Question

A drum of mass $m_1$ and radius $r_1$ rotates freely with initial angular velocity ${\omega }_0.$ A second drum of mass $m_2$ and radius $r_2(r_2>r_1)$ is mounted on the same axle and is at rest, although it is free to rotate. A thin layer of sand of mass $m$ is distributed on the inner surface of the smaller drum. At $t=0$, small perforations in the inner drum are opened. The sand starts to fly out at a constant rate of $\lambda kg/sec$ and sticks to the outer drum. Ignoring the transit time of the sand, choose the correct alternatives.

• A. Angular speed of outer drum at time $t$ is $\frac{\lambda t{\omega }_0}{m_2+\lambda t}{\left(\frac{r_1}{r_2}\right)}^2$.
• B. Difference in the final angular speeds of two drums is $0$.
• C. Difference in final angular speeds of two drums is $\left(\frac{m(r_2^2-r_1^2)+m_2{r}_2^2}{(m+m_2)r_2^2}\right){\omega }_0$
• D. Angular speed of inner drum remains constant.

Answer 1

Answer: (A), (C), (D)

Angular speed of inner drum remains constant.

As the sand leaves the inner drum through the open holes, it does not exert any force on the drum and angular momentum remains conserved.


Since, there is no interaction force on the particles falling from the inner drum to the outer drum so there will not be any change in the angular speed of the inner drum.
Applying conservation of angular momentum at initial time and at time $t$,

$m_1{r}_1^2{\omega }_0=(m_1-\lambda t)r_1^2{\omega }_0+(m_2+\lambda t)r_2^2{\omega }_2$

(assuming that the angular velocity of the inner drum doesn't change significantly)

${\omega }_2=\frac{\lambda t{r}_1^2{\omega }_0}{(m_2+\lambda t)r_2^2}$

Eventually, all of the mass $m$ (sand) gets deposited on the outer drum. So, the final angular speed of the outer drum can be found by

$m{r}_1^2{\omega }_0$ = $0+(m+m_2)r_2^2{\omega }_f$
$\therefore$ ${\omega }_f$ = $\frac{m{r}_1^2{\omega }_0}{(m+m_2)r_2^2}$

Difference in final angular speeds will be
${\omega }_0-{\omega }_f$ =$\left(\frac{m(r_2^2-r_1^2)+m_2{r}_2^2}{(m+m_2)r_2^2}\right){\omega }_0$