Question

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

• A. 5 cm
• B. 6.5 cm
• C. 1 cm
• D. 3 cm

Answer 1

The correct option is C 1 cm

$\begin{array}{c}\text{Coin slips only, when the centrifugal force due to its rotation}\\ \text{exceeds the frictional force on coin.}\\ \text{Here, since the surface of rotating turntable is equally rough}\\ \text{in both cases, frictional force will remain same (F).}\end{array}$

$\begin{array}{cc}\text{C-(i)}\to \text{Let, angular velocity of rotation}& ={\omega }_0\text{.}\\ \text{Thus,}m{\omega }_0^2\times \left(4cm\right)=F& --\text{(i)}\end{array}$

$\text{Let, coin is placed at}\left(xcm\right)\text{from centre and it just slips.}$

$\text{Here,}[\omega =2{\omega }_0]$

$Thus,m{\left(2{\omega }_0\right)}^2\times \left(xcm\right)=F-\cdots \text{(ii)}$

$\text{Equating}e{q}^n-\left(1\right)\text{and (ii), we get:}x=1cm$