A classical mechanics problem by A Former Brilliant Member

$\text{Given:}$
$\boxed{ \rightarrow \mu=\sqrt3 \\ \rightarrow\omega_0=2\sqrt{\pi} \\ \rightarrow R=\SI{0.25}{\meter} \\ \rightarrow g=\SI{10}{\meter\per\second\squared}}$

$\text{Let}$ $m$ $\text{be the mass of the cylinder, } F_1 \text{ be the frictional force acting along the floor and } F_2 \text{ be the force acting along the wall as}$ $\text{shown in the following figure.}$ $\text{Equating components of force and normal reaction horizontally and vertically and using the laws of friction we get:}$ $\\N_{ 1 }+F_{ 2 }-mg=0\quad \cdots (1) \\F_{ 1 }-N_{ 2 }=0\quad \cdots (2) \\F_{ 1 }=\mu N_{ 1 }\quad \cdots (3) \\F_{ 2 }=\mu N_{ 2 }\quad \cdots (4)$ $\text{Using (1), (2), (3) and (4) we get:}$ $\boxed{F_{ 1 }=\dfrac { \mu mg }{ 1+\mu^2 }} \text{ and } \boxed{F_2=\dfrac{\mu^2mg}{1+\mu^2}}$ $\text{Moment of inertia, I is equal to}$ $\boxed{I=\dfrac{mR^2}{2}}$ $\text{ Net torque acting on the cylinder}$ $\tau_{net}$ $\text{is}$ ( $\alpha$ $\text{is the angular acceleration):}$ $\tau_{net}=-F_1R-F_2R = I\alpha \\ \Rightarrow I\alpha=\mu mg\dfrac{1+\mu}{1+\mu^2} \\ \Rightarrow \boxed{\alpha=\dfrac{-2\mu g\left(1+\mu \right) }{R\left(1+\mu^2\right)}}$ $\alpha \text{ and } \tau_{net} \text{ are with a negative sign because the direction of }\omega_0 \text{ was assumed to be positive and the } \tau_{net} \text{ and } \alpha \text{ acting are opposing it.}$

$\text{Now, when the circular motion of the cylinder stops,}$ $\omega^2_t=0=\omega^2_0+2\alpha\theta \\ \Rightarrow \boxed{\theta=\dfrac{\omega^2_0\left(1+\mu^2\right)R}{4\mu g \left(1+\mu \right)}=\dfrac { 4\pi \times \left( 1+\left( \sqrt { 3 } \right) ^{ 2 } \right) \times 0.25 }{ 4\times 10\left( \sqrt { 3 } +\left( \sqrt { 3 } \right) ^{ 2 } \right)} } \\ \Rightarrow \boxed{\theta=\SI{0.06639}{\radian}\approx\SI{0.0665}{\radian}}$