Question

Two particles, each of mass ${}^{\prime }{m}^{\prime }$ are attached to the two ends of a light string of length $L$, which passes through a hole at the centre of a smooth table. One particle describes a circular path on the table with angular velocity ${\omega }_1$, and the other describes a conical pendulum with angular velocity ${\omega }_2$ below the table. If $l_1$ and $l_2$ are the lengths of portions of the string above and below the table, then:

• A. $\frac{l_1}{l_2}=\frac{{\omega }_2}{{\omega }_1}$
• B. $\frac{l_1}{l_2}=\frac{{\omega }_2^2}{{\omega }_1^2}$
• C. $\frac{1}{{\omega }_1^2}+\frac{1}{{\omega }_2^2}=\frac{ml}{g}$
• D. $\frac{1}{{\omega }_1^2}+\frac{1}{{\omega }_2^2}=\frac{l\cos \theta }{g}$

Answer 1

Answer: (B), (D)

The correct options are
B $\frac{l_1}{l_2}=\frac{{\omega }_2^2}{{\omega }_1^2}$
D $\frac{1}{{\omega }_1^2}+\frac{1}{{\omega }_2^2}=\frac{l\cos \theta }{g}$

The mass of each particle$=m$
Length of the string$=L$
The angular velocity of one particle$={\omega }_1$
The angular velocity of another particle$={\omega }_2$
Now, from the conservation of angular momentum,
$m{\omega }_1{l}_1^2=m{\omega }_2{l}_2^2$
$\Rightarrow {\omega }_1{l}_1^2={\omega }_2{l}_2^2⟶1$
The K.E of the two particle will be same
So, $\frac{1}{2}m{l}_1{\omega }_1^2=\frac{1}{2}m{l}_2{\omega }_2^2$
$\Rightarrow \frac{l_1}{l_2}=\frac{{\omega }_2^2}{{\omega }_1^2}$
And if the string makes an angle $\theta$ with the vertical line
then $\omega =\sqrt{\frac{g}{l\cos \theta }}$
$\Rightarrow {\omega }_1^2=\frac{g}{\frac{l}{2}\cos \theta }$
${\omega }_2^2=\frac{g}{\frac{l}{2}\cos \theta }$
So, here, $\frac{1}{{\omega }_1^2}+\frac{1}{{\omega }_2^2}=\frac{l\cos \theta }{g}$