A stationary platform P which can rotate freely about a vertical axis (figure) supports a motor M and a balance weight N. The moment of inertia of the platform with the motor and the balance weight relative to this axis is equal to I. A light frame is fixed to the motor's shaft with a uniform sphere A rotating freely with an angular velocity `omega_0` about a shaft `BB^'` coinciding with the axis `OO^'`. The moment of inertia of the sphere relative to the rotation axis is equal to `I_0`. Find
(a) the work performed by the motor in turning the shaft `BB^'` through `90^@`, through `180^@`,
(b) the moment of external forces which maintains the axis of the arrangement in the vertical position after the motor turns the shaft `BB^'` through `90^@`.

Explanation

(a) When the shaft `BB^'` is turned through `90^@` the platform must start turning with angular velocity `Omega` so that the angular momentum remains constant. Here
`(I+I_0)Omega=I_0omega_0` or, `Omega=(I_0omega_0)/(I+I_0)`
The work performed by the motor is therefore
`1/2(I+I_0)Omega^2=1/2(I_0^2omega_0^2)/(I+I_0)`
If the shaft is turned through `180^@`, angular velocity of the sphere changes sign. Thus from conservation of angular momentum,
`IOmega-I_0omega_0=I_0omega_0`
(Here `-I_0omega_0` is the complete angular momentum of the sphere i.e. we assume that the angular velocity of the sphere is just `-omega_0`). Then
`Omega=2I_0(omega_0)/(I)`
and the work done must be,
`1/2IOmega^2+1/2I_0omega_0^2-1/2I_0omega_0^2=(2I_0^2omega_0^2)/(I)`
(b) In the case (a), first part, the angular momentum vector of the sphere is precessing with angular velocity `Omega`. Thus a torque,
`I_0omega_0Omega=(I_0^2omega_0^2)/(I+I_0)` is needed.