Question

A uniform disc of mass m and radius R is projected horizontally with velocity $v_0$ on a rough horizontal floor, so that it starts off with a purely sliding motion at t = 0. After $t_0$ seconds, it acquires a purely rolling motion as shown in figure.
(a) Calculate the velocity of the centre of mass of the disc at $t_0.$
(b) Assuming the coefficient of friction to be $\mu$, calculate $t_0$. Also calculate the work done by the frictional force as a function of time and the total work done by it over a time t much longer than $t_0$

Answer 1

Between the time t = 0 to $t=t_0$. There is forward sliding, so friction f is leftwards and maximum i.e., $\mu$ mg. For time $t>t_0$, friction f will become zero, because now pure rolling has started i.e., there is no sliding (no relative motion) between the points of contact. So, for time $t<t_0$ Linear retardation, $a=\frac{f}{m}=\mu g$ and angular acceleration, Now let v be the linear velocity and co, the angular velocity of the disc at time $t=t_0$ then $v=v_0-a{t}_0=v_0-\mu g{t}_0$ ....(i) and $\omega =\alpha t_0=\frac{2\mu g{t}_0}{R}$....(ii) For pure rolling to take place $v=R\omega$ i.e. $v_0-\mu g{t}_0=2\mu g{t}_0$ $\Rightarrow t_0=\frac{v_0}{3\mu g}$ Substituting in Eq. (i), we have $v=v_0-\mu g\left(\frac{v_0}{3\mu g}\right)$ $v=\frac{2}{3}{v}_0$ Work done by friction For $t\le t_0$, linear velocity of disc at any time t is $v=v_0-\mu gt$ and angular velocity is $\omega =\alpha t=\frac{2\mu gt}{R}.$ From Work-energy theorem, work done by friction upto time t = Kinetic energy of the disc at time t-Kinetic energy of the disc at time t = 0 $\therefore W=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2-\frac{1}{2}m{v}_0^2$ $=\frac{1}{2}m[v_0-\mu gt{]}^2+\frac{1}{2}(\frac{1}{2}m{R}^2\left)\right(\frac{2\mu gt}{R}{)}^2-\frac{1}{2}m{v}_0^2$ $=\frac{1}{2}[m{v}_0^2+{\mu }^2{g}^2{t}^2-2m{v}_0\mu gt+2m{\mu }^2{g}^2{t}^2-m{v}_0^2]$ or $W=\frac{m\mu gt}{2}[3\mu gt-2v_0]$ For $t>t_0$, friction force is zero i.e., work done in friction is zero. Hence, the energy will be conserved. Therefore, total work done by friction over a time t much longer then $t_0$ is total work done upto time $t_0$ (because beyond this work done by friction is zero) which is equal to $W=\frac{m\mu gt}{2}[3\mu gt-2v_0]$ Substituting $t_0=v_0/3\mu g$, we get $W=\frac{m{v}_0}{6}[v_0-2v_0]$ $W=\frac{m{v}_0^2}{6}$