Answer :

Given,


Radius of the disc and ring, r = 10 cm = 0.1 m


Angular velocity of the disc and ring, ω0 = 10 π rad/s


Initial velocity of both the objects, u = 0 m/s


Coefficient of kinetic friction, μk = 0.2


Motion of the objects start due to frictional force,


f = ma


Where f = frictional force ,


f = μkmg


m = mass of the body


a = acceleration of the body


μkmg = ma


a = μkg ------ > (1)


We have final velocity as per first equation of motion,


v = u + at


Where,


v = final velocity


u = initial velocity


a = acceleration


t = time


v = μkg ------- > (2)


The torque generated by friction initially causes reduction in angular velocity. Thus,


T = -Iα


Where,


I = moment of inertia of the body


α = Angular acceleration


and, T = fr


Where,


f = frictional force


r = radius


T = -μkmgr


------- > (3)


As per the frist equation of rotational motion, final angular velocity,


ω = ω0 + αt


Where,


ω0 = intial angular velocity


α = angular acceleration


t = time


------ > (4)


Rolling starts when linear velocity, v = rω


-------- > (5)


From the equations 2 & 5 we get,



------- > (6)


We know that,


For the ring, I = mr2



μkgt = rω0 – μkgt


kgt = rω0



tring


tring = 0.8s


For the disc, I = 0.5mr2




kgt = rω0




tdisc


Since the tdisc < tring.


Therefore, the disc will start rolling before the ring.


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