Problem: A hoop of radius 10.3 meters and mass 6.3 kilograms is rotating at 1.74 radians/second. While subject to no external forces, the hoop contracts to a radius of 1.5 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?
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Solution: The angular momentum of the system is the product of its moment of inertia and its angular velocity (this is analogous to linear momentum, the product of mass and velocity).. If no external forces act on the system, the angular momentum will be conserved: that is, it will not change.
The system initially has moment of inertia ( 6.3 kg)( 10.3 m) ^ 2 = 668.3 kg m ^ 2. Initial angular momentum is therefore
init. ang. Momentum = ( 668.3 kg m ^ 2)( 1.74 radians/second) = 1162 kg m ^ 2/s.
The moment of inertia of the system decreases to
final moment of inertia = ( 6.3 kg)( 1.5 m) ^ 2 = 14.17 kg m ^ 2.
The new angular momentum must be equal to the former angular momentum. The new angular momentum is
new angular momentum = ( 14.17 kg m)(new angular velocity).
Since this product must be equal to the initial ang. Momentum 1162 kg m ^ 2/s, we have
new angular velocity = 1162 kg m ^ 2/s / ( 14.17 kg m ^ 2) = 82.00 radians/second.
The requested ratios are easily found. The angular velocity is found to increase by a factor of 47.15 , while the moment of inertia ratio is .02120. It is worth noting that the moment of inertia decreases by a factor of 47.15 while the angular velocity increases by this same factor.