Question

Mass of $0.5\,\text{kg}$ is attached to a string moving in a horizontal circle with angular velocity $10$ cycle/min. Keeping radius constant, tension in the string is made $4$ times by increasing angular velocity $\omega'$. The value of $\omega'$ of that mass will be

• A. $\dfrac{1}{3}$ cycle/s
• B. $\dfrac{1}{2}$ cycle/s
• C. $\dfrac{1}{5}$ cycle/s
• D. $\dfrac{1}{4}$ cycle/s

Hint:

For fixed radius, centripetal force (or tension) varies as the square of angular speed.

Answer 1

The Correct Option is A

Step 1: Relation between tension and angular velocity.
For uniform circular motion, tension provides centripetal force:
\[ T = m\omega^2 r \] With $m$ and $r$ constant,
\[ T \propto \omega^2 \]

Step 2: Use the given condition.
Tension is made $4$ times:
\[ \frac{T'}{T} = \left(\frac{\omega'}{\omega}\right)^2 = 4 \Rightarrow \omega' = 2\omega \]

Step 3: Convert initial angular speed to cycle/s.
Given $\omega = 10$ cycle/min:
\[ \omega = \frac{10}{60} = \frac{1}{6}\ \text{cycle/s} \]

Step 4: Find the new angular speed.
\[ \omega' = 2 \times \frac{1}{6} = \frac{1}{3}\ \text{cycle/s} \]

Step 5: Conclusion.
The required angular velocity is $\dfrac{1}{3}$ cycle/s.