Question

A small object placed on a rotating horizontal turn table just slips when it is placed at a distance $4\text{cm}$ from the axis of rotation. If the angular velocity of the turn-table is doubled, the object slips when its distance from the axis of rotation is (in $\text{cm}$)

Answer 1

Given:
$R_1=4\text{cm},{\omega }_1=\omega$
${\omega }_2=2\omega ,R_2=?$

The object will slip if centripetal force equals to force of friction
$mr{\omega }^2=\mu mg$
$r{\omega }^2=\mu g$
$r{\omega }^2=\text{constant}$

As $\omega$ changes from ${\omega }_1$ to ${\omega }_2$
$\left(\frac{R_1}{R_2}\right)={\left(\frac{{\omega }_2}{{\omega }_1}\right)}^2$
$\frac{4\text{cm}}{R_2}={\left(\frac{2\omega }{\omega }\right)}^2$
$\therefore R_2=1\text{cm}$