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 pure rolling motion. Calculate the velocity of the centre of mass of the disc at $t_0$.

• A. v = (2/4)$v_0$
• B. v = (2/3)$v_0$
• C. v = (3/3)$v_0$
• D. v = (4/3)$v_0$

Answer 1

The correct option is B v = (2/3)$v_0$

B. $\frac{2}{3}{v}_0$

From the Free body diagram (FBD)

In translational motion

Friction force act on the disc on the opposite direction of motion of the disc,

The friction force, $f_r=\mu N_g$,

where $\mu =$ coefficient of friction

Normal force, $N_g=mg$

where $g=$ acceleration due to gravity

In a horizontal direction,

$f_r=-ma$, $a=$ acceleration of ball. . . . . . (1)

$\mu mg=-ma$

$a=-\mu g$ . . . . . . .(2)

First equation of motion,

Velocity of center of mass, $v_{cm}$

$v=u+at$

$v_{cm}=v_0-\mu g{t}_0$. . . . . . (3)

rotational motion is start, when ball is rolling,

Torque about the centre of ball is

$\tau =I\alpha$. . . . .(4)

Moment of inertia of uniform disc, $I=\frac{m{R}^2}{2}$

From equation (4),

$\tau =f_{r}R=\alpha \frac{m{R}^2}{2}$

$=>\mu mgR=\alpha \frac{m{R}^2}{2}$

$\alpha =\frac{2\mu g}{R}$. . . . . . .(5)

First equation of motion in rotational kinetics,

$\omega ={\omega }_0+\alpha t$

$\omega =\frac{2\mu g}{R}{t}_0$. . . . . .(6)

In rolling motion, every particle moves with $v_{cm}$ velocity,

$v_{cm}=\omega R$ when rolling begins without slipping.

$v_0-\mu g{t}_0=\frac{2\mu g}{R}{t}_{0}R$ (from equation (3) and (6)

$t_0=\frac{v_0}{3\mu g}$

After time $t_0$ ball acquire pure rolling,

Now, put the value of $t_0$ in equation (3) to get $v_{cm}$

$v_{cm}=\frac{2}{3}{v}_0$