PROBLEM 16.68

A uniform sphere of radius r and mass m is placed with no initial velocity on a belt that moves to the right with a constant velocity 1.v Denoting by

kµ the coefficient of kinetic friction between the sphere and the belt, determine (a) the time 1t at which the sphere will start rolling without sliding, (b) the linear and angular velocities of the sphere at time 1.t

SOLUTION

Kinetics: ( )eff :x xF F F maΣ = Σ =

kmg maµ =

k gµ=a

( )eff :G GM M Fr IαΣ = Σ =

( ) 225kmg r mrµ α=

52

k gr

µ=α

Kinematics: kv at gtµ= = (1)

52

k gt tr

µω α= = (2)

C = Point of contact with belt

52

kC k

gv v r gt t rr

µω µ = + = +

72C kv gtµ=

(a) When sphere starts rolling ( )1 ,t t= we have

1 1 17; 2C kv v v gtµ= = 1

12 7 k

vtgµ

=

(b) Velocities when 1t t=

Eq (1): 127 k

vv gg

µµ

=

1

27

v=v

Eq (2): 15 22 7

k

k

g vr g

µωµ

=

157

vr

=ω

PROBLEM 16.80 A uniform rod of length L and mass m is supported as shown with b = 0.2 L when the cable attached to end B suddenly breaks. Determine at this instant (a) the acceleration of end B, (b) the reaction at the pin support.

SOLUTION

( ) ( )2

20.3 0.312C

mLM mg L m Lα αΣ = = +

0.3 451.730770.1733 26

g g gL L L

α = = =

(a) 360.826B

ga Lα= =

or 1813B

ga =

(b) 0xR = 450.326y y

gF R mg mLL

Σ = − = −

135 25260 52yR mg mg mg= − =

PROBLEM 16.100 A drum of 100 mm radius is attached to a disk of 200 mm radius. The diskand drum have a total weight of 5 kg and combined radius of gyration of 150 mm. A cord is attached as shown and pulled with a force P of magnitude 25 N. Knowing that the disk rolls without sliding, determine (a) the angular acceleration of the disk and the acceleration of G, (b) the minimum value of the coefficient of static friction compatible with this motion.

SOLUTION

W = (5 ¥ 9.81)N

100 mm

200 mm

25 N

a = ra = 2001000

mFH

IK a = (0.2 m) a

I = mk 2 = (5 kg) (0.15 m)2

= 0.1125 kg.m2

ÂMC = Â(MC)eff : (25 N) (0.1 m) = mar + Ia

2.5 N.m = (0.3125 kg.m2) a

a = 8 rad/s2

or aaaaa = 8 rad/s2�

(a) a = ra = (0.2 m) (8 rad/s2) = 1.6 m/s2

or a = 1.6 m/s2�

(b) Fy = 0: N + 25 N – (5 ¥ 9.81) N = 0 N = 24.05 N

ÂFx = Â (Fx)eff : F = ma = (5 kg) (1.6 m/s2)

F = 8 N

( ms)min = FN

= 8

24 05

N

N. = 0.3326 N

or ( ms)min = 0.333 �

PROBLEM 16.105

A half section of a uniform thin pipe of mass m is at rest when a force P is applied as shown. Assuming that the section rolls without sliding, determine (a) its initial angular acceleration, (b) the minimum value of the coefficient of static friction consistent with the motion.

SOLUTION

22

22

2 41GrI mr m mrπ π

= − = −

21xF F mrαπ

Σ = = −

2 2

24 2 21 1 1

2DrM P mr mrα α

π ππ Σ = = − + − −

2 2 2 2 22 2

4 2 42mr mr mr mr mrα αππ π

= − + − +

2 2 22 2mr mr απ

= −

(a) 24 1

P

mrα

π

= −

or 0.688Pmr

α =

(b)

21

2 44 1

PPF mr

mr

π

π

− = = −

0:yF N mg PΣ = = +

and ( )4s

F PN mg P

µ = =+

or ( )

0.25s

Pmg P

µ =+

PROBLEM 16.105

A half section of a uniform thin pipe of mass m is at rest when a force P is applied as shown. Assuming that the section rolls without sliding, determine (a) its initial angular acceleration, (b) the minimum value of the coefficient of static friction consistent with the motion.

SOLUTION

22

22

2 41GrI mr m mrπ π

= − = −

21xF F mrαπ

Σ = = −

2 2

24 2 21 1 1

2DrM P mr mrα α

π ππ Σ = = − + − −

2 2 2 2 22 2

4 2 42mr mr mr mr mrα αππ π

= − + − +

2 2 22 2mr mr απ

= −

(a) 24 1

P

mrα

π

= −

or 0.688Pmr

α =

(b)

21

2 44 1

PPF mr

mr

π

π

− = = −

0:yF N mg PΣ = = +

and ( )4s

F PN mg P

µ = =+

or ( )

0.25s

Pmg P

µ =+

PROBLEM 16.124 A driver starts his car with the door on the passenger’s side wide open ( 0θ = ). The 36-kg door has a centroidal radius of gyration

250k = mm, and its mass center is located at a distance 440r = mm from its vertical axis of rotation. Knowing that the driver maintains a constant acceleration of 22 m / s , determine the angular velocity of the door as it slams shut ( 90θ = ° ).

SOLUTION

Kinematics: Aa=a

where ( )/G A trα=a θ

Kinetics:

( ) ( ) ( )eff : 0 cosA A AM M I mr r ma rα α θΣ = Σ = + −

2 2 cosAmk mr ma rα α θ+ =

2 2 cosAa rk r

α θ=+

Setting ,ddωα ωθ

= and using 0.44 m, 0.25 mr k= =

( )( ) ( )2 2

0.44 mcos 1.7181 cos

0.44 m 0.25 mA

Aad a

dωω θ θθ

= =+

20 01.7181 cosf

Ad a dπω ω ω θ θ=∫ ∫

22 20

0

1 1.7181 sin 3.43622

f

A f Aa aπ

ω

ω θ ω= ⇒ = (1)

PROBLEM 16.124 CONTINUED

Given 22 m/sAa =

( )2 3.4362 2 6.8724 2.6215 rad/sf fω ω= = ⇒ =

or 2.62 rad/sfω =