A solid disc and a ring, both of radius 10 cm are placed on a horizontal table simultaneously, with initial angular speed equal to `10 π" rad s"^(-1)`. Which of the two will start to roll earlier ? The co-efficient of kinetic friction is `µ_(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 = I alpha.alpha=(tau)/(I)=(mu_(k)mgr)/(1)`
Linear velocity of CM of the body (initially at rest, `upsilon_(0)=0`) is give by
`upsilon = upsilon_(0)+at = at = mu kgt " "` ....(iv)
Angular velocity of the body after time t, i.e.,
`omega = omega_(0)+alpha t = omega_(0)-((mu_(k)mgr)/(I))t " "` ......(v)
(Here, `alpha` has been taken negative as torque due to frictional force produces retardation.)
`upsilon = r omega " "` .....(vi)
From eqns. (iv), (v) and (vi), we get
`mu_(k)g t = r [omega_(0)-((mu_(k)mgr)/(I))t]=r omega_(0)-((mu_(k)mg r^(2))/(I))t " "` ....(vii)
For a ring, as `I = mr^(2)`, from eqn. (vii)
`mu_(k) g t=r omega_(0)-((mu_(k)mgr^(2))/(mr^(2)))t=r omega_(0)-mu_(k)g t` or
`2mu_(k)g t=r omega_(0)` or `t = (r omega_(0))/(2 mu_(k)g)` or `t = (0.1xx10pi)/(2xx0.2xx9.8)=0.80s`
`("as " r = 0.1 omega_(0)=10 pi rad//s, mu_(k)=0.2, g=9.8 m//s^(2))`
or `3 mu_(k)g t = r omega_(0)` or
`t=(r omega_(0))/(3mu_(k)g)` or `t=(0.1xx10 pi)/(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 value of r and `omega_(0)`.