Auniform rod of mass 3.30×10−2 kg and length 0.450 m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. two small rings, each with mass 0.200 kg , are mounted so that they can slide along the rod. they are initially held by catches at positions a distance 5.20×10−2 m on each side from the center of the rod, and the system is rotating at an angular velocity 35.0 rev/min . without otherwise changing the system, the catches are released, and the rings slide outward along the rod and fly off at the ends.
(a) what is the angular speed of the system at the instant when the rings reach the ends of the rod?
(b) what is the angular speed of the rod after the rings leave it?

(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:

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.