Q1)
Assume the torque stays constant at 120 N-m. How long will it take for the unloaded armature to accelerate from zero up to full speed of 3000 rpm? Now I am given the formular Ix = I w-wo / t. All I know is that the rotational inertia is 2.81 kg-m^2.

Q2)
What will be the power of the motor when it is shut off while the motor is running at 3000 rpm and brakes applied. How much heat in joules will be dissipated in the brake in order to bring the motor to a stop?

Q3)
The motor has a bearing shaft diameter of 3.2 cm and the coefficient of friction of each bearing is 0.008. Find the torque of friction of the bearings.

Q4)
With regards to Q2 Instead of the brakes being applied, the motor with the power cut off is allowed to run until it is brought to stop by the torque of friction of the bearings. How long will the armature take to come to a stop. Assuming all other losses ( such as windage) are zero.

Solution Preview

Q1)
First, we need to know the angular acceleration (alpha).
=> Torque = I * alpha => 120=2.81*alpha => alpha = 42.7 (rad/s/s)

Now, w = w0 + alpha*t where w and w0 are angular velocities.
3000 rpm = 3000*2Pi/60=314.16 (rad/s)
=> 314.16=0+42.7*t
=>t=7.36 ...

Solution Summary

A response to 4 analytical questions about a motor's torque and circular motion.

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