A uniform solid cylinder of mass `m` and radius `R` is set in 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
(`a`) how long the cylinder will move with sliding,
(`b`) the total work performed by the sliding friction force acting on the cylinder.

Explanation

(a) Let us depict the forces acting on the cylinder and their point of applications for the cylinder and indicate positive direction of `x` and `varphi` as shown in the figure. From the equations for the plane motion of a solid `F_x=mw_(cx)` and `N_(cz)=I_cbeta_z`:
`kmg=mw_(cx)` or `w_(cx)=kg` (1)
`-kmgR=(mR^2)/(2)beta^z` or `beta_z=-2(kg)/(R)` (2)
Let the cylinder starts pure rolling at `t=t_0` after releasing on the horizontal floor at `t=0`. From the angular kinematical equation.
`omega_z=omega_(oz)+beta_zt`,
or `omega=omega_0-2(kg)/(R)t` (3)
From the equation of the linear kinematics,
`v_(cx)=v_(0cx)+w_(cx)t`
or `v_c=0+kg t_0` (4)
But at the moment `t=t_0`, when pure rolling starts `v_c=omegaR`
so, `kg t_0=(omega_0-2(kg)/(R)t_0)R`
Thus `t_0=(omega_0R)/(3kg)`

(b) As the cylinder pick, up speed till it starts rolling, the point of contact has a purely translatory movement equal to `1/2w_ct_0^2` in the forward directions but there is also a backward movment of the point of contant of magnitude `(omega_0tau_0-1/2betat_0^2)R`. Because of slipping the net displacement is backwards. The total work done is then,
`A_(f r)=kmg[1/2w_ct_0^2-(omega_0t_0+1/2betat_0^2)R]`
`=kmg[1/2kg t_0^2-1/2(-(2kg)/(R))t_0^2R-omega_0t_0R]`
`=(kmg)(omega_0R)/(3kg)[(omega_0R)/(6)+(omega_0R)/(3)-omega_0R]=-(momega_0^2R^2)/(6)`
The same result can also be obtained by the work-energy theorem, `A_(f r)=DeltaT`.