Question

Two identical particles are attached at the ends of a light string which passes through a hole at the center of a table. One of the particles is made to move in a circle on the table with angular velocity ${\omega }_1$ and the other is made to move in a horizontal circle as a constant pendulum with angular velocity ${\omega }_2$. If $l_1$ and $l_2$ are the length of the string over and under the table, then in order that particle under the table neither moves down nor moves up, the ratio $l_1/l_2$ is:

• A. $\frac{{\omega }_1}{{\omega }_2}$
• B. $\frac{{\omega }_2}{{\omega }_1}$
• C. $\frac{{\omega }_1^2}{{\omega }_2^2}$
• D. $\frac{{\omega }_2^2}{{\omega }_1^2}$

Answer 1

The correct option is D $\frac{{\omega }_2^2}{{\omega }_1^2}$

for horizontal equilibrium

$m{l}_2{w}_2^2=m{l}_1{w}_1^2\Rightarrow \frac{l_1}{l_2}={\left(\frac{w_2}{w_1}\right)}^2$