Question

A small object placed on a rotating horizontal turn table just slips when it is placed at a distance $4 \,cm$ 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

• A. $1\, cm$
• B. $2\, cm$
• C. $4\, cm$
• D. $8\, cm$

Answer 1

Answer: (A)

The object will slip if centripetal force $\ge$ force of friction
$mr\omega^{2} \ge\mu mg$
$r\omega^{2}\ge\mu g$
$r\omega^{2} \ge$ constant, or $ \left(\frac{r_{1}}{r_{2}}\right)$ $=\left(\frac{\omega^{2}}{\omega_{1}}\right)^{2}$
$\frac{4\,cm}{r^{2}}$ $=\left(\frac{2 \omega}{\omega}\right)^{2}$ $\therefore \, $ $r_{2}=1 \,cm$