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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 $μ$. The number of turns taken by the cylinder before stopping is given by
$\frac{{{\omega ^2}r(1 + \mu )}}{{8\pi \mu g}}$
$\frac{{{\omega ^2}R(1 + \mu^2 )}}{{8 \pi \mu g(1+ \mu)}}$
$\frac{{{\omega ^2}R(1 + \mu^2 )}}{{4 \pi \mu g(1+ \mu)}}$
$\frac{{{\omega ^2}R(1 + \mu^2 )}}{{\mu g(1+ \mu)}}$
Solution
$f_{2}=\mu N_{2}$ and $f_{1}=\mu N_{1}$
for horizontal equilibrium- $f_{1}=N_{2}$
$\Longrightarrow \mu N_{1}=N_{2}$
For vertical equilibrium-
$f_{2}+N_{1}=m g$
$\Longrightarrow \mu N_{2}+N_{1}=m g$
$\Longrightarrow N_{1}\left(\mu^{2}+1\right)=m g$
$\Longrightarrow N_{1}=\frac{m g}{\mu^{2}+1}$
Now, $f_{1}=\mu N_{1}=\frac{\mu m g}{\mu^{2}+1}$
$f_{2}=\mu N_{2}=\mu f_{1}=\frac{\mu^{2} m g}{\mu^{2}+1}$
Work done by friction $=-\left[f_{1}+f_{2}\right] \times s$ where $s=$ distance traveled by any point on the cylinder.
$s=2 \pi R N$ where $N=$ number of turns
Magnitude of work done by friction is equal to loss in energy of cylinder.
$\mathrm{K.E}=\frac{1}{2} I \omega^{2}=\frac{1}{4} m R^{2} \omega^{2}$
$\Longrightarrow\left[\frac{\mu(1+\mu) m g}{1+\mu^{2}}\right] 2 \pi R N=\frac{1}{4} m R^{2} \omega^{2}$
$\Longrightarrow N=\frac{\omega^{2} R\left(1+\mu^{2}\right)}{8 \pi \mu g(1+\mu)}$
