Question

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?​

Answer 1

Answer:

Explanation:

Given, the angular speed of turntable is 40 rev/min and the reduced moment of inertia is 2/5

times the initial moment of inertia.

Let I1

be the initial moment of inertia and I2

be the final moment of inertia, then

I1=25I2

(1)

Let ω1

be the initial angular speed and ω2

be the final angular speed.

The angular moment of the system remains constant as there is no external force acting on the body hence,

I2ω2=I1ω1ω2=I2I1ω1

Substitute the values in the above equation.

ω2=I125I1×40=52×40=100 rev/min

Therefore, angular speed of the child after he folds his hands back is 100 rev/min

.

The expression for initial rotational kinetic energy K1

is,

K1=12I1ω21

The expression for initial rotational kinetic energy K2

is,

K2=12I2ω22

Thus, the ratio of the rotational kinetic energy is,

K1K2=12I1ω2112I2ω22=I1ω21I2ω22

Substitute the values in the above expression.

K1K2=52(I1I1)(40)2(100)2=12.5K2=2.5K1

From the above expression, K2=2.5K1

. Hence, there is an increase in kinetic energy, which results in increasing of the internal energy of the boy.

Answer 2

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.