A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to `10pi rad s^(-1)`. Which of the two will start to roll earlier ? The co-efficient of kinetic friction is `mu_(k)=0.2`.

Explanation

Frictional force acting on the body,
`f=mu_(k)R=mu_(k)mg`………..`(i)`
where `R(=mg)` is the normal reaction of the table on the body. If a is the acceleration of the CM of the body
`f=ma`…………`(ii)`
From eqns. (i) and (ii)
`ma=mu_(k)mg` or `a=mu_(k)g`............`(iii)`
Torque due to frictional force i.e.,
`tau=fr=(mu_(k)mg)r=mu_(k)mgr`
As `tau=Ialpha.alpha=(tau)/(I)=(mu_(k)mgr)/(1)`
Linear velocity of CM of the body (initially at rest, `v_(0)=0`) is given by
`v=v_(0)+at=at=muk"gt"`.............`(iv)`
Angular velocity of the body after time t, i.e.
`omega=omega_(0)+alphat=omega_(0)-((mu_(k)mgr)/(I))t`...........(v)
(Here `alpha` has been taken negative as torque due to frictional force produces retardation)
`v=romega`...........(vi)
From eqs. (iv), (v) and (vi) we get
`mu_(k)"gt"=r[omega_(0)-(mu_(k)mgr)/(I)]t=romega_(0)-((mu_(k)mgr^(2))/(I))t`................(vii)
For a ring , as `I=mr^(2)`, from eqn. (vii)
`mu_(k)"gt"=romega_(0)-((mu_(k)mgr^(2))/(mr^(2)))t=romega_(0)-mu_(k)"gt"`
or `2mu_(k)"gt"=romega_(0)` or `t=(romega_(0))/(2mu_(k)g)`
or `t=(0.1xx10pi)/(2xx0.2xx9.8)=0.80s`
(as `r=0.1omega_(0)=10pirad//s`, `mu_(k)=0.2`, `g=9.8m//s^(2)`)
or `3mu_(k)"gt"=romega_(0)` or `t=(romega_(0))/(3mu_(k)g)`
or `t=(0.1xx10pi)/(3xx0.2xx9.8)=0.53s`
Since t is less in case of disc the disc begin to roll earlier than the ring for the same values of r and `omega_(0)`