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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? |
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Answer» Solution :(a) Suppose a child folds and back his hands then his moment of inertia are `I_(1)andI_(2)` RESPECTIVELY and angular speed are `omega_(1) and omega_(2)` respectively. `therefore I_(2)=(2)/(5)I_(1)` given `omega_(1)=40` revolution/MIN `=(40xx2pi)/(60)=(4)/(3)pi` RAD `s^(-1)` From the law of conservation of angular momentum `I_(1)omega_(1)=I_(2)omega_(2)` `therefore omega_(2)=(I_(1))/(I_(2))omega_(1)` `therefore omega_(2)=(5)/(2)omega_(1)` `=(5)/(2)xx40("rotation")/("min")=("100 rotation")/("min")` `=(10pi)/(3)` rad `s^(-1)` (b) Initial rotational kinetic energy `K_(1)=(1)/(2)I_(1)omega_(1)^(2)` and new rotational kinetic energy `K_(2)=(1)/(2)I_(2)omega_(2)^(2)` `:.(K_(2))/(K_(1))=(I_(2)omega_(2)^(2))/(I_(1)omega_(1)^(2))` `=(2)/(5)xx((10pi)/(3)xx(3)/(4pi))^(2)` `=(5)/(2)` `:. K_(2)=2.5K_(1)` Hence, child down (back) his hand, rotational kinetic energy increases by 2.5 times to when child fold his hand. The increase in the rotational kinetic energy is attributed to the internal energy of the body. |
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