Spring Problems Part 2-Engineering Mechanics-Assignment Solution, Exercises for Dynamics. Aligarh Muslim University

Dynamics

Description: This is assignment solution for Dynamics course. It was submitted to Deendayal Chandar at Aligarh Muslim University. It includes: Spring, Armature, Electric, Drill, Displacement, Velocity, Angular, Motor, Point, Contact, Speed
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Engineering Mechanics - Dynamics Chapter 15

e vB1y− cos θ( ) vB1x sin θ( )−( ) vB2n=

vB1x

vB1y

vB2n

vB2t

t

R

⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

Find vB1x vB1y, vB2n, vB2t, t, R,( )= vB1x

vB1y

⎛ ⎜ ⎝

⎞ ⎟ ⎠

3.00

6.00− ⎛ ⎜ ⎝

⎞ ⎟ ⎠

ft s

=

t 0.19 s=

R 0.79 ft=

vB2n

vB2t

⎛ ⎜ ⎝

⎞ ⎟ ⎠

1.70

6.36 ⎛ ⎜ ⎝

⎞ ⎟ ⎠

ft s

= vB2n

vB2t

⎛ ⎜ ⎝

⎞ ⎟ ⎠

6.59 ft s

=

*Problem 15-72

The drop hammer H has a weight WH and falls from rest h onto a forged anvil plate P that has a weight WP. The plate is mounted on a set of springs that have a combined stiffness kT. Determine (a) the velocity of P and H just after collision and (b) the maximum compression in the springs caused by the impact. The coefficient of restitution between the hammer and the plate is e. Neglect friction along the vertical guideposts A and B.

Given:

WH 900 lb= kT 500 lb ft

=

WP 500 lb= g 32.2 ft

s2 =

h 3 ft= e 0.6=

Solution:

δst WP kT

= vH1 2g h=

Guesses

vH2 1 ft s

= vP2 1 ft s

= δ 2 ft=

Given WH g

⎛ ⎜ ⎝

⎞ ⎟ ⎠

vH1 WH

g ⎛ ⎜ ⎝

⎞ ⎟ ⎠

vH2 WP g

⎛ ⎜ ⎝

⎞ ⎟ ⎠

vP2+=

e vH1 vP2 vH2−=

349

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Engineering Mechanics - Dynamics Chapter 15

1 2

kTδst 2 1

2

WP g

⎛ ⎜ ⎝

⎞ ⎟ ⎠

vP2 2+

1 2

kTδ 2 WP δ δst−( )−=

vH2

vP2

δ

⎛⎜ ⎜ ⎜ ⎝

⎞⎟ ⎟ ⎟ ⎠

Find vH2 vP2, δ,( )= vH2

vP2

⎛ ⎜ ⎝

⎞ ⎟ ⎠

5.96

14.30 ⎛ ⎜ ⎝

⎞ ⎟ ⎠

ft s

= δ 3.52 ft=

Problem 15-73

It was observed that a tennis ball when served horizontally a distance h above the ground strikes the smooth ground at B a distance d away. Determine the initial velocity vA of the ball and the velocity vB (and θ) of the ball just after it strikes the court at B. The coefficient of restitution is e.

Given:

h 7.5 ft=

d 20 ft=

e 0.7=

g 32.2 ft

s2 =

Solution:

Guesses vA 1 ft s

= vB2 1 ft s

=

vBy1 1 ft s

= θ 10 deg= t 1 s=

Given h 1 2

g t2= d vA t=

e vBy1 vB2 sin θ( )= vBy1 g t=

vA vB2 cos θ( )=

350

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Engineering Mechanics - Dynamics Chapter 15

vA

t

vBy1

vB2

θ

⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝

⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

Find vA t, vBy1, vB2, θ,( )= vA 29.30 fts= vB2 33.10 ft s

= θ 27.70 deg=

Problem 15-74

The tennis ball is struck with a horizontal velocity vA, strikes the smooth ground at B, and bounces upward at θ = θ1. Determine the initial velocity vA, the final velocity vB, and the coefficient of restitution between the ball and the ground.

Given:

h 7.5 ft=

d 20 ft=

θ1 30 deg=

g 32.2 ft

s2 =

Solution: θ θ1=

Guesses vA 1 ft s

= t 1 s= vBy1 1 ft s

= vB2 1 ft s

= e 0.5=

Given h 1 2

g t2= d vA t= vBy1 g t=

e vBy1 vB2 sin θ( )= vA vB2 cos θ( )=

vA

t

vBy1

vB2

e

⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝

⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠

Find vA t, vBy1, vB2, e,( )= vA 29.30 ft s

= vB2 33.84 ft s

= e 0.77=

Problem 15-75

The ping-pong ball has mass M. If it is struck with the velocity shown, determine how high h it rises above the end of the smooth table after the rebound. The coefficient of restitution is e.

351

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Engineering Mechanics - Dynamics Chapter 15

Given:

M 2 gm= a 2.25 m=

e 0.8= b 0.75 m=

θ 30 deg= g 9.81 m

s2 =

v 18 m s

=

Solution: Guesses v1x 1 m s

= v1y 1 m s

= v2x 1 m s

= v2y 1 m s

=

t1 1 s= t2 2 s= h 1 m=

Given v1x v cos θ( )= a v cos θ( )t1= v1y g t1 v sin θ( )+=

v2x v1x= e v1y v2y= b v2x t2= h v2y t2 g 2

⎛⎜ ⎝

⎞⎟ ⎠

t2 2−=

v1x

v1y

v2x

v2y

t1

t2

h

⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

Find v1x v1y, v2x, v2y, t1, t2, h,( )=

v1x

v1y

v2x

v2y

⎛ ⎜ ⎜ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

15.59

10.42

15.59

8.33

⎛⎜ ⎜ ⎜ ⎜ ⎝

⎞⎟ ⎟ ⎟ ⎟ ⎠

m s

= t1

t2

⎛ ⎜ ⎝

⎞ ⎟ ⎠

0.14

0.05 ⎛ ⎜ ⎝

⎞ ⎟ ⎠

s=

h 390 mm=

*Problem 15-76

The box B of weight WB is dropped from rest a distance d from the top of the plate P of weight WP, which is supported by the spring having a stiffness k. Determine the maximum compression imparted to the spring. Neglect the mass of the spring.

352

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Engineering Mechanics - Dynamics Chapter 15

Given: WB 5 lb= WP 10 lb= g 32.2 ft

s2 =

k 30 lb ft

= d 5 ft= e 0.6=

Solution:

δst WP

k = vB1 2g d=

Guesses vB2 1 ft s

= vP2 1 ft s

= δ 2 ft=

Given WB g

⎛ ⎜ ⎝

⎞ ⎟ ⎠

vB1 WB g

⎛ ⎜ ⎝

⎞ ⎟ ⎠

vB2 WP g

⎛ ⎜ ⎝

⎞ ⎟ ⎠

vP2+= e vB1 vP2 vB2−=

1 2

kδst 2 1

2

WP g

⎛ ⎜ ⎝

⎞ ⎟ ⎠

vP2 2+

1 2

kδ2 WP δ δst−( )−=

vB2

vP2

δ

⎛⎜ ⎜ ⎜ ⎝

⎞⎟ ⎟ ⎟ ⎠

Find vB2 vP2, δ,( )= vB2

vP2

⎛ ⎜ ⎝

⎞ ⎟ ⎠

1.20−

9.57 ⎛ ⎜ ⎝

⎞ ⎟ ⎠

ft s

= δ 1.31 ft=

Problem 15-77

A pitching machine throws the ball of weight M towards the wall with an initial velocity vA as shown. Determine (a) the velocity at which it strikes the wall at B, (b) the velocity at which it rebounds from the wall and (c) the distance d from the wall to where it strikes the ground at C.

Given:

M 0.5 kg= a 3 m=

vA 10 m s

= b 1.5 m=

e 0.5=θ 30 deg=

g 9.81 m

s2 =

353

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Engineering Mechanics - Dynamics Chapter 15

vBx1 1 m s

= vBx2 1 m s

=

vBy1 1 m s

= vBy2 1 m s

=

h 1 m= d 1 m=

t1 1 s= t2 1 s=

Given

vA cos θ( )t1 a= b vA sin θ( )t1+ 1 2

g t1 2− h=

vBy2 vBy1= vA sin θ( ) g t1vBy1=

d vBx2 t2= h vBy2 t2+ 1 2

g t2 2− 0=

vA cos θ( ) vBx1= e vBx1 vBx2=

vBx1

vBy1

vBx2

vBy2

h

t1

t2

d

⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

Find vBx1 vBy1, vBx2, vBy2, h, t1, t2, d,( )= vBx1

vBy1

⎛ ⎜ ⎝

⎞ ⎟ ⎠

8.81 m s

=

vBx2

vBy2

⎛ ⎜ ⎝

⎞ ⎟ ⎠

4.62 m s

=

d 3.96 m=

Problem 15-78

The box of weight Wb slides on the surface for which the coefficient of friction is μk. The box has velocity v when it is a distance d from the plate. If it strikes the plate, which has weight Wp and is held in position by an unstretched spring of stiffness k, determine the maximum compression imparted to the spring. The coefficient of restitution between the box and the plate is e. Assume that the plate slides smoothly.

354

GuessesSolution:

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Engineering Mechanics - Dynamics Chapter 15

Given:

Wb 20 lb= Wp 10 lb=

μk 0.3= k 400 lb ft

=

v 15 ft s

= e 0.8=

d 2 ft= g 32.2 ft

s2 =

Solution:

Guesses vb1 1 ft s

= vb2 1 ft s

= vp2 1 ft s

= δ 1 ft=

Given 1 2

Wb g

⎛ ⎜ ⎝

⎞ ⎟ ⎠

v2 μk Wb d− 1 2

Wb g

⎛ ⎜ ⎝

⎞ ⎟ ⎠

vb1 2=

Wb g

⎛ ⎜ ⎝

⎞ ⎟ ⎠

vb1 Wb g

⎛ ⎜ ⎝

⎞ ⎟ ⎠

vb2 Wp g

⎛ ⎜ ⎝

⎞ ⎟ ⎠

vp2+=

e vb1 vp2 vb2−= 1 2

Wp g

⎛ ⎜ ⎝

⎞ ⎟ ⎠

vp2 2 1

2 kδ2=

vb1

vb2

vp2

δ

⎛ ⎜ ⎜ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

Find vb1 vb2, vp2, δ,( )= vb1

vb2

vp2

⎛⎜ ⎜ ⎜ ⎝

⎞⎟ ⎟ ⎟ ⎠

13.65

5.46

16.38

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

ft s

= δ 0.456 ft=

Problem 15-79

The billiard ball of mass M is moving with a speed v when it strikes the side of the pool table at A. If the coefficient of restitution between the ball and the side of the table is e, determine the speed of the ball just after striking the table twice, i.e., at A, then at B. Neglect the size of the ball.

355

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Engineering Mechanics - Dynamics Chapter 15

Given:

M 200 gm=

v 2.5 m s

=

θ 45 deg=

e 0.6=

Solution:

Guesses

v2 1 m s

= θ2 1 deg= v3 1 m s

= θ3 1 deg=

Given e v sin θ( ) v2 sin θ2( )= v cos θ( ) v2 cos θ2( )= e v2 cos θ2( ) v3 sin θ3( )= v2 sin θ2( ) v3 cos θ3( )=

v2

v3

θ2

θ3

⎛⎜ ⎜ ⎜ ⎜ ⎜⎝

⎞⎟ ⎟ ⎟ ⎟ ⎟⎠

Find v2 v3, θ2, θ3,( )= v2

v3

⎛ ⎜ ⎝

⎞ ⎟ ⎠

2.06

1.50 ⎛ ⎜ ⎝

⎞ ⎟ ⎠

m s

= θ2

θ3

⎛ ⎜ ⎝

⎞ ⎟ ⎠

31.0

45.0 ⎛ ⎜ ⎝

⎞ ⎟ ⎠

deg=

v3 1.500 m s

=

*Problem 15-80

The three balls each have the same mass m. If A is released from rest at θ, determine the angle φ to which C rises after collision. The coefficient of restitution between each ball is e.

Solution:

Energy

0 l 1 cos θ( )−( )m g+ 1 2

m vA 2=

vA 2 1 cos θ( )−( )g l=

Collision of ball A with B:

m vA 0+ m v'A m v'B+= e vA v'B v'A−= v'B 1 2

1 e+( )v'B=

Collision of ball B with C:

356

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Engineering Mechanics - Dynamics Chapter 15

m v'B 0+ m v''B m v''C+= e v'B v''C v''B−= v''C 1 4

1 e+( )2vA=

Energy

1 2

m v''c 2 0+ 0 l 1 cos φ( )−( )m g+= 1

2 1 16

⎛⎜ ⎝

⎞⎟ ⎠

1 e+( )4 2( ) 1 cos θ( )−( ) 1 cos φ( )−( )=

1 e+ 2

⎛⎜ ⎝

⎞⎟ ⎠

4 1 cos θ( )−( ) 1 cos φ( )−= φ acos 1 1 e+

2 ⎛⎜ ⎝

⎞⎟ ⎠

4 1 cos θ( )−( )−

⎡ ⎢ ⎣

⎤ ⎥ ⎦

=

Problem 15-81

Two smooth billiard balls A and B each have mass M. If A strikes B with a velocity vA as shown, determine their final velocities just after collision. Ball B is originally at rest and the coefficient of restitution is e. Neglect the size of each ball.

Given:

M 0.2 kg=

θ 40 deg=

vA 1.5 m s

=

e 0.85=

Solution: Guesses vA2 1 m s

= vB2 1 m s

= θ2 20 deg=

Given MvA cos θ( ) M vB2 M vA2 cos θ2( )+=

e vA cos θ( ) vA2 cos θ2( ) vB2−=

vA sin θ( ) vA2 sin θ2( )=

vA2

vB2

θ2

⎛ ⎜ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎠

Find vA2 vB2, θ2,( )= θ2 95.1 deg= vA2

vB2

⎛ ⎜ ⎝

⎞ ⎟ ⎠

0.968

1.063− ⎛ ⎜ ⎝

⎞ ⎟ ⎠

m s

=

357

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Engineering Mechanics - Dynamics Chapter 15

The two hockey pucks A and B each have a mass M. If they collide at O and are deflected along the colored paths, determine their speeds just after impact. Assume that the icy surface over which they slide is smooth. Hint: Since the y' axis is not along the line of impact, apply the conservation of momentum along the x' and y' axes.

Given:

M 250 g= θ1 30 deg=

v1 40 m s

= θ2 20 deg=

v2 60 m s

= θ3 45 deg=

Solution:

Initial Guess:

vA2 5 m s

= vB2 4 m s

=

Given

M v2 cos θ3( ) M v1 cos θ1( )+ M vA2 cos θ1( ) M vB2 cos θ2( )+=

Mv2 sin θ3( ) M v1 sin θ1( )+ M vA2 sin θ1( ) M vB2 sin θ2( )−= vA2

vB2

⎛ ⎜ ⎝

⎞ ⎟ ⎠

Find vA2 vB2,( )= vA2

vB2

⎛ ⎜ ⎝

⎞ ⎟ ⎠

6.90

75.66 ⎛ ⎜ ⎝

⎞ ⎟ ⎠

m s

=

Problem 15-83

Two smooth coins A and B, each having the same mass, slide on a smooth surface with the motion shown. Determine the speed of each coin after collision if they move off along the dashed paths. Hint: Since the line of impact has not been defined, apply the conservation of momentum along the x and y axes, respectively.

358

Problem 15-82

docsity.com

Engineering Mechanics - Dynamics Chapter 15

Given:

vA1 0.5 ft s

=

vB1 0.8 ft s

=

α 30 deg=

β 45 deg=

γ 30 deg=

c 4=

d 3=

Solution:

Guesses vB2 0.25 ft s

= vA2 0.5 ft s

=

Given

vA1c

c2 d2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

vB1 sin γ( )− vA2− sin β( ) vB2 cos α( )−=

vA1d

c2 d2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

vB1 cos γ( )+ vA2 cos β( ) vB2 sin α( )−=

vA2

vB2

⎛ ⎜ ⎝

⎞ ⎟ ⎠

Find vA2 vB2,( )= vA2

vB2

⎛ ⎜ ⎝

⎞ ⎟ ⎠

0.766

0.298 ⎛ ⎜ ⎝

⎞ ⎟ ⎠

ft s

=

*Problem 15-84

The two disks A and B have a mass MA and MB, respectively. If they collide with the initial velocities shown, determine their velocities just after impact. The coefficient of restitution is e.

Given:

MA 3 kg=

MB 5 kg=

θ 60 deg=

vB1 7 m s

=

359

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Engineering Mechanics - Dynamics Chapter 15

vA1 6 m s

=

e 0.65=

Solution: Guesses vA2 1 m s

= vB2 1 m s

= θ2 20 deg=

Given

MA vA1 MB vB1 cos θ( )− MA vA2 MB vB2 cos θ2( )+= e vA1 vB1 cos θ( )+( ) vB2 cos θ2( ) vA2−= vB1 sin θ( ) vB2 sin θ2( )=

vA2

vB2

θ2

⎛ ⎜ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎠

Find vA2 vB2, θ2,( )= θ2 68.6 deg= vA2

vB2

⎛ ⎜ ⎝

⎞ ⎟ ⎠

3.80−

6.51 ⎛ ⎜ ⎝

⎞ ⎟ ⎠

m s

=

Problem 15-85

Two smooth disks A and B each have mass M. If both disks are moving with the velocities shown when they collide, determine their final velocities just after collision. The coefficient of restitution is e.

Given:

M 0.5 kg= c 4= vA1 6 m s

=

e 0.75= d 3= vB1 4 m s

=

Solution:

Guesses

vA2 1 m s

= vB2 1 m s

= θA 10 deg= θB 10 deg=

Given vA1 0( ) vA2 sin θA( )= vB1 c c2 d2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

vB2 sin θB( )=

M vB1 d

c2 d2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

M vA1M vA2 cos θA( ) M vB2 cos θB( )−=

e vA1 vB1 d

c2 d2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

+⎡⎢ ⎣

⎤ ⎥ ⎦

vA2 cos θA( ) vB2 cos θB( )+=

360

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Engineering Mechanics - Dynamics Chapter 15

vA2

vB2

θA

θB

⎛⎜ ⎜ ⎜ ⎜ ⎜⎝

⎞⎟ ⎟ ⎟ ⎟ ⎟⎠

Find vA2 vB2, θA, θB,( )= θA

θB

⎛ ⎜ ⎝

⎞ ⎟ ⎠

0.00

32.88 ⎛ ⎜ ⎝

⎞ ⎟ ⎠

deg= vA2

vB2

⎛ ⎜ ⎝

⎞ ⎟ ⎠

1.35

5.89 ⎛ ⎜ ⎝

⎞ ⎟ ⎠

m s

=

Problem 15-86

Two smooth disks A and B each have mass M. If both disks are moving with the velocities shown when they collide, determine the coefficient of restitution between the disks if after collision B travels along a line angle θ counterclockwise from the y axis.

Given:

M 0.5 kg= c 4= vA1 6 m s

=

θB 30 deg= d 3= vB1 4 m s

=

Solution:

Guesses

vA2 2 m s

= vB2 1 m s

= θA 10 deg= e 0.5=

Given vA10 vA2 sin θA( )= vB1 c c2 d2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

vB2 cos θB( )=

M vB1 d

c2 d2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

M vA1M vA2 cos θA( ) M vB2 sin θB( )−=

e vA1 vB1 d

c2 d2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

+⎡⎢ ⎣

⎤ ⎥ ⎦

vA2 cos θA( ) vB2 sin θB( )+=

vA2

vB2

θA

e

⎛ ⎜ ⎜ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

Find vA2 vB2, θA, e,( )= vA2

vB2

⎛ ⎜ ⎝

⎞ ⎟ ⎠

1.75−

3.70 ⎛ ⎜ ⎝

⎞ ⎟ ⎠

m s

= e 0.0113=

Problem 15-87

Two smooth disks A and B have the initial velocities shown just before they collide at O. If they have masses mA and mB, determine their speeds just after impact. The coefficient of restitution is e.

361

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Engineering Mechanics - Dynamics Chapter 15

Given:

vA 7 m s

= mA 8 kg= c 12= e 0.5=

vB 3 m s

= mB 6 kg= d 5=

Solution: θ atan d c

⎛⎜ ⎝

⎞⎟ ⎠

= θ 22.62 deg=

Guesses vA2t 1 m s

= vA2n 1 m s

=

vB2t 1 m s

= vB2n 1 m s

=

Given vB cos θ( ) vB2t= vA− cos θ( ) vA2t=

mB vB sin θ( ) mA vA sin θ( )− mB vB2n mA vA2n+=

e vB vA+( ) sin θ( ) vA2n vB2n−=

vA2t

vA2n

vB2t

vB2n

⎛ ⎜ ⎜ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

Find vA2t vA2n, vB2t, vB2n,( )=

vA2t

vA2n

vB2t

vB2n

⎛ ⎜ ⎜ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

6.46−

0.22−

2.77

2.14−

⎛⎜ ⎜ ⎜ ⎜ ⎝

⎞⎟ ⎟ ⎟ ⎟ ⎠

m s

=

vA2 vA2t 2 vA2n

2+= vA2 6.47 m s

=

vB2 vB2t 2 vB2n

2+= vB2 3.50 m s

=

*Problem 15-88

The “stone” A used in the sport of curling slides over the ice track and strikes another “stone” B as shown. If each “stone” is smooth and has weight W, and the coefficient of restitution between the “stones” is e, determine their speeds just after collision. Initially A has velocity vA1 and B is at rest. Neglect friction.

Given: W 47 lb= vA1 8 ft s

=

e 0.8= θ 30 deg=

362

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Engineering Mechanics - Dynamics Chapter 15

Solution:

Guesses vA2t 1 ft s

= vA2n 1 ft s

=

vB2t 1 ft s

= vB2n 1 ft s

=

Given vA1 sin θ( ) vA2t= 0 vB2t=

vA1 cos θ( ) vA2n vB2n+=

e vA1 cos θ( ) vB2n vA2n−=

vA2t

vA2n

vB2t

vB2n

⎛ ⎜ ⎜ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

Find vA2t vA2n, vB2t, vB2n,( )=

vA2t

vA2n

vB2t

vB2n

⎛ ⎜ ⎜ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

4.00

0.69

0.00

6.24

⎛⎜ ⎜ ⎜ ⎜ ⎝

⎞⎟ ⎟ ⎟ ⎟ ⎠

ft s

=

vA2 vA2t 2 vA2n

2+= vA2 4.06 ft s

=

vB2 vB2t 2 vB2n

2+= vB2 6.24 ft s

=

Problem 15-89

The two billiard balls A and B are originally in contact with one another when a third ball C strikes each of them at the same time as shown. If ball C remains at rest after the collision, determine the coefficient of restitution. All the balls have the same mass. Neglect the size of each ball.

Solution:

Conservation of “x” momentum:

m v 2m v' cos 30 deg( )=

v 2v' cos 30 deg( )= 1( )

Coefficient of restitution:

e v'

v cos 30 deg( ) = 2( )

Substituiting Eq. (1) into Eq. (2) yields:

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Engineering Mechanics - Dynamics Chapter 15

e v'

2v' cos 30 deg( )2 = e

2 3

=

Problem 15-90

Determine the angular momentum of particle A of weight W about point O. Use a Cartesian vector solution.

Given:

W 2 lb= a 3 ft=

b 2 ft=vA 12 ft s

=

c 2 ft=

g 32.2 ft

s2 = d 4 ft=

Solution:

rOA

c

a b+

d

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

= rv

c

b

d

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

= vAv vA rv rv

=

HO rOA WvAv( )×= HO 1.827−

0.000

0.914−

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

slug ft2

s ⋅=

Problem 15-91

Determine the angular momentum HO of the particle about point O.

Given:

M 1.5 kg=

v 6 m s

=

a 4 m=

b 3 m=

c 2 m=

d 4 m=

364

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Engineering Mechanics - Dynamics Chapter 15

Solution:

rOA

c

b

d

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

= rAB

c

a

d

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

= vA v rAB rAB

=

HO rOA MvA( )×= HO 42.0

0.0

21.0

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

kg m2⋅ s

=

*Problem 15-92

Determine the angular momentum HO of each of the particles about point O.

Given: θ 30 deg= φ 60 deg=

mA 6 kg= c 2 m=

mB 4 kg= d 5 m=

mC 2 kg= e 2 m=

vA 4 m s

= f 1.5 m=

vB 6 m s

= g 6 m=

vC 2.6 m s

= h 2 m=

a 8 m= l 5=

b 12 m= n 12=

Solution:

HAO a mA vA sin φ( ) b mA vA cos φ( )−= HAO 22.3 kg m2⋅

s =

HBO fmB vB cos θ( ) e mB vB sin θ( )+= HBO 7.18− kg m2⋅

s =

HCO hmC n

l2 n2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

vC g mC l

l2 n2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

vC−= HCO 21.60− kg m2⋅

s =

365

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Engineering Mechanics - Dynamics Chapter 15

Problem 15-93

Determine the angular momentum HP of each of the particles about point P.

Given: θ 30 deg= φ 60 deg= a 8 m= f 1.5 m=

mA 6 kg= b 12 m= g 6 m=vA 4 m s

=

c 2 m= h 2 m= mB 4 kg= vB 6

m s

= d 5 m= l 5=

mC 2 kg= vC 2.6 m s

= e 2 m= n 12=

Solution:

HAP mA vA sin φ( ) a d−( ) mA vA cos φ( ) b c−( )−=

HAP 57.6− kg m2⋅

s =

HBP mB vB cos θ( ) c f−( ) mB vB sin θ( ) d e+( )+=

HBP 94.4 kg m2⋅

s =

HCP mCn

l2 n2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

vC c h+( ) mC l

l2 n2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

vC d g+( )−=

HCP 41.2− kg m2⋅

s =

Problem 15-94

Determine the angular momentum HO of the particle about point O.

Given:

W 10 lb= d 9 ft=

v 14 ft s

= e 8 ft=

a 5 ft= f 4 ft=

b 2 ft= g 5 ft=

c 3 ft= h 6 ft=

366

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Engineering Mechanics - Dynamics Chapter 15

Solution:

rOA

f

g

h

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

= rAB

f e+

d g

h

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

=

vA v rAB rAB

= HO rOA WvA( )×= HO 16.78−

14.92

23.62−

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

slug ft2

s ⋅=

Problem 15-95

Determine the angular momentum HP of the particle about point P.

Given:

W 10 lb= d 9 ft=

v 14 ft s

= e 8 ft=

a 5 ft= f 4 ft=

b 2 ft= g 5 ft=

c 3 ft= h 6 ft=

Solution:

rPA

fc

b g+

h a

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

= rAB

f e+

d g

h

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

=

vA v rAB rAB

= HP rPA WvA( )×= HP 14.30−

9.32−

34.81−

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

slug ft2

s ⋅=

*Problem 15-96

Determine the total angular momentum HO for the system of three particles about point O. All the particles are moving in the x-y plane.

Given:

mA 1.5 kg= a 900 mm=

367

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Engineering Mechanics - Dynamics Chapter 15

vA 4 m s

= b 700 mm=

mB 2.5 kg= c 600 mm=

vB 2 m s

= d 800 mm=

mC 3 kg= e 200 mm=

vC 6 m s

=

Solution:

HO

a

0

0

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

mA

0

vA

0

⎛⎜ ⎜ ⎜⎝

⎞⎟ ⎟ ⎟⎠

⎡⎢ ⎢ ⎢⎣

⎤⎥ ⎥ ⎥⎦

×

c

b

0

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

mB

vB

0

0

⎛⎜ ⎜ ⎜⎝

⎞⎟ ⎟ ⎟⎠

⎡⎢ ⎢ ⎢⎣

⎤⎥ ⎥ ⎥⎦

×+

d

e

0

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

mC

0

vC

0

⎛⎜ ⎜ ⎜⎝

⎞⎟ ⎟ ⎟⎠

⎡⎢ ⎢ ⎢⎣

⎤⎥ ⎥ ⎥⎦

×+=

HO

0.00

0.00

12.50

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

kg m2⋅ s

=

Problem 15-97

Determine the angular momentum HO of each of the two particles about point O. Use a scalar solution.

Given:

mA 2 kg= c 1.5 m=

mB 1.5 kg= d 2 m=

e 4 m=vA 15 m s

=

f 1 m=

vB 10 m s

= θ 30 deg=

a 5 m= l 3=

b 4 m= n 4=

368

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Engineering Mechanics - Dynamics Chapter 15

Solution:

HOA mAn

n2 l2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

vA c mA l

n2 l2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

vA d−= HOA 72.0− kg m2⋅

s =

HOB mBvB cos θ( )e mB vB sin θ( ) f−= HOB 59.5− kg m2⋅

s =

Problem 15-98

Determine the angular momentum HP of each of the two particles about point P. Use a scalar solution.

Given:

mA 2 kg= c 1.5 m=

d 2 m=mB 1.5 kg=

e 4 m= vA 15

m s

= f 1 m=

vB 10 m s

= θ 30 deg=

a 5 m= l 3=

b 4 m= n 4=

Solution:

HPA mA n

n2 l2+ vA b c−( ) mA

l

n2 l2+ vA a d+( )−= HPA 66.0−

kg m2⋅ s

=

HPB mBvB cos θ( ) b e+( ) mB vB sin θ( ) a f−( )+= HPB 73.9− kg m2⋅

s =

Problem 15-99

The ball B has mass M and is attached to the end of a rod whose mass may be neglected. If the rod is subjected to a torque M = at2 + bt + c, determine the speed of the ball when t = t1. The ball has a speed v = v0 when t = 0.

369

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Engineering Mechanics - Dynamics Chapter 15

Given:

M 10 kg=

a 3 N m⋅

s2 =

b 5 N m⋅

s =

c 2 N m⋅=

t1 2 s=

v0 2 m s

=

L 1.5 m=

Solution: Principle of angular impulse momentum

M v0 L 0

t1 ta t2 b t+ c+

⌠ ⎮ ⌡

d+ M v1 L=

v1 v0 1

M L 0

t1 ta t2 b t+ c+

⌠ ⎮ ⌡

d+= v1 3.47 m s

=

*Problem 15-100

The two blocks A and B each have a mass M0. The blocks are fixed to the horizontal rods, and their initial velocity is v' in the direction shown. If a couple moment of M is applied about shaft CD of the frame, determine the speed of the blocks at time t. The mass of the frame is negligible, and it is free to rotate about CD. Neglect the size of the blocks.

Given:

M0 0.4 kg=

a 0.3 m=

v' 2 m s

=

M 0.6 N m⋅=

t 3 s=

Solution:

2a M0 v' M t+ 2a M0 v=

370

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Engineering Mechanics - Dynamics Chapter 15

v v' M t

2a M0 += v 9.50

m s

=

Problem 15-101

The small cylinder C has mass mC and is attached to the end of a rod whose mass may be

neglected. If the frame is subjected to a couple M = at2 + b, and the cylinder is subjected to force F, which is always directed as shown, determine the speed of the cylinder when t = t1. The cylinder has a speed v0 when t = 0.

Given:

mC 10 kg= t1 2 s=

a 8 N m

s2 = v0 2

m s

=

d 0.75 m= b 5 N m⋅=

e 4=

F 60 N= f 3=

Solution:

mC v0 d 0

t1 ta t2 b+

⌠ ⎮ ⌡

d+ f

e2 f 2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

F d t1+ mC v1 d=

v1 v0 1

mC d 0

t1 ta t2 b+

⌠ ⎮ ⌡

d f

e2 f 2+

⎛ ⎜ ⎝

⎞ ⎟ ⎠

F d t1+ ⎡ ⎢ ⎢ ⎣

⎤ ⎥ ⎥ ⎦

+= v1 13.38 m s

=

Problem 15-102

A box having a weight W is moving around in a circle of radius rA with a speed vA1 while connected to the end of a rope. If the rope is pulled inward with a constant speed vr, determine the speed of the box at the instant r = rB. How much work is done after pulling in the rope from A to B? Neglect friction and the size of the box.

Given:

W 8 lb=

rA 2 ft=

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Engineering Mechanics - Dynamics Chapter 15

vA1 5 ft s

=

vr 4 ft s

=

rB 1 ft=

g 32.21 ft

s2 =

Solution:

W g

⎛⎜ ⎝

⎞⎟ ⎠

rA vA1 W g

⎛⎜ ⎝

⎞⎟ ⎠

rB vBtangent=

vBtangent rA vA1 rB

⎛ ⎜ ⎝

⎞ ⎟ ⎠

= vBtangent 10.00 ft s

=

vB vBtangent 2 vr

2+= vB 10.8 ft s

=

UAB 1 2

W g

⎛⎜ ⎝

⎞⎟ ⎠

vB 2 1

2 W g

⎛⎜ ⎝

⎞⎟ ⎠

vA1 2−= UAB 11.3 ft lb⋅=

Problem 15-103

An earth satellite of mass M is launched into a free-flight trajectory about the earth with initial speed vA when the distance from the center of the earth is rA. If the launch angle at this position is φA determine the speed vB of the satellite and its closest distance rB from the center of the earth. The earth has a mass Me. Hint: Under these conditions, the satellite is subjected only to the earth’s gravitational force, F, Eq. 13-1. For part of the solution, use the conservation of energy.

Units used: Mm 103 km=

Given:

φA 70 deg=M 700 kg=

vA 10 km s

= G 6.673 10 11−× N m2⋅

kg2 =

rA 15 Mm= Me 5.976 10 24× kg=

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Engineering Mechanics - Dynamics Chapter 15

Solution: Guesses vB 10 km s

= rB 10 Mm=

Given M vA sin φA( )rA M vB rB=

1 2

M vA 2 G Me M

rA

1 2

M vB 2 G Me M

rB −=

vB

rB

⎛ ⎜ ⎝

⎞ ⎟ ⎠

Find vB rB,( )= vB 10.2 km s

= rB 13.8 Mm=

*Problem 15-104

The ball B has weight W and is originally rotating in a circle. As shown, the cord AB has a length of L and passes through the hole A, which is a distance h above the plane of motion. If L/2 of the cord is pulled through the hole, determine the speed of the ball when it moves in a circular path at C.

Given:

W 5 lb=

L 3 ft=

h 2 ft=

g 32.2 ft

s2 =

Solution: θB acos h L

⎛⎜ ⎝

⎞⎟ ⎠

= θB 48.19 deg=

Guesses TB 1 lb= TC 1 lb= vB 1 ft s

= vC 1 ft s

= θC 10 deg=

Given TB cos θB( ) W− 0= TB sin θB( ) Wg vB

2

L sin θB( ) ⎛⎜ ⎜⎝

⎞⎟ ⎟⎠

=

TC cos θC( ) W− 0= TC sin θC( ) Wg vC

2

L 2

sin θC( )

⎛⎜ ⎜ ⎜⎝

⎞⎟ ⎟ ⎟⎠

=

W g

⎛⎜ ⎝

⎞⎟ ⎠

vB L sin θB( ) Wg ⎛⎜ ⎝

⎞⎟ ⎠

vC L 2

⎛⎜ ⎝

⎞⎟ ⎠

sin θC( )=

373

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Engineering Mechanics - Dynamics Chapter 15

TB

TC

vB

vC

θC

⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

Find TB TC, vB, vC, θC,( )= TB

TC

⎛ ⎜ ⎝

⎞ ⎟ ⎠

7.50

20.85 ⎛ ⎜ ⎝

⎞ ⎟ ⎠

lb= θC 76.12 deg=

vB 8.97 ft s

= vC 13.78 ft s

=

Problem 15-105

The block of weight W rests on a surface for which the kinetic coefficient of friction is μk. It is acted upon by a radial force FR and a horizontal force FH, always directed at angle θ from the tangent to the path as shown. If the block is initially moving in a circular path with a speed v1 at the instant the forces are applied, determine the time required before the tension in cord AB becomes T. Neglect the size of the block for the calculation.

Given:

W 10 lb= μk 0.5=

FR 2 lb= T 20 lb=

FH 7 lb= r 4 ft=

v1 2 ft s

= g 32.2 ft

s2 =

θ 30 deg=

Solution:

Guesses t 1 s= v2 1 ft s

=

Given

W g

⎛⎜ ⎝

⎞⎟ ⎠

v1 r FHcos θ( )r t+ μk W rtW g

⎛⎜ ⎝

⎞⎟ ⎠

v2 r=

FR FH sin θ( )+ TW g

v2

2

r

⎛ ⎜ ⎝

⎞ ⎟ ⎠

=

t

v2

⎛ ⎜ ⎝

⎞ ⎟ ⎠

Find t v2,( )= v2 13.67 ft s

= t 3.41 s=

374

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Engineering Mechanics - Dynamics Chapter 15

Problem 15-106

The block of weight W is originally at rest on the smooth surface. It is acted upon by a radial force FR and a horizontal force FH, always directed at θ from the tangent to the path as shown. Determine the time required to break the cord, which requires a tension T. What is the speed of the block when this occurs? Neglect the size of the block for the calculation.

Given:

W 10 lb= θ 30 deg=

FR 2 lb= T 30 lb=

FH 7 lb= r 4 ft=

v1 0 ft s

= g 32.2 ft

s2 =

Solution:

Guesses t 1 s= v2 1 ft s

=

Given

W g

⎛⎜ ⎝

⎞⎟ ⎠

v1 r FHcos θ( )r t+ W g

⎛⎜ ⎝

⎞⎟ ⎠

v2 r=

FR FH sin θ( )+ TW g

v2

2

r

⎛ ⎜ ⎝

⎞ ⎟ ⎠

=

t

v2

⎛ ⎜ ⎝

⎞ ⎟ ⎠

Find t v2,( )= v2 17.76 ft s

= t 0.91 s=

Problem 15-107

The roller-coaster car of weight W starts from rest on the track having the shape of a cylindrical helix. If the helix descends a distance h for every one revolution, determine the time required for the car to attain a speed v. Neglect friction and the size of the car.

Given:

W 800 lb=

h 8 ft=

v 60 ft s

=

375

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Engineering Mechanics - Dynamics Chapter 15

r 8 ft=

Solution:

θ atan h

r ⎛⎜ ⎝

⎞⎟ ⎠

= θ 9.04 deg=

FN W cos θ( )− 0= FN W cos θ( )= FN 790.06 lb=

vt v cos θ( )= vt 59.25 ft s

=

HA tM ⌠ ⎮ ⌡

d+ H2= 0

t tFN sin θ( )r

⌠ ⎮ ⌡

d W g

⎛⎜ ⎝

⎞⎟ ⎠

h vt= FN sin θ( )r t W g

⎛⎜ ⎝

⎞⎟ ⎠

h vt=

t W vt h

FN sin θ( )g r ⎛ ⎜ ⎝

⎞ ⎟ ⎠

= t 11.9 s=

*Problem 15-108

A child having mass M holds her legs up as shown as she swings downward from rest at θ1. Her center of mass is located at point G1. When she is at the bottom position θ = 0°, she suddenly lets her legs come down, shifting her center of mass to position G2. Determine her speed in the upswing due to this sudden movement and the angle θ2 to which she swings before momentarily coming to rest. Treat the child’s body as a particle.

Given:

M 50 kg= r1 2.80 m= g 9.81 m

s2 =

θ1 30 deg= r2 3 m=

Solution:

v2b 2g r1 1 cos θ1( )−( )= v2b 2.71 ms=

r1 v2b r2 v2a= v2a r1 r2

v2b= v2a 2.53 m s

=

θ2 acos 1 v2a

2

2g r2

⎛⎜ ⎜⎝

⎞⎟ ⎟⎠

= θ2 27.0 deg=

376

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Engineering Mechanics - Dynamics Chapter 15

Problem 15-109

A small particle having a mass m is placed inside the semicircular tube. The particle is placed at the position shown and released. Apply the principle of angular momentum about point OM0 = H0), and show that the motion of the particle is governed by the differential equation θ'' + (g / R) sin θ = 0.

Solution:

ΣM0 t H0

d d

=

Rm g sin θ( ) t

m v R( )d d

=

g sin θ( ) t vd

d −= 2t

sd

d

2 −=

But, s Rθ=

Thus, g sin θ( ) R− θ''=

or, θ'' g R

⎛⎜ ⎝

⎞⎟ ⎠

sin θ( )+ 0=

Problem 15-110

A toboggan and rider, having a total mass M, enter horizontally tangent to a circular curve (θ1) with a velocity vA. If the track is flat and banked at angle θ2, determine the speed vB and the angle θ of “descent”, measured from the horizontal in a vertical xz plane, at which the toboggan exists at B. Neglect friction in the calculation.

Given:

M 150 kg= θ1 90 deg= vA 70 km hr

= θ2 60 deg=

rA 60 m= rB 57 m= r 55 m=

377

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Engineering Mechanics - Dynamics Chapter 15

Solution:

h rA rB−( )tan θ2( )=

Guesses vB 10 m s

= θ 1 deg=

Given 1 2

M vA 2 M g h+

1 2

M vB 2= M vA rA M vB cos θ( )rB=

vB

θ

⎛ ⎜ ⎝

⎞ ⎟ ⎠

Find vB θ,( )= vB 21.9 ms= θ 1.1− 10 3× deg=

Problem 15-111

Water is discharged at speed v against the fixed cone diffuser. If the opening diameter of the nozzle is d, determine the horizontal force exerted by the water on the diffuser.

Units Used:

Mg 103 kg=

Given:

v 16 m s

= θ 30 deg=

d 40 mm= ρw 1 Mg

m3 =

Solution:

Q π 4

d2v= m' ρw Q=

Fx m' v− cos θ 2

⎛⎜ ⎝

⎞⎟ ⎠

v+⎛⎜ ⎝

⎞⎟ ⎠

=

Fx 11.0 N=

*Problem 15-112

A jet of water having cross-sectional area A strikes the fixed blade with speed v. Determine the horizontal and vertical components of force which the blade exerts on the water.

Given:

A 4 in2=

378

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Engineering Mechanics - Dynamics Chapter 15

v 25 ft s

=

θ 130 deg=

γw 62.4 lb

ft3 =

Solution: Q A v= Q 0.69 ft3

s =

t md

d m'= ρQ= m' γw Q= m' 1.3468

slug s

=

vAx v= vAy 0 ft s

= vBx v cos θ( )= vBy v sin θ( )=

Fx m'g

vBx vAx−( )= Fx 55.3 lb=

Fy m' g

vBy vAy−( )= Fy 25.8 lb=

Problem 15-113

Water is flowing from the fire hydrant opening of diameter dB with velocity vB. Determine the horizontal and vertical components of force and the moment developed at the base joint A, if the static (gauge) pressure at A is PA. The diameter of the fire hydrant at A is dA.

Units Used:

kPa 103 Pa=

Mg 103 kg=

kN 103 N=

Given:

dB 150 mm= h 500 mm=

vB 15 m s

= dA 200 mm=

ρw 1 Mg

m3 =PA 50 kPa=

Solution:

AB π dB 2

⎛ ⎜ ⎝

⎞ ⎟ ⎠

2 = AA π

dA 2

⎛ ⎜ ⎝

⎞ ⎟ ⎠

2 = m' ρw vBπ

dB 2

⎛ ⎜ ⎝

⎞ ⎟ ⎠

2 = vA

m' ρw AA

=

379

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Engineering Mechanics - Dynamics Chapter 15

Ax m' vB= Ax 3.98 kN=

Ay− 50π dA 2

⎛ ⎜ ⎝

⎞ ⎟ ⎠

2 + m' 0 vA−( )= Ay m' vA PAπ

dA 2

⎛ ⎜ ⎝

⎞ ⎟ ⎠

2 += Ay 3.81 kN=

M m' h vB= M 1.99 kN m⋅=

Problem 15-114

The chute is used to divert the flow of water Q. If the water has a cross-sectional area A, determine the force components at the pin A and roller B necessary for equilibrium. Neglect both the weight of the chute and the weight of the water on the chute.

Units Used:

Mg 103 kg= kN 103 N=

Given:

Q 0.6 m3

s = ρw 1

Mg

m3 =

A 0.05 m2= h 2 m=

a 1.5 m= b 0.12 m=

Solution:

t md

d m'= m' ρw Q=

vA Q A

= vB vA=

ΣFx m' vAx vBx−( )= Bx Axm' vAx vBx−( )= ΣFy m' vAy vBy−( )= Ay m' 0 vB−( )−⎡⎣ ⎤⎦= Ay 7.20 kN=

ΣMA m' d0A vA d0B vB−( )= Bx 1 h

m' b vA a b−( )vA+⎡⎣ ⎤⎦= Bx 5.40 kN=

Ax Bx m' vA−= Ax 1.80− kN=

380

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Engineering Mechanics - Dynamics Chapter 15

Problem 15-115

The fan draws air through a vent with speed v. If the cross-sectional area of the vent is A, determine the horizontal thrust on the blade. The specific weight of the air is γa.

Given:

v 12 ft s

=

A 2 ft2=

γa 0.076 lb

ft3 =

g 32.20 ft

s2 =

Solution:

m' t md

d = m' γa v A= m' 0.05669

slug s

=

T m' v 0−( )

g = T 0.68 lb=

*Problem 15-116

The buckets on the Pelton wheel are subjected to a jet of water of diameter d, which has velocity vw. If each bucket is traveling at speed vb when the water strikes it, determine the power developed by the wheel. The density of water is γw.

Given:

d 2 in= θ 20 deg=

vw 150 ft s

= g 32.2 ft

s2 =

vb 95 ft s

=

γw 62.4 lbf

ft3 =

381

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Engineering Mechanics - Dynamics Chapter 15

Solution: vA vw vb−= vA 55 ft s

=

vBx vA− cos θ( ) vb+= vBx 43.317 ft s

=

ΣFx m' vBx vAx−( )=

Fx γw g

⎛ ⎜ ⎝

⎞ ⎟ ⎠

π d2

4

⎛ ⎜ ⎝

⎞ ⎟ ⎠

vA vBxvA−( )−⎡⎣ ⎤⎦= Fx 266.41 m

s2 lb⋅=

P Fx vb= P 4.69 hp=

Problem 15-117

The boat of mass M is powered by a fan F which develops a slipstream having a diameter d. If the fan ejects air with a speed v, measured relative to the boat, determine the initial acceleration of the boat if it is initially at rest. Assume that air has a constant density ρa and that the entering air is essentially at rest. Neglect the drag resistance of the water.

Given:

M 200 kg=

h 0.375 m=

d 0.75 m=

v 14 m s

=

ρa 1.22 kg

m3 =

Solution:

Q A v= Q π 4

d2v= Q 6.1850 m3

s =

t md

d m'= m' ρa Q= m' 7.5457

kg s

=

ΣFx m' vBx vAx−( )=

F ρa Q v= F 105.64 N=

ΣFx M ax= F M a=

382

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Engineering Mechanics - Dynamics Chapter 15

a F M

= a 0.53 m

s2 =

Problem 15-118

The rocket car has a mass MC (empty) and carries fuel of mass MF. If the fuel is consumed at a constant rate c and ejected from the car with a relative velocity vDR, determine the maximum speed attained by the car starting from rest. The drag resistance due to the atmosphere is FD = kv2 and the speed is measured in m/s.

Units Used:

Mg 103 kg=

Given:

MC 3 Mg= MF 150 kg=

vDR 250 m s

= c 4 kg s

=

k 60 N s2

m2 ⋅=

Solution:

m0 MC MF+= At time t the mass of the car is m0 c t

Set F k v2= , then kv2 m0 c t−( ) t vd

d vDRc−=

Maximum speed occurs at the instant the fuel runs out. t MF c

= t 37.50 s=

Thus, Initial Guess: v 4 m s

=

Given

0

v

v 1

c vDR k v 2−

⌠⎮ ⎮ ⎮⌡

d

0

t

t 1

m0 c t

⌠ ⎮ ⎮ ⌡

d=

v Find v( )= v 4.06 m s

=

383

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Engineering Mechanics - Dynamics Chapter 15

Problem 15-119

A power lawn mower hovers very close over the ground. This is done by drawing air in at speed vA through an intake unit A, which has cross-sectional area AA and then discharging it at the ground, B, where the cross-sectional area is AB. If air at A is subjected only to atmospheric pressure, determine the air pressure which the lawn mower exerts on the ground when the weight of the mower is freely supported and no load is placed on the handle. The mower has mass M with center of mass at G. Assume that air has a constant density of ρa.

Given:

vA 6 m s

=

AA 0.25 m 2=

AB 0.35 m 2=

M 15 kg=

ρa 1.22 kg

m3 =

Solution: m' ρa AA vA= m' 1.83 kg s

=

+↑ ΣFy m' vBy vAy−( )= P AB M gm' 0 vA−( )−⎡⎣ ⎤⎦=

P 1

AB m' vA M g+( )= P 452 Pa=

*Problem 15-120

The elbow for a buried pipe of diameter d is subjected to static pressure P. The speed of the water passing through it is v. Assuming the pipe connection at A and B do not offer any vertical force resistance on the elbow, determine the resultant vertical force F that the soil must then exert on the elbow in order to hold it in equilibrium. Neglect the weight of the elbow and the water within it. The density of water is γw.

Given:

d 5 in= θ 45 deg=

384

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Engineering Mechanics - Dynamics Chapter 15

P 10 lb

in2 = γw 62.4

lb

ft3 =

v 8 ft s

=

Solution:

Q v π 4

d2⎛⎜ ⎝

⎞⎟ ⎠

=

m' γw g

Q=

Also, the force induced by the water pressure at A is

A π 4

d2=

F P A= F 196.35 lb=

2F cos θ( ) F1m' v− cos θ( ) v cos θ( )−( )=

F1 2 F cos θ( ) m' v cos θ( )+( )=

F1 302 lb=

Problem 15-121

The car is used to scoop up water that is lying in a trough at the tracks. Determine the force needed to pull the car forward at constant velocity v for each of the three cases. The scoop has a cross-sectional area A and the density of water is ρw.

385

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Engineering Mechanics - Dynamics Chapter 15

Solution:

The system consists of the car and the scoop. In all cases

ΣFs m t vd

d VDe

t me

d d

−=

F 0 Vρ A V−= F VA=

Problem 15-122

A rocket has an empty weight W1 and carries fuel of weight W2. If the fuel is burned at the rate c and ejected with a relative velocity vDR, determine the maximum speed attained by the rocket starting from rest. Neglect the effect of gravitation on the rocket.

Given: W1 500 lb= W2 300 lb= c 15 lb s

= vDR 4400 ft s

= g 32.2 ft

s2 =

Solution: m0 W1 W2+

g =

The maximum speed occurs when all the fuel is consumed, that is, where t W2 c

= t 20.00 s=

ΣFx m t vd

d vDR

t me

d d

−=

At a time t, M m0 c g

t−= , where c g t

me d d

= . In space the weight of the rocket is zero.

0 m0 c t−( ) t vd

d vDRc−=

Guess vmax 1 ft s

=

Given

0

vmax v1

⌠ ⎮ ⌡

d

0

t

t

c g

vDR

m0 c g

t

⌠⎮ ⎮ ⎮ ⎮ ⎮⌡

d=

vmax Find vmax( )= vmax 2068 ft s

=

386

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Engineering Mechanics - Dynamics Chapter 15

Problem 15-123

The boat has mass M and is traveling forward on a river with constant velocity vb, measured relative to the river. The river is flowing in the opposite direction at speed vR. If a tube is placed in the water, as shown, and it collects water of mass Mw in the boat in time t, determine the horizontal thrust T on the tube that is required to overcome the resistance to the water collection.

Units Used:

Mg 103 kg=

Given:

M 180 kg= Mw 40 kg=

vb 70 km hr

= t 80 s=

ρw 1 Mg

m3 =vR 5

km hr

=

Solution: m' Mw

t = m' 0.50

kg s

=

vdi vb= vdi 19.44 m s

=

ΣFi m t vd

d vdi m'+=

T vdi m'= T 9.72 N=

*Problem 15-124

The second stage of a two-stage rocket has weight W2 and is launched from the first stage with velocity v. The fuel in the second stage has weight Wf. If it is consumed at rate r and ejected with relative velocity vr, determine the acceleration of the second stage just after the engine is fired. What is the rocket’s acceleration just before all the fuel is consumed? Neglect the effect of gravitation.

Given:

W2 2000 lb= Wf 1000 lb= r 50 lb s

=

v 3000 mi hr

= vr 8000 ft s

= g 32.2 ft

s2 =

387

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Engineering Mechanics - Dynamics Chapter 15

Solution:

Initially,

ΣFs m t vd

d vdi

t me

d d

⎛ ⎜ ⎝

⎞ ⎟ ⎠

−=

0 W2 Wf+

g ⎛ ⎜ ⎝

⎞ ⎟ ⎠

a vr r g

−= a vr r

W2 Wf+ ⎛ ⎜ ⎝

⎞ ⎟ ⎠

= a 133 ft

s2 =

Finally,

a1 vr r

W2 ⎛ ⎜ ⎝

⎞ ⎟ ⎠

= a1 200 ft

s2 =0

W2 g

⎛ ⎜ ⎝

⎞ ⎟ ⎠

a1 vr r g

⎛⎜ ⎝

⎞⎟ ⎠

−=

Problem 15-125

The earthmover initially carries volume V of sand having a density ρ. The sand is unloaded horizontally through A dumping port P at a rate m' measured relative to the port. If the earthmover maintains a constant resultant tractive force F at its front wheels to provide forward motion, determine its acceleration when half the sand is dumped. When empty, the earthmover has a mass M. Neglect any resistance to forward motion and the mass of the wheels. The rear wheels are free to roll. Units Used:

Mg 103 kg=

kN 103 N=

Given:

A 2.5 m2= ρ 1520 kg

m3 =

m' 900 kg s

= V 10 m3=

F 4 kN=

M 30 Mg=

Solution:

When half the sand remains,

M1 M 1 2

Vρ+= M1 37600 kg=

388

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Engineering Mechanics - Dynamics Chapter 15

t md

d m'= ρv A= v

m' ρ A

= v 0.24 m s

=

Σ F m t vd

d t md

d vDR−= F M1 a m' v−=

a F m' v+

M1 = a 0.11

m

s2 =

a 112 mm

s2 =

Problem 15-126

The earthmover initially carries sand of volume V having density ρ. The sand is unloaded horizontally through a dumping port P of area A at rate of r measured relative to the port. Determine the resultant tractive force F at its front wheels if the acceleration of the earthmover is a when half the sand is dumped. When empty, the earthmover has mass M. Neglect any resistance to forward motion and the mass of the wheels. The rear wheels are free to roll.

Units Used:

kN 103 N=

Mg 1000 kg=

Given:

V 10 m3= r 900 kg s

=

ρ 1520 kg

m3 = a 0.1

m

s2 =

A 2.5 m2= M 30 Mg=

Solution:

When half the sand remains, M1 M 1 2

Vρ+= M1 37600 kg=

t md

d r= r ρv A= v

r ρ A

= v 0.237 m s

=

F m t vd

d t md

d v−= F M1 a rv−= F 3.55 kN=

389

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Engineering Mechanics - Dynamics Chapter 15

Problem 15-127

If the chain is lowered at a constant speed v, determine the normal reaction exerted on the floor as a function of time. The chain has a weight W and a total length l.

Given:

W 5 lb ft

=

l 20 ft=

v 4 ft s

=

Solution:

At time t, the weight of the chain on the floor is W M g v t( )=

t vd

d 0= Mt M v t( )=

t Mt

d d

M v=

Σ Fs M t vd

d vDt

t Mt

d d

+=

R M g v t( )− 0 v M v( )+=

R M g v t v2+( )= R W g

g v t v2+( )=

*Problem 15-128

The rocket has mass M including the fuel. Determine the constant rate at which the fuel must be burned so that its thrust gives the rocket a speed v in time t starting from rest. The fuel is expelled from the rocket at relative speed vr. Neglect the effects of air resistance and assume that g is constant.

Given:

M 65000 lb= vr 3000 ft s

=

v 200 ft s

=

g 32.2 ft

s2 =

t 10 s=

390

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Engineering Mechanics - Dynamics Chapter 15

Solution:

A System That Losses Mass: Here,

W m0 t me

d d

t−⎛⎜ ⎝

⎞ ⎟ ⎠

g=

Applying Eq. 15-29, we have

+↑Σ Fz m tv d d

vDE t me

d d

−= integrating we find

v vDE ln mo

m0 t me

d d

t

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

g t−=

with m0 M= vDE vr=

v vr ln m0

m0 t me

d d

t

⎛ ⎜ ⎜ ⎝

⎞ ⎟ ⎟ ⎠

g t( )−=

t me

d d

A= m0

e

v g t+ vr

m0+ ⎛⎜ ⎜ ⎜⎝

⎞⎟ ⎟ ⎟⎠

1 t

= A m0

e

v g t+ vr

m0+ ⎛⎜ ⎜ ⎜⎝

⎞⎟ ⎟ ⎟⎠

1 t

= A 43.3 slug

s =

Problem 15-129

The rocket has an initial mass m0, including the fuel. For practical reasons desired for the crew, it is required that it maintain a constant upward acceleration a0. If the fuel is expelled from the rocket at a relative speed ver, determine the rate at which the fuel should be consumed to maintain the motion. Neglect air resistance, and assume that the gravitational acceleration is constant.

Solution:

a0 t vd

d =

+↑ ΣFs = m t vd

d ver

t me

d d

mgmao ver t md

d −=

ver dm m

a0 g+( )dt=

391

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Engineering Mechanics - Dynamics Chapter 15

Since ver is constant, integrating, with t = 0 when m = m0 yields

ver ln m m0

⎛ ⎜ ⎝

⎞ ⎟ ⎠

a0 g+( )t= m m0

e

a0 g+

ver

⎛ ⎜ ⎝

⎞ ⎟ ⎠

t

=

The time rate fuel consumption is determined from Eq.[1]

t md

d m

a0 g+

ver =

t md

d m0

a0 g+

ver

⎛ ⎜ ⎝

⎞ ⎟ ⎠

e

a0 g+

ver

⎛ ⎜ ⎝

⎞ ⎟ ⎠

t

=

Note : ver must be considered a negative quantity.

Problem 15-130

The jet airplane of mass M has constant speed vj when it is flying along a horizontal straight line. Air enters the intake scoops S at rate r1. If the engine burns fuel at the rate r2 and the gas (air and fuel) is exhausted relative to the plane with speed ve, determine the resultant drag force exerted on the plane by air resistance. Assume that air has a constant density ρ. Hint: Since mass both enters and exits the plane, Eqs. 15-29 and 15-30 must be combined.

Units Used:

Mg 1000 kg=

kN 103 N=

Given:

M 12 Mg= r2 0.4 kg s

=

vj 950 km hr

= ve 450 m s

=

r1 50 m3

s = ρ 1.22

kg

m3 =

Solution:

ΣFs m t vd

d t me

d d

vDE( )− t mi

d d

vDi( )+=

t vd

d 0= vDE Ve= vDi vj=

t mi

d d

r1ρ=

392

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Engineering Mechanics - Dynamics Chapter 15

A r1ρ= t me

d d

r2 A+= B r2 A+=

Forces T and FD are incorporated as the last two terms in the equation,

FD ve B vj A−= FD 11.5 kN=

Problem 15-131

The jet is traveling at speed v, angle θ with the horizontal. If the fuel is being spent at rate r1 and the engine takes in air at r2 whereas the exhaust gas (air and fuel) has relative speed ve, determine the

acceleration of the plane at this instant. The drag resistance of the air is FD = kv2 The jet has weight W. Hint: See Prob. 15-130.

Given:

v 500 mi hr

= ve 32800 ft s

=

θ 30 deg= k1 0.7 lb s2

ft2 =

r1 3 lb s

= W 15000 lb=

r2 400 lb s

= g 32.2 ft

s2 =

Solution:

t mi

d d

r2 g1

= A1 r2= t me

d d

r1 r2+

g1 = B r1 r2+= v1 v=

ΣFs m t vd

d vDe

t me

d d

vDi t mi

d d

+=

W− sin θ( ) k1 v12− W a ve Bv1 A1+=

a W− sin θ( ) k1 v12− ve

B g

+ v1 A1 g

− ⎛ ⎜ ⎝

⎞ ⎟ ⎠

g

W = a 37.5

ft

s2 =

393

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Engineering Mechanics - Dynamics Chapter 15

*Problem 15-132

The rope has a mass m' per unit length. If the end length y = h is draped off the edge of the table, and released, determine the velocity of its end A for any position y, as the rope uncoils and begins to fall.

Solution:

Fs m t vd

d vDi

t mi

d d

+= At a time t, m m' y= and t mi

d d

m' t yd

d = m' v= .

Here, vDi v= , t vd

d g= .

m' g y m' y t vd

d v m' v( )+=

g y y t vd

d v2+= Since v

t yd

d = , then dt

dy v

=

g y v y y vd

d v2+=

Multiply both sides by 2ydy

2g y2 dy 2v y2 dv 2y v2 dy+=

y2g y2 ⌠⎮ ⎮⌡

d v2 y21 ⌠ ⎮ ⌡

d= 2 3

g y3 C+ v2 y2=

C 2−

3 g h3=v 0= at y h=

2 3

g h3 C+ 0=

2 3

g y3 2 3

g h3− v2 y2= v 2 3

g y3 h3−

y2

⎛⎜ ⎜⎝

⎞⎟ ⎟⎠

=

Problem 15-133

The car has a mass m0 and is used to tow the smooth chain having a total length l and a mass per unit of length m'. If the chain is originally piled up, determine the tractive force F that must be supplied by the rear wheels of the car, necessary to maintain a constant speed v while the chain is being drawn out.

394

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Engineering Mechanics - Dynamics Chapter 15

Solution:

+⎯⎯→ ΣFs = m t vd

d vDi

t mi

d d

+

At a time t, m m0 c t+=

Where, c t mi

d d

= m' t xd

d = m'v=

Here, vDi v= t vd

d 0=

F m0 m'v t−( ) 0( ) v m' v( )+= m'v2= F m'v2=

Problem 15-134

Determine the magnitude of force F as a function of time, which must be applied to the end of the cord at A to raise the hook H with a constant speed v. Initially the chain is at rest on the ground. Neglect the mass of the cord and the hook. The chain has a mass density ρ.

Given:

v 0.4 m s

= ρ 2 kg m

= g 9.81 m

s2 =

Solution:

t vd

d 0= y v t=

mi m y= m v t=

t mi

d d

m v=

+↑ ΣFs m tv d d

vDi t mi

d d

⎛ ⎜ ⎝

⎞ ⎟ ⎠

+=

F m g v t− 0 v m v+= F m g v t v m v+=

F ρg v t v2+=

f1 ρg v= f1 7.85 N s

= f2 ρv 2= f2 0.320 N=

F f1 t f2+=

395

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Engineering Mechanics - Dynamics Chapter 16

Problem 16-1

A wheel has an initial clockwise angular velocity ω and a constant angular acceleration α. Determine the number of revolutions it must undergo to acquire a clockwise angular velocity ωf. What time is required?

Units Used: rev 2π rad=

Given: ω 10 rad s

= α 3 rad

s2 = ωf 15

rad s

=

Solution: ωf 2 ω2 2α θ+= θ

ωf 2 ω2−

2α = θ 3.32 rev=

ωf ω α t+= t ωf ω−

α = t 1.67 s=

Problem 16-2

A flywheel has its angular speed increased uniformly from ω1 to ω2 in time t. If the diameter of the wheel is D, determine the magnitudes of the normal and tangential components of acceleration of a point on the rim of the wheel at time t, and the total distance the point travels during the time period.

Given: ω1 15 rad s

= ω2 60 rad s

= t 80 s= D 2 ft=

Solution: r D 2

=

ω2 ω1 α t+= α ω2 ω1

t = α 0.56

rad

s2 =

at αr= at 0.563 ft

s2 =

an ω2 2r= an 3600

ft

s2 =

θ ω2

2 ω1 2

2α = θ 3000 rad=

d θr= d 3000 ft=

Problem 16-3

The angular velocity of the disk is defined by ω = at2 + b. Determine the magnitudes of the velocity and acceleration of point A on the disk when t = t1.

396

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Engineering Mechanics - Dynamics Chapter 16

Given:

a 5 rad

s3 =

b 2 rad s

=

r 0.8 m=

t1 0.5 s=

Solution: t t1=

ω a t2 b+= ω 3.25 rad s

=

α 2a t= α 5.00 rad

s2 =

v ωr= v 2.60 m s

=

a αr( )2 ω2r( )2+= a 9.35 m s2

=

*Problem 16-4

The figure shows the internal gearing of a “spinner” used for drilling wells. With constant angular acceleration, the motor M rotates the shaft S to angular velocity ωM in time t starting from rest. Determine the angular acceleration of the drill-pipe connection D and the number of revolutions it makes during the start up at t.

Units Used: rev 2π=

Given:

ωM 100 rev min

= rM 60 mm=

rD 150 mm= t 2 s=

Solution:

ωM αM t=

αM ωM

t = αM 5.24

rad

s2 =

αM rM αDrD=

397

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Engineering Mechanics - Dynamics Chapter 16

αD αM rM rD

⎛ ⎜ ⎝

⎞ ⎟ ⎠

= αD 2.09 rad

s2 =

θ 1 2

αDt 2= θ 0.67 rev=

Problem 16-5

If gear A starts from rest and has a constant angular acceleration αA, determine the time needed for gear B to attain an angular velocity ωB.

Given:

αA 2 rad

s2 = rB 0.5 ft=

rA 0.2 ft=ωB 50 rad s

=

Solution:

The point in contact with both gears has a speed of

vp ωB rB= vp 25.00 ft s

=

Thus,

ωA vp rA

= ωA 125.00 rad s

=

So that ω αc t= t ωA αA

= t 62.50 s=

Problem 16-6

If the armature A of the electric motor in the drill has a constant angular acceleration αA, determine its angular velocity and angular displacement at time t. The motor starts from rest.

Given:

αA 20 rad

s2 = t 3 s=

Solution:

ω αc t= ω αA t= ω 60.00 rad s

=

398

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