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>
The net force is 270 N
Explanation:
We can solve this problem by using Newton's second law, which states that the net force on an object is equal to the product between its mass and its acceleration:
where
F is the force
m is the mass
a is the acceleration
In this problem, we have
m = 90.0 kg
Substituting, we find the net force on the object:
Learn more about Newton's second law:
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