A bob is whirled in a horizontal plane by means of a string with an initial speed of $\omega$ rpm. The tension

Question

A bob is whirled in a horizontal plane by means of a string with an initial speed of $\omega$ rpm. The tension in the string is $T$. If speed becomes $2\omega$ while keeping the same radius, the tension in the string becomes:

Accepted Answer

1. **Identify the Principle:** For a bob whirled in a horizontal circle, the tension ($T$) in the string provides the necessary centripetal force ($F_c$).
2. **Formula:** The centripetal force is given by $F_c = m \omega^2 r$, where $m$ is the mass, $\omega$ is the angular velocity, and $r$ is the radius [Source 51, 52].
   Therefore, $T = m \omega^2 r$.
3. **Analyze the Change:**
   - Initial Tension: $T_1 = T = m \omega^2 r$.
   - Final Angular Velocity: $\omega' = 2\omega$.
   - Radius remains constant.
   - Final Tension: $T_2 = m (\omega')^2 r = m (2\omega)^2 r = m (4\omega^2) r = 4 (m \omega^2 r)$.
4. **Conclusion:** Since $m \omega^2 r = T$, the new tension $T_2 = 4T$.

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