Question
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
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