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a flywheel of mass 5 kg has a radius of 0.2m. it is making 240 rpm. the aveerage torque necessary to bring it to rest in 20 s is _____

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Final answer:

The average torque necessary to bring the flywheel to rest in 20 s is approximately -0.12π N*m.

Step-by-step explanation:

To find the average torque necessary to bring the flywheel to rest, we can use the equation:

T = (I Angular acceleration) / t

Where T is the torque, I is the moment of inertia, Angular acceleration is the rate at which the angular velocity changes, and t is the time taken to bring the flywheel to rest.

First, let's find the moment of inertia using the formula:

I = 0.5 * mass * radius^2

Where mass is the mass of the flywheel and radius is the radius of the flywheel.

Substituting the given values, we have:

I = 0.5 * 5 kg * (0.2 m)^2 = 0.1 kg*m^2

Next, let's find the angular acceleration:

Angular acceleration = (final angular velocity - initial angular velocity) / t

Substituting the given values, we have:

Angular acceleration = (0 - 2π * 240rpm) / (20 s)

Angular acceleration = -24π rad/s^2

Finally, we can calculate the torque:

T = (0.1 kg*m^2 * -24π rad/s^2) / (20 s) = -0.12π N*m

Therefore, the average torque necessary to bring the flywheel to rest in 20 s is approximately -0.12π N*m.

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