 0

∑ Imp1→2

Fy ∆t = 15 N ⋅ s

   = F∆t = ( − 18.56 N ⋅ s ) i + (15 N ⋅ s ) j

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

F∆t = 23.9 N ⋅ s

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 13.12

To determine the fraction of energy lost, T1 = 12 m p v12 = 12 (10 kg )( 3 m s ) 2 = 45 J

(

)

T1 = 12 m p + mc v22 = 12 (10 kg + 25 kg )( 0.742 m s ) 2 = 9.63 J T1 − T2 45 J − 9.63 J = = 0.786 T1 45 J

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Seventh Edition

Vector Mechanics for Engineers: Dynamics 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 Central Impact

• Oblique Impact: Impact for which one or both of the bodies move along a line other than the line of impact.

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13 - 51

Seventh Edition

Vector Mechanics for Engineers: Dynamics 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. • A period of restitution follows during which the bodies either regain their original shape or remain permanently deformed. • Wish to determine the final velocities of the two bodies. The total momentum of the two body system is preserved, m A v A + m B v B = m B v′B + m B v ′B • A second relation between the final velocities is required. © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

13 - 52

Seventh Edition

Vector Mechanics for Engineers: Dynamics Direct Central Impact e = coefficient of restitution • Period of deformation: m A v A − ∫ Pdt = m Au

=

∫ Rdt = u − v′A ∫ Pdt v A − u

0 ≤ e ≤1 • Period of restitution:

m Au − ∫ Rdt = m A v′A

• A similar analysis of particle B yields

v′B − u e= u − vB

• Combining the relations leads to the desired second relation between the final velocities.

v′B − v′A = e( v A − v B )

• Perfectly plastic impact, e = 0: v′B = v′A = v′

m Av A + mB v B = ( m A + mB ) v′

• Perfectly elastic impact, e = 1: Total energy and total momentum conserved.

v′B − v′A = v A − v B

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Oblique Central Impact • Final velocities are unknown in magnitude and direction. Four equations are required.

• 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 = ( v′A ) t

( v B ) t = ( v′B ) t

m A ( v A ) n + m B ( v B ) n = m A ( v′A ) n + m B ( v′B ) n

( v′B ) n − ( v′A ) n = e[ ( v A ) n − ( v B ) n ]

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13 - 54

Seventh Edition

Vector Mechanics for Engineers: Dynamics Oblique Central Impact • Block constrained to move along horizontal surface.   • Impulses from internal forces F and − F along the n axis and from external force Fext exerted by horizontal surface and directed along the vertical to the surface. • Final velocity of ball unknown in direction and magnitude and unknown final block velocity magnitude. Three equations required.

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Oblique Central Impact

• Tangential momentum of ball is conserved.

( v B ) t = ( v′B ) t

• Total horizontal momentum of block and ball is conserved.

m A ( v A ) + m B ( v B ) x = m A ( v ′A ) + m B ( v′B ) x

• Normal component of relative velocities of block and ball are related by coefficient of restitution.

( v′B ) n − ( v′A ) n = e[ ( v A ) n − ( v B ) n ]

• Note: Validity of last expression does not follow from previous relation for the coefficient of restitution. A similar but separate derivation is required. © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Problems Involving Energy and Momentum • Three methods for the analysis of kinetics problems: - Direct application of Newton’s second law - Method of work and energy - Method of impulse and momentum • Select the method best suited for the problem or part of a problem under consideration.

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 13.14 SOLUTION: • Resolve ball velocity into components normal and tangential to wall. • Impulse exerted by the wall is normal to the wall. Component of ball momentum tangential to wall is conserved. 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.

• Assume that the wall has infinite mass so that wall velocity before and after impact is zero. Apply coefficient of restitution relation to find change in normal relative velocity between wall and ball, i.e., the normal ball velocity.

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 13.14 SOLUTION: • Resolve ball velocity into components parallel and perpendicular to wall. vn = v cos 30° = 0.866v vt = v sin 30° = 0.500v • Component of ball momentum tangential to wall is conserved. vt′ = vt = 0.500v t n

• Apply coefficient of restitution relation with zero wall velocity. 0 − vn′ = e( vn − 0 ) vn′ = −0.9( 0.866v ) = −0.779v    v ′ = −0.779v λn + 0.500v λt  0.779  v′ = 0.926v tan −1  = 32.7°  0.500 

© 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 13.15 SOLUTION: • Resolve the ball velocities into components normal and tangential to the contact plane. • Tangential component of momentum for each ball is conserved. 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.

• Total normal component of the momentum of the two ball system is conserved. • The normal relative velocities of the balls are related by the coefficient of restitution. • Solve the last two equations simultaneously for the normal velocities of the balls after the impact.

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 13.15 SOLUTION: • Resolve the ball velocities into components normal and tangential to the contact plane.

( v A ) n = v A cos 30° = 26.0 ft

s

( vB ) n = −vB cos 60° = −20.0 ft

s

( vA )t

= v A sin 30° = 15.0 ft s

( vB ) t

= vB sin 60° = 34.6 ft s

• Tangential component of momentum for each ball is conserved.

( v′A ) t = ( v A ) t

= 15.0 ft s

( v′B ) t = ( vB ) t = 34.6 ft

s

• Total normal component of the momentum of the two ball system is conserved. m A ( v A ) n + mB ( vB ) n = m A ( v′A ) n + mB ( v′B ) n m( 26.0) + m( − 20.0 ) = m( v′A ) n + m( v′B ) n

( v′A ) n + ( v′B ) n = 6.0 © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 13.15 • The normal relative velocities of the balls are related by the coefficient of restitution. ( v′A ) n − ( v′B ) n = e[ ( v A ) n − ( vB ) n ] = 0.90[ 26.0 − ( − 20.0 ) ] = 41.4

• Solve the last two equations simultaneously for the normal velocities of the balls after the impact. ( v′A ) n = −17.7 ft s ( v′B ) n = 23.7 ft s    v A′ = −17.7λt + 15.0λn

n

 15.0  v′A = 23.2 ft s tan −1  = 40.3°  17.7     vB′ = 23.7λt + 34.6λn t

 34.6  v′B = 41.9 ft s tan −1  = 55.6° 23 . 7  

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 13.16 SOLUTION: • Determine orientation of impact line of action. • The momentum component of ball A tangential to the contact plane is conserved. • The total horizontal momentum of the two ball system is conserved. Ball B is hanging from an inextensible • The relative velocities along the line of cord. An identical ball A is released action before and after the impact are from rest when it is just touching the related by the coefficient of restitution. cord and acquires a velocity v0 before • Solve the last two expressions for the striking ball B. Assuming perfectly velocity of ball A along the line of action elastic impact (e = 1) and no friction, and the velocity of ball B which is determine the velocity of each ball horizontal. immediately after impact. © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 13.16 r = 0 .5 2r θ = 30° sin θ =

SOLUTION: • Determine orientation of impact line of action.

• The momentum component of ball A tangential to the contact plane is conserved.    mv A + F∆t = mv A′

mv0 sin 30° + 0 = m( v′A ) t

( v′A ) t

= 0.5v0

• The total horizontal (x component) momentum of the two ball system is conserved.

    mv A + T∆t = mv A′ + mvB′ 0 = m( v′A ) t cos 30° − m( v′A ) n sin 30° − mv′B 0 = ( 0.5v0 ) cos 30° − ( v′A ) n sin 30° − v′B 0.5( v′A ) n + v′B = 0.433v0

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 13.16 • The relative velocities along the line of action before and after the impact are related by the coefficient of restitution. ( v′B ) n − ( v′A ) n = e[ ( v A ) n − ( vB ) n ] v′B sin 30° − ( v′A ) n = v0 cos 30° − 0 0.5v′B − ( v′A ) n = 0.866v0

• Solve the last two expressions for the velocity of ball A along the line of action and the velocity of ball B which is horizontal.

( v′A ) n = −0.520v0

v′B = 0.693v0

   v A′ = 0.5v0λt − 0.520v0λn 0.52  β = tan −1  = 46.1°  0.5  α = 46.1° − 30° = 16.1° v′B = 0.693v0 ← v′A = 0.721v0

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 13.17 SOLUTION: • Apply the principle of conservation of energy to determine the velocity of the block at the instant of impact. • Since the impact is perfectly plastic, the block and pan move together at the same velocity after impact. Determine that velocity from the requirement that the total momentum of the block and pan is conserved. A 30 kg block is dropped from a height of 2 m onto the the 10 kg pan of a • Apply the principle of conservation of spring scale. Assuming the impact to energy to determine the maximum be perfectly plastic, determine the deflection of the spring. maximum deflection of the pan. The constant of the spring is k = 20 kN/m.

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 13.17 SOLUTION: • Apply principle of conservation of energy to determine velocity of the block at instant of impact. T1 = 0

V1 = WA y = ( 30 )( 9.81)( 2 ) = 588 J

T2 = 12 m A ( v A ) 22 = 12 ( 30 )( v A ) 22

V2 = 0

T1 + V1 = T2 + V2 0 + 588 J = 12 ( 30 )( v A ) 22 + 0

( v A ) 2 = 6.26 m s

• Determine velocity after impact from requirement that total momentum of the block and pan is conserved. m A ( v A ) 2 + mB ( vB ) 2 = ( m A + mB ) v3

( 30)( 6.26) + 0 = ( 30 + 10) v3

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v3 = 4.70 m s

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Seventh Edition

Vector Mechanics for Engineers: Dynamics Sample Problem 13.17 • Apply the principle of conservation of energy to determine the maximum deflection of the spring. T3 = 12 ( m A + mB ) v32 = 12 ( 30 + 10 )( 4.7 ) 2 = 442 J V3 = Vg + Ve =0+

1 kx 2 2 3

=

1 2

(20 × 10 )(4.91× 10 ) 3

−3 2

= 0.241 J

T4 = 0 Initial spring deflection due to pan weight: x3 =

WB (10 )( 9.81) −3 = = 4 . 91 × 10 m 3 k 20 × 10

V4 = Vg + Ve = ( WA + WB )( − h ) + 12 kx42

(

) = −392( x4 − 4.91 × 10−3 ) + 12 ( 20 × 103 ) x42 = −392( x4 − x3 ) + 12 20 × 103 x42

T3 + V3 = T4 + V4

(

) (

)

442 + 0.241 = 0 − 392 x4 − 4.91 × 10−3 + 12 20 × 103 x42 x4 = 0.230 m h = x4 − x3 = 0.230 m − 4.91 × 10−3 m © 2003 The McGraw-Hill Companies, Inc. All rights reserved.

h = 0.225 m 13 - 68