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.

Options

A. `t=(Romega_(0))^(2)/(mu_(k)g(1+(mR^(2))/(I)))`
B. `t=(Romega_(0))/(mu_(k)g(1+(mR^(2))/(I)))`
C. `t=(Romega_(0))/(mu_(k)g(1+(mR^(2))/(I))^(1/2))`
D. `t=(Romega_(0))/(mu_(k)g(1+(R^(2))/(I)))`

Answer

B

Explanation

(a) Before being Brought in contact with the the was in in pure rotational motion hence ` v_(CM)=0`

(b) when the disc is placed in contact with the table due to friction velocity of a point on the rim decreases ,
(c) when the rotating disc is placed in correct with the table due to friction centre of mass acquiries some linear velocity .
friction is responsible for the effects in (b) and ( c)
(e) when rolling starts `v_(CM)-omegaR.`

where `omega` is angular speed of the disc when rolling just starts.
(f) Acceleration produced in centre of mass due to friction
`a_(CM)=(F)/(m)=(mu_(k)mg)/(m)=mu_(k)g.`

Angular retardation produced by the torque due to fricton .
`alpha=(tau)/(I)=(mu_(k)mgR)/(I) " "[:' tau=(mu_(k)N)R=mu_(k)mgR]`
`therefore v_(CM=u_(CM))+a_(CM^(t))`
` implies V_(CM)=mu_(k)"gt" ( :'U_(CM)=0)`
` and omega=omega_(0)+at`
` implies omega=omega_(0)-(mu_(k)mgR)/(I)t`
For rolling without sliping ,`(V_(CM))/(R)=omega`
`implies (V_(CM))/(R)=omega_(0)-(mu_(k)mgR)/(I)t`
`(mu_(k)g t)/(R)=omega_(0)-(mu_(k)mgR)/(I)t`
`t=(Romega_(0))/(mu_(k)g(1+(mR^(2))/(I)))`
note in this problem , frictional force help in seting pure rolloing motion .