Question

A uniform cylinder of radius R is spinned with angular velocity $\omega$ about its axis and then placed into a corner. The coefficient of friction between the cylinder and planes is $\mu$. The number of turns taken by the cylinder before stopping is given by

• A. $\frac{{\omega }^{2}r(1+\mu )}{8\pi \mu g}$
• B. $\frac{{\omega }^{2}R(1+{\mu }^2)}{8\pi \mu g(1+\mu )}$
• C. $\frac{{\omega }^{2}R(1+{\mu }^2)}{4\pi \mu g(1+\mu )}$
• D. $\frac{{\omega }^{2}R(1+{\mu }^2)}{\pi \mu g(1+\mu )}$

Answer 1

The correct option is B $\frac{{\omega }^{2}R(1+{\mu }^2)}{8\pi \mu g(1+\mu )}$

$f_2=\mu N_2$ and $f_1=\mu N_1$

for horizontal equilibrium-$f_1=N_2$

$⟹\mu N_1=N_2$

For vertical equilibrium-

$f_2+N_1=mg$

$⟹\mu N_2+N_1=mg$

$⟹N_1({\mu }^2+1)=mg$

$⟹N_1=\frac{mg}{{\mu }^2+1}$

Now, $f_1=\mu N_1=\frac{\mu mg}{{\mu }^2+1}$

$f_2=\mu N_2=\mu f_1=\frac{{\mu }^{2}mg}{{\mu }^2+1}$

Work done by friction $=-[f_1+f_2]\times s$

where $s=$ distance traveled by any point on the cylinder.

$⟹s=2\pi RN$ where $N=$ number of turns

Magnitude of work done by friction is equal to loss in energy of cylinder.

K.E=$\frac{1}{2}I{\omega }^2=\frac{1}{4}m{R}^2{\omega }^2$

$⟹\left[\frac{\mu (1+\mu )mg}{1+{\mu }^2}\right]2\pi RN=\frac{1}{4}m{R}^2{\omega }^2$

$⟹N=\frac{{\omega }^{2}R(1+{\mu }^2)}{8\pi \mu g(1+\mu )}$

Answer-(B)