Problem: A disk is subjected to a torque of 3.9 meter Newtons. As a result, it is observed to accelerate from 9 radians per second to 12.9 radians per second during an observation lasting 2 seconds. What is the moment of inertia of the disk?
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Solution: The information given implies an angular acceleration of 1.95 radians/second ^ 2. Newtons second law in rotational form says that `tau = I `omega , where I is the moment of inertia , `tau is the torque and `omega is the angular acceleration.
Since we know the angular acceleration and the torque, it is easy to find the moment of inertia. We obtain
I = `tau / `omega = ( 3.9 meter Newtons)/( 1.95 radians/second ^ 2) = 2 kg m ^ 2/s ^ 2/(rad/s ^ 2) = 2 kg m ^ 2.
Generalized Response: From the initial and final angular velocities and the time interval we see that the angular acceleration is
angular acceleration = `alpha = (`omegaf - `omega0) / `dt.
If this acceleration is the result of a net torque `tau, then since by Newton's Second Law we have `tau = I `alpha, we see that
I = `tau / `alpha.
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