Question

A cylinder rests on a horizontal rotating disc, as shown in the figure. Find at what angular velocity, $\omega$, the cylinder falls of the disc, if the distance between the axes of the disc and cylinder is $R$, and the coefficient of friction $\mu$$>$$D/h$, where $D$ is the diameter of the cylinder and $h$ is its height.

• A. $\sqrt{\frac{3Dg}{hR}}$
• B. $\sqrt{\frac{3Dg}{2hR}}$
• C. $\sqrt{\frac{Dg}{hR}}$
• D. None of these

Answer 1

The correct option is C $\sqrt{\frac{Dg}{hR}}$

Under equilibrium, the torque about the center of mass is zero.

$⟹f\frac{h}{2}=N\frac{D}{2}$

$⟹\mu N\frac{h}{2}=N\frac{D}{2}$

$⟹\mu =\frac{D}{h}$

This is the limiting condition for the coefficient of friction to avoid toppling.

Since the coefficient of friction $\mu >\frac{D}{h}$, to balance the force of friction to avoid toppling of the cylinder, the force that acts on the block will be $f=\mu N=\frac{ND}{h}$

Balancing forces in horizontal direction,

$m{\omega }^{2}R=\mu N=\frac{ND}{h}$ ; Here $N=mg$

$⟹\omega =\sqrt{\frac{Dg}{hR}}$