A uniform circular disc placed on a rough horizontal surface has … - Vidyadip

MCQ

$A$ uniform circular disc placed on $a$ rough horizontal surface has initially a velocity $v_0$ and an angular velocity $\omega_0$ as shown in the figure. The disc comes to rest after moving some distance in the direction of motion. Then $\frac{{{v_0}}}{{r{\omega _0}}}$ is

✓
$\frac{1}{2}$
B
$1$
C
$\frac{3}{2}$
D
$2$

✓

Answer

Correct option: A.

$\frac{1}{2}$

a
Slipping will take place and to oppose the relative motion between the disc and the ground, friction will act in the shown direction.

So we can write $f=m a$ So $a=\frac{f}{m}$

$f \cdot r=I \alpha$

For the disc $I=\frac{1}{2} m r^{2}$

$\alpha=\frac{2 f}{m r}$

If the disc stops moving after travelling for a while, both $v_{0}$ and $\omega_{0}$ become zero at the same time.

If time $t$ required to stop, we can write

$v_{0}-a t=0$ and $\omega_{0}-\alpha t=0$

$\therefore \frac{v_{0}}{a}=\frac{\omega_{0}}{\alpha}$

$\therefore \frac{v_{0} m}{f}=\frac{\omega_{0} r m}{2 f}$

$\therefore \frac{v_{0}}{\omega_{0} r}=\frac{1}{2}$

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