A bowler throws a bowling ball of radius R = 11 cm down a lane. The ball slides

Question

A bowler throws a bowling ball of radius R = 11 cm down a lane. The ball slides on the lane, with initial speed vcm,0 = 7.5 m/s and initial angular speed 0 = 0. The coefficient of kinetic friction between the ball and the lane is 0.20. The kinetic frictional force fk acting on the ball (Fig. 12-35) causes a linear acceleration of the ball while producing a torque that causes an angular acceleration of the ball. When speed vcm has decreased enough and angular speed has increased enough, the ball stops sliding and then rolls smoothly.

Please answer this question with both numerical and explanations. The last 2 answers I have recieved were wrong!

(a) What then is vcm in terms of ?
m·
(b) During the sliding, what is the ball's linear acceleration?
m/s2
(c) During the sliding, what is the ball's angular acceleration?
  rad/s2
(d) How long does the ball slide?
s
(e) How far does the ball slide?
m
(f) What is the speed of the ball when smooth rolling begins?
m/s

Explanation / Answer

for ball,

forces on it = mass * acceleration
f=ma (b) a=kg=1.254m/s2 until the ball starts rolling
where f is frictional force=mgk
where k is coefficient of friction.

now after time t its Vcom be V and initial be U
so V=U-gkt

for torque,
f(R)=I (c) =51.227 rad /sec2 until the ball starts rolling
where I=2/5 m(R)^2

after time t let angular speed be .
so, =kmgRt/2/5m(R)^2
=5kgt/2R

for rolling V=R=5kgt/2=U - kgt
so t=2U/7kg
=1.14 s ..............(d)

so final V=6.428 m/s .........(f)

distance travelled=U^2-V^2/2a = 8.7m.......(e)


can you clarify first part of question?

Edit: b & c part are cut in answer i will write again.
(b) a=kg=2.254m/s2 until the ball starts rolling
(c) =51.227 rad /sec2 until the ball starts rolling

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