(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?
Answers
i.e. angular momentum= Iw = constant .
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Given,
w1 = 40 rev/min
Let initial moment of inertia of child is I .
Then, I1 = I
I2 = (2/5)I
Use, law of conservation of angular momentum,
I1w1 = I1w2
w2 = (I1/I2)w1 = 5/2 × 40 = 100
w2 = 100 rev/min
(B) final kinetic energy of rotation/initial kinetic energy of rotation = {(1/2)I1w1²}/{(1/2)I2w2²} = (2/5I)(100)²/(I)×(40)²
= 2/5 × 10000/1600 = 5/2
Hence, final kinetic energy of rotation = 5/2( initial kinetic energy of rotation) it means final KE of rotation of the child is greater then initial . it is obtained from muscular energy of the child when he fold back his hands.
Answer:
(a) 100 rev/min
Initial angular velocity, ω1= 40 rev/min
Final angular velocity = ω2
The moment of inertia of the boy with stretched hands = I1
The moment of inertia of the boy with folded hands = I2
The two moments of inertia are related as:
I2 = (2/5) I1
Since no external force acts on the boy, the angular momentum L is a constant.
Hence, for the two situations, we can write:
I2ω2 = I1 ω1
ω2 = (I1/I2) ω1
= [ I1 / (2/5)I1 ] × 40 = (5/2) × 40 = 100 rev/min
(b) Final K.E. = 2.5 Initial K.E.
Final kinetic rotation, EF = (1/2) I2 ω22
Initial kinetic rotation, EI = (1/2) I1 ω12
EF / EI = (1/2) I2 ω22 / (1/2) I1 ω12
= (2/5) I1 (100)2 / I1 (40)2
= 2.5
∴ EF = 2.5 E1
The increase in the rotational kinetic energy is attributed to the internal energy of the boy.