Question

A solid cylinder of mass, $m$, and radius, $r$, is rotating at an angular velocity, $\omega$ when a non-rotating hoop of equal mass and radius drops onto the cylinder.
In terms of its initial angular velocity, $\omega$, what is its new angular velocity, ${\omega }^{\prime }$?

• A. ${\omega }^{\prime }=\frac{\omega }{3}$
• B. ${\omega }^{\prime }=\frac{3\omega }{4}$
• C. ${\omega }^{\prime }=\omega$
• D. ${\omega }^{\prime }=3\omega$
• E. ${\omega }^{\prime }=\frac{2\omega }{3}$

Answer 1

Answer: (A)

${\omega }^{\prime }=\frac{\omega }{3}$

Moment of inertia of solid cylinder $I_{cylinder}=\frac{1}{2}m{r}^2$

Thus initial angular momentum $L_i=\frac{1}{2}m{r}^{2}w$

Final moment of inertia of the system $I^{\prime }=I_{cylinder}+I_{hoop}=\frac{1}{2}m{r}^2+m{r}^2=\frac{3}{2}m{r}^2$

Final angular momentum $L_f=I^{\prime }{w}^{\prime }=\frac{3}{2}m{r}^2{w}^{\prime }$

Using conservation of angular momentum : $L_i=L_f$

$\therefore$ $\frac{1}{2}m{r}^{2}w=\frac{3}{2}m{r}^2{w}^{\prime }$ $⟹w^{\prime }=\frac{w}{3}$