Dissipated Heat in Joules for Braking Motor with Given Parameters

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The discussion focuses on calculating the heat dissipated in joules when a motor is braked while running at 3000 rpm. The key formula used is the kinetic energy equation W = 1/2 * I * ω², where I is the moment of inertia. Participants clarify that the moment of inertia should be calculated for a solid cylinder, using I = 1/2 * m * r², rather than the initial incorrect approach. The conversion of 3000 rpm to angular speed is also emphasized as crucial for accurate calculations. Ultimately, the correct methodology hinges on understanding the motor's design and applying the appropriate formulas.
Dangousity
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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




Homework Equations



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?
 
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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

Sorry about the messy paper
 

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Dangousity said:
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

Sorry about the messy paper
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
 
Andrew Mason said:
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?
 
Dangousity said:
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
 
Andrew Mason said:
No. You have to treat the armature as a solid cylinder.

AM

So I = 1/2 m r2 instead.
 
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