5/23/2018 Chp 17 Self Study

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PROBLEM 17.11

attached to cords that are wrapped on the pulleys as The coefficient of kinetic friction between block B asurface is 0.25. Knowing that the system is released froin the position shown, determine (a) the velocity of cyl

as it strikes the ground, (b) the total distance that bmoves before coming to rest.

SOLUTION

Let Av = speed of blockA, Bv = speed of blockB, = angular speed of pulley.

Kinematics. rA = 0.25 m, rB = 0.15 m

(a) CylinderAfalls to ground.

Work of blockA: 1 2U = WAsA= (12.5)(9.81)(0.9) = 110.363J

Normal contact force acting on blockB: N= WB= (10)(9.81) = 98.1 N

Friction force on blockB: (0.25f kF N= = )(98.1) = 24.525 N

Work of friction force: 1 2 f BU F s = = (24.525)(0.54) = 13.2435J

Total work: 1 2U = 110.363 13.2435 = 97.1J

The double pulley shown has a mass of 15 kg and has a c

dal radius of gyration of 160 mm. CylinderAand bloc

vA = rAw = 0.25w

vB = rBw = 0.15w

sA = rAq = 0.25q

sB = rBq = 0.15q

sB = (0.9) = 0.54 m

sA = 0.9 m

0.15

0.25

5/23/2018 Chp 17 Self Study

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PROBLEM 17.11 (Continued)

Kinetic energy: 2 2 21 21 1 1

0;2 2 2

A A B BT T m v I m v= = + +

2 2 2 2 2 22

1 1 1

2 2 2

1

2

A A C B BT m r m k m r = + +

= (12.5)(0.25)2+ (15)(0.16)2+ (10)(0.15)2 w2

= 0.6951w2

Principle of work and energy.

1 1 2 :T U T+ =

Velocity of cylinderA: vA= (0.25)(11.82) A = 2.96 mv

(b) BlockBcomes to rest.

For blockBand pulley C. 2 23 41 1

; 0:2 2

B BT I m v T = + =

2 2 2 23

1 1

2 2C B BT m k m r = +

= 42.54J

Work of friction force: 3 4U

Principle of work and energy.

3 3 4 4: 42.54 24.525s B= 0T U T+ =

s B= 1.735 m

Total distance for blockB. :B Bd s s= + d = 0.54 + 1.735 d = 2.2

0 + 97.1 = 0.6951w2

w= 11.82 rad/s

12

= (15)(0.16)2+ (10)(0.15)2 (11.82)2

= Ffs B= 24.525s B

5/23/2018 Chp 17 Self Study

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PROBLEM 17.16

A slender 4-kg rod can rotate in a vertical plane about a pivot atB. Aof constant k =400 N/m and of unstretched length 150 mm is attacthe rod as shown. Knowing that the rod is released from rest in the pshown, determine its angular velocity after it has rotated through 90.

SOLUTION

Position 1:

Spring:

Unstretched

Length

1

2 21

(150 mm ) 370 150 220 mm 0.22 m

1 1(400 N/m)(0.22 m) 9.68 J

2 2e

x CD

V kx

= = = =

= = =

Gravity: 2

1

(4 kg)(9.81m/s )(0.18 m) 7.063 J

9.68 J 7.063 J 16.743 J

g

e g

V Wh mgh

V V V

= = = =

= + = + =

Kinetic energy: 1 0T =

Position 2:

Spring: 2

2 22

230 mm 150 mm 80 mm 0.08 m

1 1(400 N/m)(0.08 m) 1.28 J

2 2e

x

V kx

= = =

= = =

Gravity:

2

0

1.28 J

g

e g

V Wh

V V V

= =

= +

=

5/23/2018 Chp 17 Self Study

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PROBLEM 17.16 (Continued)

Kinetic energy: 2 2 2

2 2 2

2 22 2 2

2 22 2

22 2

(0.18 m)

1 1(4 kg)(0.6 m) 0.12 kg m

12 12

1 1

2 2

1 1(4 kg)(0.18 ) (0.12)

2 2

0.1248

v r

I mL

T mv I

T

= =

= = =

= +

= +

=

Conservation of energy:

1 1 2 2

22

22

2

0 16.743 J 0.1248 1.28 J

123.9

11.131 rad/s

T V T V

+ = +

+ = +

=

= 2 11.13 rad=

5/23/2018 Chp 17 Self Study

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PROBLEM 17.17

A slender 4-kg rod can rotate in a vertical plane about a pivot aspring of constant k =400 N/m and of unstretched length 150 attached to the rod as shown. Knowing that the rod is released frin the position shown, determine its angular velocity after it has through 90.

SOLUTION

Position 1:

Spring:

Unstretched

Length

1

2 21

(150 mm ) 370 150 220 mm 0.22 m

1 1(400 N/m)(0.22 m) 9.68 J

2 2e

x CD

V kx

= = = =

= = =

Gravity: 2

1

(4 kg)(9.81 m/s )( 0.22 m) 7.063 J

9.68 J 7.063 J 2.617 J

g

e g

V Wh mgh

V V V

= = = =

= + = =

Kinetic energy: 1 0T =

Position 2:

Spring: 2

2 22

230 mm 150 mm 80 mm 0.08 m

1 1(400 N/m)(0.08 m) 1.28 J

2 2e

x

V kx

= = =

= = =

Gravity:

2

0

1.28 J

g

e g

V Wh

V V V

= =

= +

=

5/23/2018 Chp 17 Self Study

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PROBLEM 17.17 (Continued)

Kinetic energy: 2 2 2

2 2 2

2 22 2 2

2 22 2

22 2

(0.18 m)

1 1(4 kg)(0.6 m) 0.12 kg m

12 12

1 1

2 2

1 1(4 kg)(0.18 ) (0.12)

2 2

0.1248

v r

I mL

T mv I

T

= =

= = =

= +

= +

=

Conservation of energy:

1 1 2 2

22

22

2

0 2.617 J 0.1248 1.28 J

10.713

3.273 rad/s

T V T V

+ = +

+ = +

=

= 2 3.27 rad=

5/23/2018 Chp 17 Self Study

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PROBLEM 17.28

A collar B, of mass m and of negligible dimension, is attached to the rihoop of the same mass m and of radius rthat rolls without sliding on a horsurface. Determine the angular velocity 1 of the hoop in terms ofgand

B is directly above the centerA, knowing that the angular velocity of the h31whenB is directly belowA.

SOLUTION

The point of contact with ground is the instantaneous center.

Position 1. PointBis at the top.

1 1 1

2 2 2

1

2 2 2 21 1 1

2 21

2

1 1 1

2 2 2

1 1 1(2 ) ( ) ( )

2 2 2

3

B A

B A

v r v r

T mv mv I

m r m r mr

mr

= = =

= + +

= + +

=

Position 2. PointBis at the bottom.

2 2

2 2 22 2

2 2 22 2

2 22

2 21

2 21

0

1 1 1

2 2 21 1

0 ( ) ( )2 2

(3 )

9

B A

B A

v v r

T mv mv I

m r mr

mr

mr

mr

= = =

= + +

= + +

=

=

=

Work. 1 2 ( ) 2U mg h mgr = =

Principle of work and energy.2 2 2

1 1 2 2 1 1: 3 2 9T U T mr mgr m + = + =

21

3

g

r = 1 0.577 =

5/23/2018 Chp 17 Self Study

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PROBLEM 17.38

The ends of a 4.5-kg rodABare constrained to move along slots vertical plane as shown. A spring of constant k = 600 N/m is attaendAin such a way that its extension is zero when q = 0. If threleased from rest when 0, = determine the angular velocityrod and the velocity of endBwhen 30 . =

SOLUTION

Moment of inertia. Rod. 21

12I mL=

Position 1. 1 1 1

1 1

1 1

0 0 0

elevation above slot. 0

elongation of spring. 0

v

h h

e e

= = =

= =

= =

2 21 1 1

21 1 1

1 10

2 2

10

2

T mv I

V ke Wh

= + =

= + =

Position 2. 30 =

2

2

2

2 2 22 2 2

cos30

(1 cos30 )

1sin30

2 4

1 1 1(1 cos30 )

2 2 4

e L L

e L

Lh L

V ke Wh k L WL

+ =

=

= =

= + =

Kinematics. Velocities atAandBare directed as shown. Point Cis the instantaneous center of rotation

geometry,2

.Lb =

2 22

2

2

2 2

2( cos30 )

1 1

2 2

1 1 1

2 2 2 12

1

6

B

Lv b

v L

T mv I

Lm mL

WL

g

= =

=

= +

= +

=

5/23/2018 Chp 17 Self Study

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PROBLEM 17.38 (Continued)

Conservation of energy.

2 2 2 21 1 2 2

1 1 1: 0 0 (1 cos30 )

6 2 4

WT V T V L kL WL

g+ = + + = +

2 23 (1 cos30 )g 3k

2L m =

Data: 4.5 kgm

g

L

k

=

= 9.81 m/s2

= 0.6 m

= 600 N/m

2 (3)(9.81)

(2)(0.6) 4.5

=

=4.1648

4.16 rad=

Bv = (0.6)(cos30)(4.1648) B = 2.16 mv

(3)(600)(1 cos 30)2

5/23/2018 Chp 17 Self Study

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PROBLEM 17.45

The 4-kg rod AB is attached to a collar of negligible mass at A and to a flat B. The flywheel has a mass of 16 kg and a radius of gyration of 18Knowing that in the position shown the angular velocity of the flywheel is 6clockwise, determine the velocity of the flywheel when PointBis directly bel

SOLUTION

Moments of inertia.

RodAB:2

2

2

1

12

1(4 kg)(0.72 m)

12

0.1728 kg m

AB AB ABI m L=

=

=

Flywheel: 2

2

2

(16 kg)(0.18 m)

0.5184 kg m

CI mk=

=

=

Position 1. As shown. 1=

1

1 1

0.24sin 19.471

0.72

1(0.72)cos 0.33941 m

2

(4)(9.81)(0.33941)

13.3185 J

AB

h

V W h

= =

= =

=

=

=

Kinematics. 1 10.24Bv r = =

BarABis in translation. 0,AB Bv v = =

2 2 21 1

2 21 1

2

1

1 1 1

2 2 2

1 1(4)(0.24 ) 0 (0.5184)

2 2

0.3744

AB AB AB CT m v I I

= + +

= + +

=

5/23/2018 Chp 17 Self Study

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PROBLEM 17.45 (Continued)

Position 2. PointBis directly below C.

2

2 2

1

2

1(0.72) 0.24

2

0.12 m

(4)(9.81)(0.12)

4.7088 J

AB

AB

h L r

V W h

=

=

=

=

=

=

Kinematics. 2 20.24Bv r = =

2

2

2 2 22 2

2 2 22 2 2

22

0.333330.72

1 0.122

1 1 1

2 2 2

1 1 1(4)(0.12 ) (0.1728)(0.33333 ) (0.5184)

2 2 2

0.2976

BAB

B

AB AB AB C

v

v v

T m v I I

= =

= =

= + +

= + +

=

Conservation of energy. 2 21 1 2 2 1 2: 0.3744 13.3185 0.2976 4.7088T V T V + = + + = +

Angular speed data: 1 60 rpm 2 rad/s = =

Solving Equation (1) for 2 , 2 8.8655 rad/s = 2 84.7 rp=

5/23/2018 Chp 17 Self Study

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PROBLEM 17.55

Two disks of the same thickness and same material are attacheshaft as shown. The 3-kg diskAhas a radius 100 mm,Ar = andhas a radius 125 mm.Br = Knowing that the angular velocity system is to be increased from 200 rpm to 800 rpm duringinterval, determine the magnitude of the couple Mthat must be ato diskA.

SOLUTION

Mass of diskB.

2

2125 mm

3 kg

100 mm4.6875 kg

BB A

A

rm m

r

=

=

=

Moment of inertia.

2 2

2

1 1(3 kg)(0.1m) (4.6875 kg)(0.125 m)

2 2

0.05162 kg m

A BI I I= +

= +

=

Angular velocities. 1

2

2200 rpm 20.944 rad/s

60

2800 rpm 83.776 rad/s

60

= =

= =

Principle of impulse and momentum.

Syst. Momenta1 + Syst. Ext. Imp.12 = Syst. Momenta2

Moments aboutB: 1 2I Mt I + =

Couple M. 2 1( )I

Mt

=

20.05162 kg m

(83.776 rad/s 20.944 rad/s)3 s

= 1.081 NM =

5/23/2018 Chp 17 Self Study

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PROBLEM 17.69

A wheel of radius rand centroidal radius of gyration k is released from rthe incline shown at time 0.t = Assuming that the wheel rolls without sdetermine (a) the velocity of its center at time t, (b) the coefficient offriction required to prevent slipping.

SOLUTION

Kinematics. Rolling motion. Instantaneous center at C.

Gv v r

v

r

= =

=

Moment of inertia. 2I mk=

Kinetics.

1Syst. Momenta + 1 2Syst. Ext. Imp. = 2Syst. Momenta

moments about C:

2

0 ( ) sin

( sin )

mgt r mvr I

mk vmgr t mrv

r

+ = +

= +

(a) Velocity of Point G.2

2 2

sinr gt

r k

=

+

v

components parallel to incline:

2

2 2

2

2 2

0 sin

sinsin

sin

mgt Ft mv

mr gt Ft mgt

r k

k mgt

r k

+ =

=

+

=

+

5/23/2018 Chp 17 Self Study

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PROBLEM 17.69 (Continued)

components normal to incline:

0 cos 0

cos

Nt mgt

Nt mgt

+ =

=

(b) Required coefficient of static friction.

2

2 2

sin

( ) cos

s

F

N

Ft

Nt

k mgt

r k mgt

=

=

+

2

2

tas

k

r

+

5/23/2018 Chp 17 Self Study

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PROBLEM 17.71

The double pulley shown has a mass of 3 kg and a radius of gyration of 10Knowing that when the pulley is at rest, a force P of magnitude 2applied to cord B, determine (a) the velocity of the center of the pulle1.5 s, (b) the tension in cord C.

SOLUTION

For the double pulley, 0.150 m

0.080 m

0.100 m

C

B

r

r

k

=

=

=

Principle of impulse and momentum.

1Syst. Momenta + 1 2Syst. Ext. Imp. = 2Syst. Momenta

Kinematics. Point Cis the instantaneous center. Cv r =

Moments about C:

2

0 ( )

( )

C B C C

C C

Pt r r mgtr I mvr

mk m r r

+ + = +

= +

( )2 2

2 2

( )

(24)(1.5) (0.230) (3)(9.81)(1.5) (0.150)

3(0.100 0.150 )

17.0077 rad/s

C B C

C

Pt r r mgtr

m k r

+ =

+

=

+

=

5/23/2018 Chp 17 Self Study

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PROBLEM 17.71 (Continued)

(a) (0.150)(17.0077) 2.55115 m/sv = = 2.55 m=v

Linear components: 0 Pt mgt Qt mv+ + =

(3)(2.55115)(3)(9.81) 24

1.5

mvQ mg P

t= +

= +

(b) Tension in cord C. 10.5Q =

5/23/2018 Chp 17 Self Study

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PROBLEM 17.89

Collar Chas a mass of 8 kg and can slide freely on rod AB,in turn can rotate freely in a horizontal plane. The assemrotating with an angular velocity of 1.5 rad/s when a located betweenAand Cis released, projecting the collar alorod with an initial relative speed 1.5 m/s.rv = Knowing tcombined mass moment of inertia aboutB of the rod and sp

21.2 kg m , determine (a) the minimum distance between thand Point B in the ensuing motion, (b) the corresponding avelocity of the assembly.

SOLUTION

Kinematics. v r

=

Moments of inertia. 2

21.2 8

B CI I m r

r

= +

= +

Conservation of angular momentum.

1 2 1 1 1 2 2 2( ) ( ) : ( ) ( )B B B C B CH H I m v r I m v r = + = +

2 21 1 2 2 1 1 2 2

1 12

2

( ) ( ) or B C B CI m r I m r I I

I

I

+ = + =

=

Potential energy. 1 20 0V V= =

Kinetic energy.2 2 2

2 2

1 1 1

2 2 2

1 1

2 2

C r C B

C r

T m v m v I

m v I

= + +

= +

Position 1. Just after spring is released. 1

1

1

1

( )r r

r r

v v

I I

=

=

=

=

Position 2. Distance ris minimum. 2

2

2

( ) ( ) 0r r

r r

v v

I I

=

= =

=

=

5/23/2018 Chp 17 Self Study

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PROBLEM 17.89 (Continued)

Conservation of energy.

2 2 21 1 2 2 1 1 1 2 2

1 1 1: ( ) 0 0

2 2 2C rT V T V m v I I + = + + + = +

2

2 21 1

1 2 1 12

211 1

2

( )

1

C r

I

m v I I I

II

I

=

=

Data. 1

1

1

21

2

600 mm 0.6 m

15 rad/s

( ) 1.5 m

1.2 (8)(0.6)

4.08 kg m

r

r

v

I

= =

=

=

= +

=

Equation (2): 2 21

2

1

2

22 2

(8)(1.5) 4.08 1 (1.5)

2.9608

4.081.378 1.2 8

2.9608

I

I

I

I

I r

=

=

= = = +

(a) 2 0.14917 mr = 2 149.2 r =

(b) Equation (1): 24.08

(1.5)1.378

= 2 4.44 r =

5/23/2018 Chp 17 Self Study

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PROBLEM 17.96

SOLUTION

Bar: L = 0.8 m m = 7.5 kg

I mL

Bullet: 0m = 0.04 kg

Support location: h = 0.3 m

Kinematics.

2

B

G

v L h

L

=

=

Kinetics.

Syst. Momenta1 + Syst. Ext. Imp.12 = Syst. Momenta2

Moments about C: 0 0 ( ) ( )2

B

Lm v L h m v L h mv h I = + +

(0.04)(550)(0.5) = (0.04)(0.5w)(0.5) + (7.5)(0.1w)(0.1) + (0.4w)

2= = (7.5)(0.8)2= 0.4 kg m2

1 1

12 12

( ) w= (0.8 0.3)w= 0.5w

v h w= (0.4 0.3)w= 0.1w

A bullet weighing 40 g is fired with a horizontal velocity of 5into the lower end of a slender 7.5-kg bar of length L = 80Knowing that h = 300 mmand that the bar is initially at rest, det(a) the angular velocity of the bar immediately after the bullet beembedded, (b) the impulsive reaction at C, assuming that thebecomes embedded in 0.001 s.

0 G

5/23/2018 Chp 17 Self Study

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PROBLEM 17.96 (Continued)

(a) 11 = 0.485w or w = 22.68 w = 22.7 radB

G

v

v

= (0.5)(22.68) = 11.34 m/s

= (0.1)(22.68) = 2.268 m/s

Horizontal components:

0 0 0 0 0 0( ) : ( ) ( )B G Bm v C t m v mv C t m v v mv + = =

( )C t = (0.04)(550 11.34) (7.5)(2.268)

= 4.536 N s

(b)4.536

0.001

C tC

t

= =

= 4.54 kNC