A sphere of mass `M` and radius `r` shown in figure slips on a rough horizontal plane. At some instant it has translational velocity `V_(0)` and rotational velocity about the centre `(v_(0))/(2r)`. Find the translational velocity after the sphere starts pure rolling.
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Explanation

Velocity of the centre = `V_(0)` and the angular velocity about the centre `= (v_(0))/(2 r)`. Thus `v_(0) gt omega_(0) r`.
The sphere slips forward and thus the friction by the plane on the sphere will act backward. As the friction is kinetic, its value is `mu N = mu Mg` and the sphere will be decelerated by `a_(cm) = f//M`. Hence
`v(t) = v_(0) - (f)/(M) t`. ....(i)
This friction will also have a torque `Gamma fr` about the centre. This torque is clockwise and in the direction of `omega_(0)`. Hence the angular acceleration about the centre will be
`alpha = f ( r)/((2//5) Mr^(2)) = (5 f)/(2 Mr)`
and the clockwise angular velocity at time `t` will be
`omega(t) = omega_(0) +(5f)/(2 Mr) t = (v_(0))/(2 r) + (5f)/(2 Mr) t.`
Pure rolling starts when `v(t) = r omega(t)`
i.e., `v(t) = (v_(0))/(2) + (5 f)/(2 M) t.` ....(ii)
Eliminating `t` from (i) and (ii),
`(5)/(2) v(t) + v(t) = (5)/(2) V_(0) + (v_(0))/(2)`
or `v(t) = (2)/(7) xx 3 V_(0) = (6)/(7) V_(0)`
Thus, the sphere rolls with translational velocity `6 v_(0)//7` in the forward direction.