A disc of radius `R` is rotating with an angular speed `omega_(0)` about a horizontal axis. It is placed on a horizontal table. The coefficient of kinetic friction is `mu_(k)`.
(a) What was the velocity of its centre of mass before being brought in contact with the table ?
(b) What happens to the linear velocity of a point on its rim when placed in contact with the table ?
(c ) What happens to the linear speed of the centre of mass when disc is placed in contact with the table ?
(d) Which force i sresponsible for the effects in (b) and (c ).
(e) What condition should be satisfied for rolling to begin ?
(f) Calculate the time taken for the rolling to begin.

Explanation

(a) Before the disc is brought in contact with the table, it is simple rotating about its horizontal axis. Therefore, its centre of mass is at rest, i.e. `v_(cm) = Zero`.
(b)Linear velocity of any point on the rim of the disc would decrease. When disc is placed in contact with the table.
(c) When rotating disc is placed in contact with the table, its centre of mass acquires some velocity - which was zero before contact. Hence, velocity of centre of mass increases.
(d)In (a) above, force of friction between the rotating disc and table is responsible for decreasing linear velocity of any point on the rim of the disc. In (c), the disc rolls over because of friction. Hence, velcoity of centre of mass of disc increases due to friction.
(e) Rolling would begin only when `v_(cm) = R omega`.
(f) Acceleration produced in centre of mass due to friction `a_(cm) = (F)/(m) = (mu_(k)mg)/(m) = mu_(k)g` ..(i)
Angular acceleration produced by the torque due to friction, `alpha = (tau)/(I) = ((mu_(k)mg)R)/(I)`
from `v_(cm) = u_(cm) + a_(cm) t`
`v_(cm) = 0 + (mu_(k)g) t = mu_(k) g t` ...(ii)
From `omega = omega_(0) + alpha t = omega_(0) + ((mu_(k)mg)Rt)/(I)` ..(iii)
For rolling without slipping
`(v_(cm))/(R ) = omega`
Using (ii) and (iii), we get
`(mu_(k) g t)/(R ) = omega_(0) + ((mu_(k)mg)Rt)/(I)`
`(-(mu_(k) mg)Rt)/(I) + (mu_(k) g t)/(R) = omega_(0)`
`mu_(k)g(-(mR)/(I)+(1)/(R))t = omega_(0)`
`mu_(k)g(-(mR^(2))/(I)+1)(t)/(R) = omega_(0)`
`t = (R omega_(0))/(mu_(k) g(1 -(mR^(2))/(I)))`