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 th
1 answer:
Answer:
<em>0.634 rev</em>
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
<em>integrating the angular speed ω with respect to time t leaves us with</em>
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 = =<em> 0.634 rev</em>
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Answer:
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Explanation:
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I hope this answer helps.