Question 11: Work, Energy and Power & Circular Motion

Question

A coin is placed on a disc rotating with an angular velocity $\omega$. The co-efficient of friction between the disc and the coin is $\mu$. The maximum distance of the coin from the centre of the disc up to which it will rotate with the disc is

Accepted Answer

Centripetal force needed:
$$F_c = m\,\omega^2\,r$$
Maximum static friction available:
$$F_{\rm fric\,max} = \mu\,m\,g$$
Set them equal for the limiting case:
$$\mu\,m\,g = m\,\omega^2\,r_{\max}$$
Solve for $r_{\max}$:
$$r_{\max} = \frac{\mu\,g}{\omega^2}$$
Hence the correct choice is Option B: $\displaystyle \frac{\mu g}{\omega^2}$.

Answer

$\sqrt{\frac{\mu}{\omega^2}}$

Answer

$\frac{\mu g}{\omega^2}$

Answer

$\sqrt{\frac{\mu g}{\omega^2}}$

Answer

$\frac{\mu g}{\omega}$

Work, Energy and Power & Circular Motion

A coin is placed on a disc rotating with an angular velocity \omega. The co-efficient of friction between the disc and the coin is \mu. The maximum distance of the coin from the centre of the disc up to which it will rotate with the disc is

Choose the correct answer: