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(a) A child stands at the centre of a tu...

(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?

Text Solution

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(a) 100 rev/min (use angular momentum conservation).
(b) The new kinetic energy is 2.5 times the initial kinetic energy of rotation. The child uses his internal energy to increase his rotational kinetic energy.
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Knowledge Check

  • A child stands at the centre of a turn table with his two arms out stretched. The turntable is set rotating with an angular speed of 40 rev/min. Now, the child folds his hands back and thereby reduces his moment of inertia to (2)/(5) times the initial value. The new kinetic energy of rotation is x times the initial kinetic energy of rotation. The value of x is :

    A
    2.5
    B
    5
    C
    1
    D
    6.756

  • The ratio of angular speed of a second-hand to the hour hand of a watch is

    A
    `60:1`
    B
    `72:1`
    C
    `720:1`
    D
    `3600:1`

  • The ratio of angular speeds of minute-hand and hour hand of a watch is :

    A
    `1:12`
    B
    `6:1`
    C
    `12:1`
    D
    `1:6`

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