A uniform solid cylinder of mass $m$ and radius $r$ is set rotation about its axis with an angular velocity ${\omega }_0$then lowered with its lateral surface onto a horizontal plane and released. The coefficient of friction between the cylinder and the plane is equal to $\mu$. Find the total work performed by the sliding friction force acting on the cylinder.

$$\begin{gathered} \mathrm{Acceleration} \mathrm{for} \mathrm{translation}, a =\frac{f}{m} \\ =\frac{\mu mg}{m} \\ =\mu g \\ \mathrm{Retardation} \mathrm{for} \mathrm{rotation} \alpha = \frac{\tau }{l} \\ =\frac{fR}{l} \\ =\frac{\mu mgR}{{\displaystyle \frac{m{R}^2}{2}}} \\ =2\mu g/R \\ \mathrm{If} t \mathrm{be} \mathrm{the} \mathrm{required} \mathrm{time} \mathrm{and} v \mathrm{and} \mathrm{\omega } \mathrm{are} \mathrm{linear} \mathrm{and} \mathrm{angular} \mathrm{velocities} \\ \mathrm{after} \mathrm{pure} \mathrm{rolling}, \mathrm{then} \\ \mathrm{\omega }= {\mathrm{\omega }}_0-\mathrm{\alpha } \mathrm{t} \\ \mathrm{and} v = 0 + at \\ \mathrm{After} \mathrm{getting} \mathrm{pure} \mathrm{rolling}, \mathrm{v} = \mathrm{\omega } \mathrm{R}. \\ \mathrm{After} \mathrm{solving} \mathrm{equations}, \mathrm{we} \mathrm{get} \\ \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{frictions} W = k_i– k_f \\ = \frac{1}{2}I{{\omega }_0}^2-\left[\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2\right] \\ = -\frac{m{{\omega }_0}^2{R}^2}{6} \end{gathered}$$