A rope exerts a 280 N force while pulling an 80 Kg skier upward along a hill inclined at 12o. The rope pulls parallel to the hil
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(a) 2.75 rev/min
The moment of inertia of the rod rotating about its center is:
where
is its mass
L = 0.450 m is its length
Substituting,
The moment of inertia of the two rings at the beginning is
where
m = 0.200 kg is the mass of each ring
is their distance from the center of the rod
Substituting,
So the total moment of inertia at the beginning is
The initial angular velocity of the system is
The angular momentum must be conserved, so we can write:
(1)
where is the moment of inertia when the rings reach the end of the rod; in this case, the distance of the ring from the center is
so the moment of inertia of the rings is
and the total moment of inertia is
Substituting into (1), we find the final angular speed:
(b) 103.0 rev/min
When the rings leave the rod, the total moment of inertia is just equal to the moment of inertia of the rod, so:
So using again equation of conservation of the angular momentum:
We find the new final angular speed: