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During the launch from a board, a diver's angular speed about her center of mass changes from zero to 8.10 rad/s in 240 ms. Her rotational inertia about her center of mass is 11.6 kg·m2. During the launch, what are the magnitudes of (a) her average angular acceleration and (b) the average external torque on her from the board?

1 Answer

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Answer:

Part a)


\alpha = 33.75 rad/s^2

Part b)


\tau = 391.5 Nm

Step-by-step explanation:

Part a)

Average angular acceleration is given as rate of change in angular speed

so it is given as


\alpha = (\omega_f - \omega_i)/(\Delta t)


\alpha = (8.10 - 0)/(240* 10^(-3))


\alpha = 33.75 rad/s^2

Part b)

average external torque is given as


\tau = I\alpha

here we know that


I = 11.6 kg m^2


\tau = 11.6 * 33.75


\tau = 391.5 Nm

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