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

Answer:

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

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  

                $\frac{1}{2}mv^2 +\frac{1}{2} I\omega^2=mgh$

Here , $I=\frac{2}{5} MR^2 , v= R \omega$

  $\frac{1}{2}mv^2\frac{1}{2}(\frac{2}{5}MR^2)(\frac{v}{R^2} )=mgh$

  $\frac{1}{2} [v^2+\frac{2}{5}v^2]= gh$

$\frac{7}{10} v^2=gh$

$0.7v^2=gh$

 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,

$\omega=\frac{v}{R}$

   = (3.408 m/s)/(0.038 m)

$\omega$   =  89.7 rad/s

Hence , the angular speed is 89.7 rad/s