Question

A wheel (to be considered as a ring) of mass $m$ and radius $R$, rolls without sliding on a horizontal surface, with constant velocity $V_0$. It encounters a stop of height $R/2$ at which it ascends without sliding. The angular velocity of the ring just after it comes in contact with the step is,

• A. $\frac{3V_0}{4R}$
• B. $\frac{V_0}{2R}$
• C. $\frac{V_0}{R}$
• D. $\frac{5V_0}{4R}$

Answer 1

The correct option is A $\frac{3V_0}{4R}$

As no external torque is acting on the ring, applying conservation of angular momentum conservation about point $P$, we get

$L_i=L_f$


About point $P$,

$L_i=|{\overrightarrow{r}}_{com}\times \overrightarrow{p}|+I_{com}.\omega$

$=r\times \left(mv\right)+I_{com}.\omega$

Here, $I_{com}=m{R}^2$

and, $r=(R-\frac{R}{2})=\frac{R}{2}$

$L_i=\frac{R}{2}m{V}_0+m{R}^2.\omega \{\because \omega =\frac{V_0}{R}\}$

$=\frac{m{V}_{0}R}{2}+M{R}^2\left(\frac{V_0}{R}\right)$

$\therefore L_i=\frac{3m{V}_{0}R}{2}$

$L_f=I_P.\omega$ (Here, velocity of wheel $V_0=0$)

$I_P=$ moment of inertia of wheel point $P$.

According to parallel axis theorem,
$I_P=m{R}^2+m{R}^2=2m{r}^2$

$\therefore L_f=2m{R}^2.\omega$

As, $(L_i=L_f)$

$\Rightarrow \frac{3m{V}_{0}R}{2}=2m{R}^2\omega$

$\therefore \omega =\frac{3V_0}{4R}$

Hence, $\left(A\right)$ is the correct answer.