A ball of mass 0.1 kg is whirled in a horizontal circle of radius 1 m by means of a string at an initial speed

Question

A ball of mass 0.1 kg is whirled in a horizontal circle of radius 1 m by means of a string at an initial speed of 10 rpm. Keeping the radius constant, the tension in the string is reduced to one quarter of its initial value. The new speed is:

Accepted Answer

1. **Identify the Force:** For a ball whirled in a horizontal circle, the tension ($T$) in the string provides the necessary centripetal force ($F_c$) to maintain the circular motion.
2. **Formula:** The centripetal force is given by $F_c = m \omega^2 R$, where $m$ is mass, $\omega$ is angular velocity, and $R$ is radius [Source 44, 64].
   $$ T = m \omega^2 R $$
3. **Proportionality:** Since the mass ($m$) and radius ($R$) are kept constant, the tension is directly proportional to the square of the angular speed.
   $$ T \propto \omega^2 $$
4. **Calculation:**
   - Initial Tension: $T_1 \propto \omega_1^2$
   - Final Tension: $T_2 \propto \omega_2^2$
   - Given: $T_2 = \frac{1}{4} T_1$
   $$ \frac{T_2}{T_1} = \frac{\omega_2^2}{\omega_1^2} $$
   $$ \frac{1}{4} = \left( \frac{\omega_2}{10} ight)^2 $$
   Taking the square root of both sides:
   $$ \frac{1}{2} = \frac{\omega_2}{10} $$
   $$ \omega_2 = 5 	ext{ rpm} $$

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