Systems of Particles and Rotational Motion
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Question : 13
Total: 33
(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?
(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?
Solution:
(a) Given that, initial angular speed, ω 1 = 40 ω 2 = ?
Suppose that initial moment of inertia of the child isI 1
I 2 =
I 1
As no external torque acts in the process, therefore, according to the principle of conservation of angular momentum,
I 1 ω 1 = I 2 ω 2
⇒ ω 2 =
ω 1 =
× 40 = 100 r p m
=
= (
) (
) 2
=
× (
) 2 =
= 2.5
= 2.5 ⇒ K 2 = 2.5 K 1
The new kinetic energy is 2.5 times initial kinetic energy of rotation. The child uses his internal energy to increase his rotational kinetic energy.
Suppose that initial moment of inertia of the child is
As no external torque acts in the process, therefore, according to the principle of conservation of angular momentum,
(b)
=
The new kinetic energy is 2.5 times initial kinetic energy of rotation. The child uses his internal energy to increase his rotational kinetic energy.
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