Question

Two particles each of mass $5\text{kg}$ are attached to the two ends of a light string of length $15\text{cm}$ which passes through a hole at the center of a table. One particle describes a circle on the table with angular velocity ${\omega }_1=2\text{rad/sec}$ and the other describes a circle as a conical pendulum with angular velocity ${\omega }_2=3\text{rad/sec}$ below the table as shown in the figure. If $l_1$ and $l_2$ are the lengths of the portion of the string above and below the table, then the ratio of $l_1$ and $l_2$ is

• A. $\frac{3}{4}$
• B. $\frac{4}{3}$
• C. $\frac{4}{9}$
• D. $\frac{9}{4}$

Answer 1

The correct option is D $\frac{9}{4}$

FBD (1) of mass $5\text{kg}$ describing a conical pendulum motion:


FBD (2) of mass $5\text{kg}$ describing a horizontal circular motion:

From FBD (1), we get
$T\cos \theta =mg$ ... (1)
$T\sin \theta =mr{\omega }_2^2$ ... (2)

From FBD (2), we get
$T=m{l}_1{\omega }_1^2$ ... (3)
From figure, $r=l_2\sin \theta$
$\therefore T=m{l}_2{\omega }_2^2$ (from (1)) and $T=m{l}_1{\omega }_1^2$ (from (3))
$\Rightarrow \frac{l_1}{l_2}=\frac{{\omega }_2^2}{{\omega }_1^2}=\frac{9}{4}$