A metal ball whose mass is 7.8 kilograms is rolling around a track of radius 10.4 meters, sweeping out its radial angle at the rate of .5 radians/second. It maintains its angular momentum while the radius of the track shrinks to 8.400001 meters.
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 7.8 kg rotating at a distance of 10.4 meters from its axis of rotation, initially has moment of inertia
Initial angular momentum is therefore
The moment of inertia of the system decreases as the radius decreases until it reaches a final moment of inertia
The final angular momentum must be equal to the initial angular momentum.
- new angular momentum = ( 550.3681 kg m)(new angular velocity).
Since this product must be equal to the initial ang. momentum 421.8241 kg m ^ 2/s, we have
The requested ratios are easily found.
The angular momentum of the system is
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
and the final angular momentum will be
By conservation of angular momentum the two angular momenta are equal so
or
We thus have
"