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scoundrel [369]
3 years ago
8

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?
Physics
1 answer:
Artemon [7]3 years ago
6 0

Answer:

Part a)

\alpha = 33.75 rad/s^2

Part b)

\tau = 391.5 Nm

Explanation:

Part a)

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

so it is given as

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

\alpha = \frac{8.10 - 0}{240\times 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 \times 33.75

\tau = 391.5 Nm

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