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.

Answer 1

Hey mate,

◆ Answer- 100 rev/min

◆ Explaination-
# Given-
ωi = 40 rev/min
ωf = ?
If = 2/5 Ii

# Solution-
Let, Ii be M.I. of boy with outstretched hands and If be M.I. of boy with folded hands.

By law of conservation of angular momentum,
Li = Lf
Iiωi = Ifωf
ωf = (Ii/If)ωi
ωf = [Ii/(2/5)Ii]×40
ωf = 100 rev/min

Angular speed of the child is 100 rev/min.

Hope, this helps you...

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