Ball on a Turntable

Let the angular velocity of ball be $\omega$ , and that of the turntable be $\Omega \hat { z }$ . Since the ball is moving in a circular motion, There must be a force action on the ball, which in this case is frictional force $f_{f}$ . This force cause change in linear momentum

${ f }_{ f }=\frac { dp }{ dt } =m\frac { dv }{ dt } ............(1)$

As well as change in angular momentum.

$-{ f }_{ f }\times (l\hat { z } )=I\frac { d\omega }{ dt } ........(2)$

Where $l$ is the radius of the ball. Before we go any further let us express the balls velocity in the lab frame. If we let ${ r }$ be the position of the ball from the center of the turntable. Its velocity can simply be expressed as the sum of the turntables velocity and the velocity of the ball.

$v=(\Omega \hat { z) } \times r\quad +\quad \omega \times (l\hat { z } )........3$

Substitute equation $(1)$ into equation $(2)$

$\quad \quad m\frac { dv }{ dt } \times (-l\hat { z } )=I\frac { d\omega }{ dt } \\ \Rightarrow \quad \frac { d\omega }{ dt } =-\left( \frac { lm }{ I } \right) \hat { z } \times \frac { dv }{ dt } .....4$

Take the derivative of equation $(3)$

$\frac { dv }{ dt } =\Omega \hat { z } \times \frac { dr }{ dt } +\frac { d\omega }{ dt } \times \left( l\hat { z } \right) \\ \\$

Plugging the value of equation (4) to this equation and taking the cross product we get

$\frac { dv }{ dt } =\Omega \hat { z } \times v+\left( \frac { m{ l }^{ 2 } }{ I } \right) \frac { dv }{ dt } \\ \Rightarrow \quad a=\frac { dv }{ dt } =\left( \frac { \Omega }{ 1+\left( \frac { m{ l }^{ 2 } }{ I } \right) } \right) \hat { z } \times v$

Momentum of inertia of a sphere $I=\frac { 2 }{ 5 } m{ l }^{ 2 }$ , thus

$a=\left( \frac { 2 }{ 7 } \Omega \right) \hat { z } \times v$

We know an object moving in a circular motion as an acceleration of $a=\Omega \hat { z } \times v$ , So basically the ball is moving at an angular frequency of $\frac { 2 }{ 7 }$ times the frequency of the turntable.

$\left\lfloor 100(\frac { 2 }{ 7 } ) \right\rfloor =28$