A uniform disc of mass `m` and radius `r` is projected horizontally with velocity `v_(0)` on a rough horizontal floor so that it starts off with a purely sliding motion at `t=0`. At `t=t_(0)` seconds it acquires a purely rolling motion.
a. Calculate the velocity of the centre of mass of the disc at `t=t_(0)`.

b. Assuming coefficient of friction to be `mu` calculate `t_(0)`.
The work done by the frictional force as a function of time.
Total work done by the friction over a time t much longer than `t_(0)`.

Explanation

F.B.D. of the disc.
When the disc is projected it starts sliding and hence there is a relative motion between the points of contact and surface. Therefore, frictional force acts on the disc in the direction opposite to the motion.
(a) Now for translational motion
`vec(a)_(c.m.) = (vec(f))/(m)`
`f = mu N` (as it slides)`
`= mu m g`
`rArr" "a_(c.m.) =- mu "g, negative sign indicates that a"_(c.m.) "is opposite to v"_(c.m.)`
`rArr" "v_(c.m.(t)) = v_(0) - mu g t_(0)`
`rArr" "t_(0) = ((v_(0) - v))/(mu g), "where v"_(c.m.(t_(0))) = v" "...(i)`
For rotational motion about centre
`tau_(f) + tau_(mg) = I_(c.m.) alpha" "rArr" "mu g r = (mr^(2))/(2) alpha`
`alpha = (2mu g)/(r) " "...(ii)`
Therefore, `omega_((1_(0))) = 0 + (2 mu g)/(r)t,`
Using `omega_(1) = omega_(0) + alpha t`
`rArr" "omega = (2 (v_(0) - v))/(r)" "...(iii)`
Using (i)
`(v_(cm))_(t_(0)) = omega r`
`rArr" "v = 2(v_(0) - v)" ""using (iii)"`
`rArr" "v = (2)/(3) v_(0)`
(b) Putting the value of v in equation (i)
we get `t_(0) = (v_(0))/(3 mu g)`
(c) Work done by the frictional force is equal to change in K.E.
`rArr" "W_("friction") = (1)/(2) m(v_(0) - mu g t)^(2-) + (1)/(2) ((mr^(2))/(2))((2 mu g t)/(r))^(2) - (1)/(2) mv_(0)^(2)`
`= m ((3)/(2) mu^(2) g^(2) t^(2) - v_(0) mu g t),"for t" le t_(0)`
(d) For time `t gt t_(0)`, work done by the friction is zero.
`rArr` For longer time total work done
`m ((3)/(2) mu^(2) g^(2) t^(2) - v_(0) mu g t)`
`rArr" "W = - m v_(0)^(2)//6`