Question

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

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

Answer 1

The correct option is A $1cm$

$R_1=4cm$ , ${\omega }_1=\omega$, ${\omega }_2=2\omega$

$R_2=?$

The object will slip if $centripetalforce\ge forceoffriction$

$mR{\omega }^2$ ≥ $\mu mg$

$R{\omega }^2$ ≥ $\mu g=constant$

Hence,

$\left[\frac{R_1}{R_2}\right]=[\frac{{\omega }_2}{{\omega }_1}{]}^2$

∴r2=1cm$\left[\frac{4}{R_2}\right]$ $=$ $[\frac{2\omega }{\omega }{]}^2$

$R_2=1cm$

this is the required solution.

Hence,

The correct option is A.