Question

A particle of mass $m$ is attached to one end of a weightless and inextensible string of length $L$. The particle is on a smooth horizontal table. The string passes through a hole in the table and to its other end is attached a small particle of equal mass $m$. The system is set in motion with the first particle describing a circle on the table with constant angular velocity ${\omega }_1$ and the second particle moving in the horizontal circle as a conical pendulum with constant angular velocity ${\omega }_2$. Then ${\omega }_2^2:{\omega }_1^2=$
[Take $r_1=1\text{m},L=3\text{m}$]

• A. $1:2$
• B. $1:4$
• C. $1:\sqrt{2}$
• D. $2:3$

Answer 1

The correct option is A $1:2$

Here tension in the string remains same as there is no friction between the thread and the edge of the hole in the table. For the circular motion of upper mass $m$, we have
$T=m{\omega }_1^2{r}_1...\left(i\right)$



For lower mass $m$,


$T\sin \theta =m{\omega }_2^2{r}_2...\left(ii\right)$
$T\cos \theta =mg...\left(iii\right)$
If length of the thread is taken as $L$, as shown in figure, radius $r_2$ can be given as,
$r_2=(L-r_1)\sin \theta ...\left(iv\right)$

From eq. (i), (ii), & (iv)
$m{\omega }_1^2{r}_1\sin \theta =m{\omega }_2^2(L-r_1)\sin \theta$
$\Rightarrow \frac{r_1}{(L-r_1)}=\frac{{\omega }_2^2}{{\omega }_1^2}$
$\Rightarrow \frac{1}{3-1}=\frac{{\omega }_2^2}{{\omega }_1^2}$
$\therefore \frac{{\omega }_2^2}{{\omega }_1^2}=\frac{1}{2}$