A small motor is mounted on the axis of a space probe with its rotor (the rotating part of the motor) parallel to the axis of the probe. Its function is to control the rotational position of the probe about the axis. The moment of inertia of the probe is [07] times that of the rotor. Initially, the probe and rotor are at rest. The motor is turned on and after some period of time, the probe is seen to have rotated by positive 32.6 degrees. Through how many revolutions has the rotor turned

0.634 rev

Explanation:

It is stated that,

The moment of inertia of the probe = 7 times the moment of inertia of the rotor

After running for some time, the probe is seen to have rotated through positive 32.6°.

According to the laws of conservation of angular momentum, the angular momentum of the probe must be equal to the angular momentum of the rotor.

==> Irωr = Ipωp ....... equ 1

integrating the angular speed ω with respect to time t leaves us with

Ir∅r = Ip∅p ...... equ 2

where,

Ir = moment of inertia of the rotor

ωr = angular speed of the rotor

Ip = moment of inertia of the rotor

ωp = angular speed of the probe

∅r = angular position  of the rotor

∅p = angular position of the probe

equ 2 can be rewritten as

Ip/Ir = ∅r/∅p

from the statement, Ip/Ir = 7

therefore,

7 = ∅r/∅p = ∅r/32.6

∅r = 7 x 32,6 = 228.2°

converting to rev =   = 0.634 rev

grav potential energy