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

Here `v_(cm) lt Romega_(0)`, friction acts in forward direction. Let pure rolling starts after time `t_(0)`.
Since weight, normal reaction and friction pass through `A`, hence the angular momentum about `A` is conserved.
Conservation of angular momentum about `A`
`I_(c.m.)omega_(0)=mvR+I_(c.m.)omega`
`(1)/(2)mR^(2)omega_(0)=mvR+(1)/(2)mR^(2)((v)/(R))=(3)/(2)mvR`
`v=(omega_(0)R)/(3)`
(`a`) `f=f_(max)=muN=mumg`
Acceleration `=(f)/(m)=(mumg)/(m)=mug`
`v=0+at_(0)implies(omega_(0)R)/(3)=mug t_(0)impliest_(0)=(omega_(0)R)/(3mug)`
(`b`) `K_(i)=(1)/(2)I_(c.m.)omega_(0)^(2)=(1)/(2)xx(1)/(2)mR^(2)omega_(0)^(2)=(1)/(4)mR^(2)omega_(0)^(2)`
`K_(f)=(1)/(2)mv^(2)(1+(k^(2))/(R^(2)))=(1)/(2)m((omega_(0)R)/(3))^(2)(1+(1)/(2))`
`=(1)/(12)mR^(2)omega_(0)^(2)`
`K_(i)=K_(f)+W_(f)`
Work done against friction
`W_(f)=K_(i)-K_(f)=((1)/(4)-(1)/(12))mR^(2)omega_(0)^(2)=(1)/(6)mR^(2)omega_(0)^(2)`
Work done by friction `W'_(f)= -W_(f)= -(1)/(6)mR^(2)omega_(0)^(2)`