Question

A coin placed on a rotating turn table just slips if it is placed at a distance of 20 cm from the centre. If the angular velocity of the turn table is doubled, the coin will just slip at a distance of

• A. 40 cm
• B. 20 cm
• C. 10 cm
• D. 5 cm

Answer 1

The correct option is D

5 cm

The coin will just slips on the rotating table if and only if the value of static frictional force reaches to limiting frictional force which is constant for the given system of coin and rotating table.
As, $F_L=mr{\omega }^2\Rightarrow m{r}_1{\omega }_1^2=m{r}_2{\omega }_2^2$

$\Rightarrow r_2=r_1(\frac{{\omega }_1}{{\omega }_2}{)}^2$
$\Rightarrow r_2=\left(20\right)(\frac{\omega }{2\omega }{)}^2=5cm.$