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 pi rad//s`. Which of the two will start to roll earlier ? The coefficient of kinetic friction is `mu_(k) = 0.2`.

Explanation

When a disc or ring starts rotatory motion on a horizontal surface, initial translational velocity of centre of mass is zero.

The frictional force causes the centre of mass to accelerate linearly but frictional torque causes angular retardation. As force of normal reaction N=mg, hence frictional force `f=u_(k) N=u_(k)` mg.
For linear motion `f=u_(k)`. mg=ma
and for rotational motion, `tau=f. R=mu_(k) mg. R=-I alpha`
Let perfect rolling motion starts at time t, when `v=R omega`
From (i) `" " a=mu_(k).g`
`therefore " " v=u+at=0+mu_(k).g.t`
From (ii) `" " alpha=-(mu_(k).mgR)/(I)=-(mu_(k).mgR)/(K^(2))=-(mu_(k).gR)/(K^(2))`
`therefore " " omega=omega_(0)+alphat=omega_(0)-(mu_(k)lgR)/(K^(2))t`
Since `" " v=Romega, " hence" mu_(k).g.t=R[omega_(0)=mu_(k).(gR)/(K^(2))t]`
`implies " " t^(2)=(Romega_(0))/(mu_(k).g(1+(R^(2))/(K^(2)))`
For disc, `K^(2)=(R^(2))/(2), "hence" t=(omega_(0)R)/(mu_(k).g(1+(R^(2))/(R^(2)//2)))=(omega_(0)R)/(3mu_(k).g)`
For ring, `K^(2)=R^(2), "hence" t=(omega_(0)R)/(mu_(k).g(1+(R^(2))/(R^(2))))=(omega_(0)R)/(2mu_(k).g)`
Thus, it is clear that disc will start to roll earlier. The actual time at which disc starts rolling will be
`t=(omega_(0)R)/(2mu_(k).g)=((10pi)xx(0.1))/(3xx(0.2)xx9.8)=0.538`.