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Academic session 2026-27, Chapter 6 in the current curriculum.
Welcome to the NCERT Solutions for Class 11 Physics - Chapter System of Particles and Rotational Motion. This page offers a step-by-step solution to the specific question from Exercise 1, Question 13:
(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?
. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.(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?
(a) 100 rev/min
Initial angular velocity, ω1= 40 rev/min
Final angular velocity = ω2
The moment of inertia of the boy with stretched hands = I1
The moment of inertia of the boy with folded hands = I2
The two moments of inertia are related as:
I2 = 2/5 I1
Since no external force acts on the boy, the angular momentum L is a constant.
Hence, for the two situations, we can write:
I2 ω2 = I1 ω1
ω2 = I1 ω1 / I2
= I1 / 2/5 I2 x 40 = 5/2 x 40
= 100 rev/min
(b) Final K.E. = 2.5 Initial K.E.
Final kinetic rotation,EF = 1/2 I2 ω22
Initial kinetic rotation, EI = 1/2 I1 ω12
EF / EI = 1/2 I2 ω22 / 1/2 I1 ω12
= 2/5 I1 / I1 x 1002 / 402
= 5/2 = 2.5
∴ EF = 2.5 EI
The increase in the rotational kinetic energy is attributed to the internal energy of the boy.
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
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Welcome to the NCERT Solutions for Class 11 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 13: (a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set....
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