Question

A bowling ball (r=22cm) with center of mass G and is projected down the lane that has a friction coefficient of 0.2. The ball's center has an initial horizontal velocity 11 m/s and backspin w. What angular velocity should be given to the 5-kg ball so that it stops spinning and translating at the same instant?

Answer 1

Answer:124.95 rad/s

Step-by-step explanation:

Given

radius of ball
`r=22 cm`

coefficient of friction
`\mu =0.2`

ball center has a velocity of
`v=11 m/s`

back spin angular velocity is
`\omega _0`

mass of ball
`m=5 kg`

as there is no motion in Y direction therefore net Force is zero

`\sum F_y=0`

`W-N=0`

`W=N`

Friction will provide the torque

`f_r=\mu N`

motion is in x direction therefore

`f_r=ma`

`\mu N=\mu N=ma`

`a=(0.2* 5* 9.8)/(5)`

`a=1.96 m/s^2`

time taken to stop its linear velocity

`v=u+at`

`0=11-1.96* t`

`t=(11)/(1.96)=5.61 s`

torque
`f_r\cdot r=I\alpha`

`I=(2)/(5)mr^2`

`f_r\cdot r=(2)/(5)mr^2\cdot \alpha`

`\alpha =22.27 rad/s^2`

using

`\omega =\omega _0+\alpha t`

`0=\omega -22.27* 5.61`

`\omega _0=124.95 rad/s`