Dissipated Heat in Joules for Braking Motor with Given Parameters

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Homework Help Overview

The discussion revolves around calculating the heat dissipated in joules when a motor, running at 3000 rpm, has its brakes applied. The motor's parameters include a mass of 250 kg, a diameter of 30 cm, and a torque of 120 N-m. Participants are exploring the relationship between kinetic energy and the moment of inertia in this context.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to apply the kinetic energy formula W=1/2 * I * ω² to find the heat dissipated. Questions arise regarding the moment of inertia and the conversion of rpm to angular speed. There is also a discussion about the correct formula for the moment of inertia of the motor armature.

Discussion Status

The discussion is ongoing, with participants providing insights into the necessary calculations and clarifications about the moment of inertia. Some guidance has been offered regarding the treatment of the armature as a solid cylinder, but no consensus has been reached on the correct values or methods to use.

Contextual Notes

Participants note that additional information about the motor's design is needed to accurately determine the moment of inertia. There are also concerns about the completeness of the original problem statement.

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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