Question

A merry-go-round consists of a uniform ring of mass 2M and radius R attached to a central post with four light, uniform rods, each of mass M/4 and length R. Identical twins sit opposite one another where two of the rods join the ring. The mass of each child is M. At time t = 0, the system is rotating with angular speed ω0. The twins then simultaneously pull themselves toward one another along their respective rods, until each twin is a distance R/2 from the center.

a) Calculate the moment of inertia of the merry-go-round about the central post. The mass of the post is negligible.

b) Neglecting the spatial size of each child, determine the angular speed ω of the system when the twins have reached the distance R/2 from the center.

c) Does the total mechanical energy of the system remain constant, increase, or decrease? Why? If applicable, calculate the change in the total mechanical energy.

Answer 1

Moment of inertia of ring is mr2. Moment of inertial of rod about one end is mL2/3

There is a ring of mass 2M & radius R and four rods of length R and mass M/4

Moment of inertia of merry-go-round is I = 2MR2 + 4 (M/4) R2/3

I = 7MR2 / 3 (without children)

each child of mass M are seated at circumference (at R). The moment of inertia of child is MR2. two children hence 2MR2

The moment of inertial of merry-go-round with twins is I0 = 13 MR2 / 3

B) If there is no external torque then the angular momentum remains constant.

Initial angular momentum is Ioωo = (13/3) MR2ωo

When the twins are at R/2 from centre then the moment of inertia of each child is MR2 / 4

two children hence MR2 / 2

The final moment of inertial of merry-go-round with twins is I1 = 17 MR2 / 6

Final angular momentum is I1ω1 = (17/6) MR2ω1

Initial angular momentum = final angular momentum

(13/3) MR2ωo = (17/6) MR2ω1

ω1 = (26/17) ωo

c) The total mechanical energy of the system increases. The rotational kinetic energy is given by L2 / 2I

Since L is constant and I decreases then rotational kinetic energy increases.

The twins are moved from initial position at R from centre to R/2 from centre. They have to do some work to change their position. That work will be converted into rotational KE.

Initial energy K1 = (1/2)(13/3) MR2ωo2 = (13/6) MR2ωo2 = 2.17 MR2ωo2

final energy K2 = (1/2)(17/6) MR2ω12 = (169/51) MR2ωo2 = 3.31 MR2ωo2

The increase in energy is 1.14  MR2ωo2