Asked by Anonymous
A wheel has a rotational inertia of 3.0 kg⋅m2. At the instant the wheel has a counterclockwise angular velocity of 10.0 rad/s, an average counterclockwise torque of 6.0 N⋅m is applied, and continues for 4.0 s. What is the magnitude of the change in angular momentum of the wheel?
Answers
Answered by
Anonymous
I = 3
omega = 10 + alpha * t
torque = I * alpha = 3 * alpha
so
6 = 3 * alpha
alpha = 2 rad/s^2
so
omega = 10 + 2 * 4= 18 rad/s
change in I omega = 3 (18 - 10) = 24 kg m^2/sec
omega = 10 + alpha * t
torque = I * alpha = 3 * alpha
so
6 = 3 * alpha
alpha = 2 rad/s^2
so
omega = 10 + 2 * 4= 18 rad/s
change in I omega = 3 (18 - 10) = 24 kg m^2/sec
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.