Question

If the ball is released from rest at a height of 0.63 m above the bottom of the track on the no-slip side, what is its angular speed when it is on the frictionless side of the track

Answer 1

When the ball is on the frictionless side of the track , the angular speed is 89.7 rad/s.

Step-by-step explanation:

Consider the ball is a solid sphere of radius 3.8 cm and mass 0.14 kg .

Given , mass, m=0.14 kg

Ball is released from rest at a height of, h= 0.83 m

Solid sphere of radius, R = 3.8 cm

=0.038 m

From the conservation of energy

ΔK = ΔU

`(1)/(2)mv^2 +(1)/(2) I\omega^2=mgh`

Here ,
`I=(2)/(5) MR^2 , v= R \omega`

`(1)/(2)mv^2(1)/(2)((2)/(5)MR^2)((v)/(R^2) )=mgh`

`(1)/(2) [v^2+(2)/(5)v^2]= gh`

`(7)/(10) v^2=gh`

`0.7v^2=gh`

v=
`\sqrt{[gh/(0.7)]`

=
`\sqrt{ [(9.8 m/s^2)(0.83 m) / (0.7) ]`

= 3.408 m/s

Hence, angular speed when it is on the frictionless side of the track,

`\omega=(v)/(R)`

= (3.408 m/s)/(0.038 m)

`\omega` = 89.7 rad/s

Hence , the angular speed is 89.7 rad/s