# Breaking Heat in Joules

## Homework Statement

Motor running, brakes applied. how much heat in joules will be dissipated?
mass 250 kg
diameter 30 cm
3000 rpm
torque 120 N-m

W=1/2 *I *w2

## The Attempt at a Solution

I keep going the route of:

(1)/(2)*250*.3*315

1*125*.3*315

125*.3*315

37.5*315

11812.5 J

Can anyone show me what I am doing wrong?

Andrew Mason
Homework Helper
You will have to explain the question. There is insufficient information here. Give us the complete wording of the question.

AM

If the power to the motor is shut off while the motor is running at 3000 rpm, and a brake is applyed, how much heat in Joules will be dissipated in the brakes in order to bring the motor to a stop.

Figuring W=1/2*I*w2

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Andrew Mason
Homework Helper
If the power to the motor is shut off while the motor is running at 3000 rpm, and a brake is applyed, how much heat in Joules will be dissipated in the brakes in order to bring the motor to a stop.

Figuring W=1/2*I*w2

Ok. We needed the information in the attached paper about the design of the motor in order to determine the moment of inertia. What is the moment of inertia of the motor armature then?

You seem to have the right idea here. The energy dissipated in the brake is equal to the kinetic energy of the rotating motor which is $\frac{1}{2}I\omega^2$. You just have to determine I and convert 3000 rpm into angular speed and then plug in the values (using appropriate units, of course).

AM

You just have to determine I and convert 3000 rpm into angular speed and then plug in the values (using appropriate units, of course).

AM

Accord to my instruction "I" is I=Mr2
M=250kg
r2=225cm = 2.25 meter

Is this correct?

Andrew Mason
Homework Helper
Accord to my instruction "I" is I=Mr2
M=250kg
r2=225cm = 2.25 meter

Is this correct?
No. You have to treat the armature as a solid cylinder. Look up the moment of inertia of a solid cylinder.

AM

No. You have to treat the armature as a solid cylinder.

AM

So I = 1/2 m r2 instead.