A uniform cylinder of radius R is spinned wit | vedclass

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Question

$f_{2}=\mu N_{2}$ and $f_{1}=\mu N_{1}$

for horizontal equilibrium- $f_{1}=N_{2}$

$\Longrightarrow \mu N_{1}=N_{2}$

For vertical equilibrium-

Vedclass · Accepted Answer

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

Vedclass · Answer

$f_{2}=\mu N_{2}$ and $f_{1}=\mu N_{1}$

for horizontal equilibrium- $f_{1}=N_{2}$

$\Longrightarrow \mu N_{1}=N_{2}$

For vertical equilibrium-

$f_{2}+N_{1}=m g$

$\Longrightarrow \mu N_{2}+N_{1}=m g$

$\Longrightarrow N_{1}\left(\mu^{2}+1ight)=m g$

$\Longrightarrow N_{1}=\frac{m g}{\mu^{2}+1}$

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 $=-\left[f_{1}+f_{2}ight] 	imes s$ 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.

$\mathrm{K.E}=\frac{1}{2} I \omega^{2}=\frac{1}{4} m R^{2} \omega^{2}$

$\Longrightarrow\left[\frac{\mu(1+\mu) m g}{1+\mu^{2}}ight] 2 \pi R N=\frac{1}{4} m R^{2} \omega^{2}$

$\Longrightarrow N=\frac{\omega^{2} R\left(1+\mu^{2}ight)}{8 \pi \mu g(1+\mu)}$

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

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