Q. 133.7( 18 Votes )

# 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 for this increase in kinetic energy?

Answer :

A

Given

Angular speed of child, ω = 40 rev min^{-1}

Angular momentum of child = I ω

Where I is moment of inertia of child

ω is angular velocity of child

When the child folds his hand his moment of inertia decreases, but angular momentum is conserved. Therefore, angular speed increases. We need to find the new angular speed (ω’)

Let new moment of inertia be I’

And new angular velocity be ω’

Given, I’ = 2/5×I

Since Angular momentum is conserved,

I ω = I’ ω’

ω’ = I ω / I’

= 5/2 × ω

= 5/2 × 40

= 100 rev min^{-1}

= 5/3 rev s^{-1}

=1.66 rev s^{-1}

B

Kinetic energy of rotation,

KE_{rot} = 1/2 I ω^{2}

Let moment of inertia of child before folding hands be I

Given, angular velocity before folding hand, ω = 100 rev min^{-1}

∴ KE = 1/2 Iω^{2}

= 1/2 I (40)^{2}

= 800I

Let moment of inertia after folding hands be I’, Then,

I’ = 2/5 I

Angular speed after folding hands, ω’ = 10π /3

∴ KE’ = 1/2 I’ω’^{2}

= 1/2 (2I/5) × (100)^{2}

= 2000I

KE’/KE = 2000I/ 800I

= 5/2

i.e. KE’ is greater than KE. When the child folded the hands, Kinetic energy increased

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