Question:medium

When a ceiling fan is switched off, its angular velocity falls to \((\frac{1}{3})^{\text{rd}\) while it makes 24 rotations. How many more rotations will it make before coming to rest?

Show Hint

Use same deceleration for both stages and compare rotations carefully.
Updated On: May 14, 2026
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Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
The fan decelerates due to frictional torque. Assuming a constant angular retardation (\(\alpha\)), we can use equations of rotational motion.
The square of the angular velocity is linearly related to the angular displacement.
Step 2: Key Formula or Approach:
Kinematic equation: \(\omega^2 = \omega_0^2 + 2\alpha\theta\).
Let 1 rotation = \(2\pi\) radians. Let \(n\) be the number of rotations, so \(\theta = 2\pi n\).
The formula can be written as \(\omega^2 = \omega_0^2 + K \cdot n\), where \(K\) is a constant.
Step 3: Detailed Explanation:
Let initial angular velocity be \(\omega_0\).
After \(n_1 = 24\) rotations, \(\omega_1 = \omega_0 / 3\).
Applying the formula for the first part: \[ (\omega_0/3)^2 = \omega_0^2 + K(24) \] \[ \frac{\omega_0^2}{9} - \omega_0^2 = 24K \implies -\frac{8\omega_0^2}{9} = 24K \implies K = -\frac{\omega_0^2}{27} \] Now, let it make \(n_2\) more rotations to come to rest from \(\omega_1\): \[ 0 = (\omega_0/3)^2 + K(n_2) \] \[ 0 = \frac{\omega_0^2}{9} - \left(\frac{\omega_0^2}{27}\right)n_2 \] Divide by \(\omega_0^2/9\): \[ 0 = 1 - \frac{n_2}{3} \implies \frac{n_2}{3} = 1 \implies n_2 = 3 \] Alternatively, for the whole journey from \(\omega_0\) to 0: \[ 0 = \omega_0^2 + K(24 + n_2) \implies \omega_0^2 = \left(\frac{\omega_0^2}{27}\right)(24+n_2) \implies 27 = 24 + n_2 \implies n_2 = 3 \] Step 4: Final Answer:
The fan will make 3 more rotations before coming to rest.
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