During the launch from a board, a divers angulad speed about her center of mass changes from zero to 5.4 rad/s in 450 ms. Her rotational inertia about her center of mass is 12 kg times m^2. During launch, what was the magnitude of her average angular accerlertation?

During the launch, what was the magnitude of the average external torque on her from the board?

You have to use the basic definitions here:

angular acceleration= changeangularvelocity/time

torque=Inertia*angularacceleation.

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To find the magnitude of the average angular acceleration, we can use the formula:

Average Angular Acceleration (α) = Change in Angular Speed / Change in Time

Given that the change in angular speed is 5.4 rad/s and the change in time is 450 ms (which can be converted to 0.45 s), we can substitute these values into the formula:

α = 5.4 rad/s / 0.45 s = 12 rad/s^2

Therefore, the magnitude of the average angular acceleration during the launch is 12 rad/s^2.

To find the magnitude of the average external torque on the diver from the board, we can use the formula:

Torque (τ) = Moment of Inertia (I) * Angular Acceleration (α)

Given that the moment of inertia is 12 kg times m^2 and the average angular acceleration is 12 rad/s^2, we can substitute these values into the formula:

τ = 12 kg times m^2 * 12 rad/s^2 = 144 Nm

Therefore, the magnitude of the average external torque on the diver from the board during the launch is 144 Nm.