Verified Solution • 2025–26 Session moderate

Q6.12: NCERT Physics - Part 1 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 6.12 Solution: (a) A child stands at the centre of a turntable with his two…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly assume that rotational kinetic energy is conserved when the moment of inertia changes, leading to the wrong conclusion that the angular speed remains constant. You must recognize that while angular momentum is conserved in the absence of external torque, the internal muscular work done to pull the arms inward increases the system's kinetic energy.

(a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value? Assume that the turntable rotates without friction. (b) Show that the child's new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

Step-by-Step Verified Solution

**Answer:** (a) ω2=100\omega_2 = 100 rev/min (b) KE increases because the child does work in pulling his arms inward. **Step-by-Step Explanation:** (a) Using conservation of angular momentum:
I1ω1=I2ω2  ⟹  ω2=I1I2ω1=52×40=100 rev/minI_1\omega_1 = I_2\omega_2 \implies \omega_2 = \frac{I_1}{I_2}\omega_1 = \frac{5}{2} \times 40 = 100 \text{ rev/min}
(b) K1=12I1ω12K_1 = \frac{1}{2}I_1\omega_1^2, K2=12I2ω22=12(25I1)(52ω1)2=52K1K_2 = \frac{1}{2}I_2\omega_2^2 = \frac{1}{2}\left(\frac{2}{5}I_1\right)\left(\frac{5}{2}\omega_1\right)^2 = \frac{5}{2}K_1. The increase in KE comes from the muscular work done by the child in folding his arms [citation:34].

All Questions in this Chapter

Q6.1 Q6.2 Q6.3 Q6.4 Q6.5 Q6.6 Q6.7 Q6.8 Q6.9 Q6.10 Q6.11 Q6.12 Q6.13 Q6.14 Q6.15 Q6.16 Q6.17

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