Question

A 25 kg child stands 2.5 m from the center of a frictionless merry‐go‐round, which has a 200 kg*m^2 moment of inertia and is spinning with one revolution every two seconds. The child then moves inward to a radius of 1.5 m. a) What is the initial angular velocity of the merry‐go‐round?
b) What is the new angular velocity of the merry‐go‐round, after the child moves?
c) By how much did the kinetic energy of the merry‐go‐round increase? Where did this energy come from?

Answer 1

Answer:

Explanation:

a ) Time period  T = 2 s

Angular velocity ω = 2π / T

=  2π / 2 = 3.14 rad /s

Initial moment of inertia I₁ = 200 + mr²

= 200 + 25 x 2.5²

=356.25

Final moment of inertia

I₂ = 200 + 25 X 1.5 X 1.5

= 256.25

b ) We apply law of conservation of momentum

I₁ X ω₁ =  I₂ X ω₂

ω₂ = I₁ X ω₁ / I₂

Putting the values

$w_2=\frac{356.25\times3.14}{256.25}$

ω₂ = 4.365 rad s⁻¹

c ) Increase in rotational kinetic energy

=1/2 I₂ X ω₂² -  1/2 I₁ X ω₁²

.5 X 256.25 X 4.365² - .5 X 356.25 X 3.14²

= 684.95 J

This energy comes from work done against the centripetal pseudo -force.