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PROBLEM 17.122 Solve Problem 17.121, assuming that the gymnasts change places so that gymnast A jumps onto the plank while gymnast B stands at C. PROBLEM 17.121 The plank CDE has a mass of 15 kg and rests on a small pivot at D. The 55-kg gymnast A is standing on the plank at C when the 70-kg gymnast B jumps from a height of 2.5 m and strikes the plank at E. Assuming perfectly plastic impact and that gymnast A is standing absolutely straight, determine the height to which gymnast A will rise.

SOLUTION I =

Moment of inertia.

1 1 mP (2 L)2 = mP L2 12 3

(v)1 = 2 gh1

Velocity of jumper at E.

(1)

Principle of impulse-momentum.

Syst. Momenta1

Kinematics: Moments about D:

+

Syst. Ext. Imp.1→2

=

Syst. Momenta 2

vC = Lω vD = Lω mE v1 L + 0 = mE vE L + mC vC L + I ω 1 = mE L2ω + mC L2ω + mP L2ω 3 mE v1 ω= mE + mC + 13 mP L vC = Lω =

Gymnast (flier) rising.

hC =

mE v1 mE + mC + 13 mP

vC2 2g

(2)

(3)

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1905

PROBLEM 17.122 (Continued)

Data:

mE = mA = 55 kg mC = mB = 70 kg mP = 15 kg h1 = 2.5 m

From Equation (1)

v1 = (2)(9.81)(2.5) = 7.0036 m/s (55)(7.0036) 55 + 70 + 5 = 2.9631 m/s

From Equation (2)

vC =

From Equation (3)

h2 =

(2.9631) 2 (2)(9.81)

= 0.447 m

h2 = 447 mm 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1906

PROBLEM 17.123 A small plate B is attached to a cord that is wrapped around a uniform 8-lb disk of radius R = 9 in. A 3-lb collar A is released from rest and falls through a distance h = 15 in. before hitting plate B. Assuming that the impact is perfectly plastic and neglecting the weight of the plate, determine immediately after the impact (a) the velocity of the collar, (b) the angular velocity of the disk.

SOLUTION The collar A falls a distance h. From the principle of conservation of energy. v1 = 2 gh

Impact analysis:

e=0

Kinematics. Collar A and plate B move together. The cord is inextensible. v2 = R ω

or

ω2 =

v2 R

Let m = mass of collar A and M = mass of disk. Moment of inertia of disk:

I =

1 MR 2 2

Principle of impulse and momentum. I ω1 = 0

Syst. Momenta1 +

Moments about C:

Syst. Ext. Imp.1→2

=

m v1 R = I ω2 + m v2 R

Syst. Momenta 2

(1)

1 v  MR 2  2  + m v2 R 2 R 1 m v1 = Mv2 + m v2 2 2m v2 = v1 2m + M

m v1 R =

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1907

PROBLEM 17.123 (Continued)

m = 3 lb/g

Data:

M = 8 lb/g h = 15 in. = 1.25 ft R = 9 in. = 0.75 ft v1 = (2)(32.2)(1.25) = 8.972 ft/s

(a)

(b)

Velocity of A.

Angular velocity.

v2 =

(2)(3) 3 v1 = v1 (2)(3) + 8 7

v2 =

3 (8.972) = 3.8452 ft/s 7

ω2 =

3.8452 0.75

v 2 = 3.85 ft/s  ω 2 = 5.13 rad/s



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1908

PROBLEM 17.124 Solve Problem 17.123, assuming that the coefficient of restitution between A and B is 0.8. PROBLEM 17.123 A small plate B is attached to a cord that is wrapped around a uniform 8-lb disk of radius R = 9 in. A 3-lb collar A is released from rest and falls through a distance h = 15 in. before hitting plate B. Assuming that the impact is perfectly plastic and neglecting the weight of the plate, determine immediately after the impact (a) the velocity of the collar, (b) the angular velocity of the disk.

SOLUTION WD = 8 lb WD 8 = = 0.2484 lb ⋅ s 2 /ft g 32.2 R = 9 in. = 0.75 ft 1 I D = mD R 2 = 0.06988 lb ⋅ s 2 ⋅ ft 2 WA = 3 lb mD =

WA 3 = = 0.09317 lb ⋅ s 2 /ft g 32.2 h = 15 in. = 1.25 ft

mA =

Collar A falls through distance h. Use conservation of energy. T1 = 0 V1 = WA h 1 mA v A2 2 V2 = 0 T2 =

T1 + V1 = T2 + V2 : 0 + WA h = v A2 =

1 mA v A2 + 0 2 2m A h = 2 gh WA

= (2)(32.2)(1.25) = 80.5 ft 2 /s 2 vA = 8.972 ft/s

Impact. Neglect the mass of plate B. Neglect the effect of weight over the duration of the impact.

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1909

PROBLEM 17.124 (Continued)

ω′ = ω

Kinematics.

v′B = R ω = 0.75ω ′

Conservation of momentum.

m A v A R + 0 = m Av′A R + I Dω ′ + mB vB′ R

Moments about D:

(0.09317)(8.972)(0.75) = (0.09317)(0.75)v′A + 0.06988 ω ′

Coefficient of restitution.

(1)

vB′ − v′A = e(v A − vB ) 0.75ω ′ − v′A = 0.8(8.972 − 0)

(2)

Solving Eqs. (1) and (2) simultaneously (a)

Velocity of A.

v′A = −0.25648 ft/s

(b)

Angular velocity.

ω ′ = 9.228 rad/s

vA′ = 0.256 ft/s 

ω′ = 9.23 rad/s



PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1910

PROBLEM 17.125 Two identical slender rods may swing freely from the pivots shown. Rod A is released from rest in a horizontal position and swings to a vertical position, at which time the small knob K strikes rod B which was at rest. If h = 12 l and e = 12 , determine (a) the angle through which rod B will swing, (b) the angle through which rod A will rebound.

SOLUTION m = mAC = mBD

Let

I AC = I =

Moment of inertia.

I BD =

1 mL2 12

1 mL2 12

Rod AB falls to vertical position.

Position 0.

V1 = 0

T1 = 0

Position 1.

V2 = − mg

L 2

L (ω AC )1 2 1 1 T1 = m(v AC )12 + I (ω AC )12 2 2 1 2 = mL (ω AC )12 6 1 1 T0 + V0 = T1 + V1: 0 + 0 = mL2 (ω AC )12 − mgL 6 2 3g (ω AC )12 = L (v AC )1 =

Conservation of energy.

(1)

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1911

PROBLEM 17.125 (Continued)

Impact.

+

Syst. Momenta1

Syst. Ext. Imp.1→ 2 = Syst. Momenta2

L (ω AC )1 2 L (v AC ) 2 = (ω AC ) 2 2 (v AC )1 =

Kinematics

Moments about C:

m(v AC )1

L L + I (ω AC )1 − L  Kdt = m(v AC ) 2   + I (ω AC ) 2 2 2 1 2 1 mL (ω AC )1 − L  Kdt = mL2 (ω AC ) 2 3 3

(2)

Syst. Momenta1 + Syst. Ext. Imp.1→ 2 = Syst. Momenta2 Kinematics Moments about D:

(vBD ) 2 =

L (ωBD ) 2 2

0 + ( L − h)  Kdt = m(vBD ) 2

L + I (ω BD )2 2

1 ( L − h)  Kdt = mL2 (ωBD ) 2 3

(3)

Multiply Eq. (2) by (L − h) and Eq. (3) by L and then add to eliminate  Kdt. 1 2 1 1 mL ( L − h)(ω AC )1 = mL2 ( L − h)(ω AC ) 2 + mL3 (ωBD ) 2 3 3 3

(1)

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1912

PROBLEM 17.125 (Continued)

L(ω AC )2 − ( L − h)(ωBD ) 2 = −eL(ω AC )1

Condition of impact:

(2)

For h = 12 L and e = 0.5 Eqs. (1) and (2) become 1 3 1 1 mL (ω AC )1 = mL3 (ω AC ) 2 + mL3 (ω BD ) 2 6 6 3 −

Dividing Eq. (3) by

1 6

1 = 0.5L(ω AB )1 2

(3) (4)

mL3 and transposing terms gives (ω AC ) 2 + 2(ωBD )2 = (ω AC )1

(5)

Dividing Eq. (4) by L /2 and transposing terms gives 2(ω AC ) 2 − (ωBD ) 2 = −(ω AC )1

(6)

Solving Eqs. (5) and (6) simultaneously for (ω AC ) 2 and (ωBD ) 2 gives

(a)

(ω AC ) 2 = −0.2(ω AC )1

(7)

(ωBD ) 2 = 0.6(ω AC )1

(8)

Angle of swing θ B for rod B.

Apply the principle of conservation of energy to rod B. T2 + V2 = T3 + V3

Position (2): Just after impact. Position (3): At maximum angle of swing.

Potential energy. Use the pivot point D as the datum. L 2 L V3 = − mg cos θ B 2

V2 = − mg

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1913

PROBLEM 17.125 (Continued)

Kinetic energy. 1 1 I D (ωBD ) 22 = mL3 (ω BD ) 22 2 6 T3 = 0

T2 =

1 2 L L mL (ω BD ) 22 − mg = 0 − mg cos θ B 6 2 2 1L 1 − cos θ B = (ω BD ) 22 3g 1L = [0.6(ω AB )1 ]2 3g  L   3g  1 =   (0.36)     = 0.36 3  g  L 

cos θ B = 0.64

(b)

θ B = 50.2° 

Angle of rebound θA for rod A.

Apply the principle of conservation of energy to rod A. T2 + V2 = T4 + V4

Position (2): Just after impact. Position (4): At maximum angle of rebound.

Potential energy. Use the pivot Point C as the datum. V2 = − mg

L 2

V4 = −mg

L cos θ A 2

Kinetic energy. T2 =

1 1 I C (ω AB ) 22 = mL2 (ω AC )22 2 6

T4 = 0

1 2 L L mL (ω AC )22 − mg = 0 − mg cos θ A 6 2 2 1L 1 − cos θ A = (ω AC )22 3g 1L = [−0.2(ω AB )1 ]2 3g  L   3g  1 =   (0.04)     = 0.04 3    g  L 

cos θ A = 0.96

θ A = 16.26° 

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1914

PROBLEM 17.126 A 2-kg solid sphere of radius r = 40 mm is dropped from a height h = 200 mm and lands on a uniform slender plank AB of mass 4 kg and length L = 500 mm which is held by two inextensible cords. Knowing that the impact is perfectly plastic and that the sphere remains attached to the plank at a distance a = 40 mm from the left end, determine the velocity of the sphere immediately after impact. Neglect the thickness of the plank.

SOLUTION Masses and moments of inertia. Sphere:

mS = 2 kg, r = 40 mm = 0.040 m IS =

Plank AB:

2 2 mS r 2 =   (2 kg)(0.04 m)2 = 1.28 × 10−3 kg ⋅ m 2 5 5

m AB = 4 kg, L = 500 mm = 0.5 m I AB =

1  1  mAB L2 =   (4 kg)(0.5 m)2 = 83.333 × 10−3 kg ⋅ m 2 12  12 

Velocity of sphere at impact. vS = 2 gh = (2)(9.81 m/s)(0.200 m) = 1.9809 m/s

Before impact. Linear momentum:

mS v S = (4 kg)(1.9809 m/s) = 7.9236 kg ⋅ m/s

with its line of action lying at distance

L 2

− a from the midpoint of the plank.

L − a = 0.25 m − 0.04 m = 0.21 m. 2

After impact. Assume that both cables are taut so vA is perpendicular to the cable at A and vB is perpendicular to the cable at B.

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1915