Set 8 Problem number 18

• Important Note on Notation

• Problem

• Solution

• Generalized Solution

Problem

A hoop of radius 14.9 meters and mass 3.3 kilograms is rotating at 1.96 radians/second. While subject to no external forces, the hoop contracts to a radius of 5 meters without losing any of its mass.

• What will be its angular velocity at the new radius?
• What will be the ratio of its angular velocities and of its moments of inertia after contracting as compared with before?

Solution

The Impulse-Momentum Theorem for rotation states that the product of torque and time is equal to the change in the quantity I `omega.  We call I `omega the angular momentum of an object whose moment of inertia is I and angular velocity `omega.

If no external forces act on the system, then any torque exerted on one part of the system by the other will be matched by an equal and opposite torque and the total angular momentum of the system will be conserved.

The present system, which initially consists of a mass of 3.3 kg rotating at a distance of 14.9 meters from its axis of rotation,  initially has moment of inertia

• initial moment of inertia = ( 3.3 kg)( 14.9 m) ^ 2 = 732.6331 kg m ^ 2.

Initial angular momentum is therefore

• init. ang. momentum = ( 732.6331 kg m ^ 2)( 1.96 radians/second) = 1435.961 kg m ^ 2/s.

The moment of inertia of the system decreases as the radius decreases until it reaches a final moment of inertia

• final moment of inertia = ( 3.3 kg)( 5 m) ^ 2 = 82.5 kg m ^ 2.

The final angular momentum must be equal to the initial angular momentum.

• The new angular momentum is

• new angular momentum = ( 82.5 kg m)(new angular velocity).

Since this product must be equal to the initial ang. momentum 1435.961 kg m ^ 2/s, we have

• new angular velocity = 1435.961 kg m ^ 2/s / ( 82.5 kg m ^ 2) = 17.40558 radians/second.

The requested ratios are easily found.

• The angular velocity is now found to increase by a factor of 8.880401 , while the moment of inertia ratio is .1126075 .
• It is worth noting that the moment of inertia decreases by a factor of 8.880401 while the angular velocity increases by this same factor.

Generalized Solution

The angular momentum of the system is

• angular momentum = I `omega

Initially the moment of inertia is I = m * r1^2.

After the change in radius the system's final moment of inertia will be I2 = m * r2^2.

If the initial angular velocity is `omega1, the initial angular momentum will be

• initial angular momentum = I1 * `omega1

and the final angular momentum will be

• final angular momentum = I2 * `omega2.

By conservation of angular momentum the two angular momenta are equal so

• I1 * `omega1= I2 * `omega2

or

• m * r1^2 * `omega1 = m * r2^2 * `omega2.

We thus have

• `omega2 / `omega1 = (r1 / r2):  the angular velocity of an isolated system consisting of a ring of constant mass is inversely proportional to the radius.