Answer the following questions:-

 (i) 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 \( \dfrac{2}{5}\) times the initial value? Assume that the turntable rotates without friction

 

(ii) 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?

Asked by Abhisek | 1 year ago |  73

1 Answer

Solution :-

(i) Given,

Initial angular velocity, ω1= 40 rev/min
let the final angular velocity = ω2
Let the boy’s moment of inertia with hands stretched out = I1
Let the boy’s moment of inertia with hands folded in = I2
We know :
I2 = (\( \dfrac{2}{5}\)) I1
As no external forces are acting on the boy, the angular momentum will be constant.
Thus,  we can write:
Iω2  =  I1 ω1
ω2 = (\( \dfrac{I_2}{I_1}\)) ω1
= [ (\( \dfrac{I_1}{ \dfrac{2}{5}} \))I1 ] × 40  =  ( \( \dfrac{5}{2}\)) × 40  =  100 rev/min

(ii)  Final kinetic energy of rotation, EF = (\( \dfrac{1}{2}\)) I2 ω22

Initial kinetic energy of rotation, EI =  (\( \dfrac{1}{2}\)) I1 ω12

\( \dfrac{E_F}{E_I}\)= (\( \dfrac{1}{2}I_2 ω_2^2 .\dfrac{1}{2}I_1 ω1_2 \)

= (\( \dfrac{2}{5}\)) I1 \( \dfrac{ (100)^2}{I_1}\) (40)2

= 2.5

∴ EF = 2.5 EI

It is clear that there is an  increase in the kinetic energy of rotation and it can be attributed to the internal energy used by the boy to fold his hands.

Answered by Pragya Singh | 1 year ago

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