Question

A coin placed on uniformly rotating turntable just slips if it is placed at a distance of 4 cm from the centre. If the angular velocity of the turntable is doubled, it will just slip at a distance of

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

Answer 1

Answer: (A)

. 1 cm

Since it just slips so maximum friction will be equal to the centripetal force in both the case, i.e.;
$f_L=m{r}_1{\omega }_1^2=m{r}_2{\omega }_2^2$
$\Rightarrow r_1{\omega }_1^2=r_2{\omega }_2^2$
$\Rightarrow r_2=r_1{\left(\frac{{\omega }_1}{{\omega }_2}\right)}^2$
$\Rightarrow r_2=4{\left(\frac{\omega }{2\omega }\right)}^2$
$r_2=1cm$.