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  • (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) A child stands at the centre of a tu | Class 11 Physics Chapter System of Particles and Rotational Motion, System of Particles and Rotational Motion NCERT Solutions

Q13.

(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:

I=  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 ω/ I2

= I1 / 2/5 I 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, E=  1/2 I1 ω12

EF / EI   =  1/2 I2 ω22   / 1/2 I1 ω12

= 2/5 I1 / Ix 1002 / 402

= 5/2 = 2.5

∴ E= 2.5 EI 

The increase in the rotational kinetic energy is attributed to the internal energy of the boy.

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Important Questions & Answers

Why is this answer important for exams?

This question is important because it tests key concepts from the NCERT syllabus and is frequently asked in CBSE exams.

Which NCERT concept is used in this question?

This question is based on core NCERT concepts explained in the chapter and should be revised thoroughly before exams.

What is the correct answer to: (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 =...

How do you solve (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?

Step-by-step explanation:
• (a) 100 rev/min

• Initial angular velocity, ω1= 40 rev/min

• Final angular velocity = ω2

What common mistakes should be avoided in this question?

Students often lose marks by skipping steps, writing incomplete explanations, or misunderstanding keywords used in the question.

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Baba bakxhod
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