Angular Momentum of a flywheel problem

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Homework Help Overview

The problem involves calculating the average torque acting on a flywheel and determining the angle through which it turns, given its rotational inertia and changes in angular momentum over a specified time period.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss finding angular acceleration and its relationship to angular velocity and displacement using kinematic equations. There is a focus on the connection between torque and angular momentum.

Discussion Status

Some participants have provided hints regarding the use of kinematics to find angular velocity and displacement. There is an ongoing exploration of the correct approach, with no explicit consensus reached on the solution.

Contextual Notes

Participants express uncertainty about the calculations and the steps needed to arrive at the correct answer, indicating potential gaps in information or understanding of the concepts involved.

Seraph404
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Homework Statement



The angular momentum of a flywheel having a rotational inertia of 0.140 kg m[tex]^{2}[/tex] about its central axis decreases from 3.00 to 0.800 kg m[tex]^{2}[/tex]/s in 1.50 s. a) What is the magnitude of the average torque acting on the flywheel about its central axis during this period? b) Assuming a constant angular acceleration, through what angle does the flywheel turn?


Homework Equations



[tex]\tau[/tex] = I[tex]\alpha[/tex] ?


The Attempt at a Solution



a) Part a is easy. Torque equals final momentum minus initial momentum over the time. The answer is -1.47 N

b) Part b is what I need help with. As a hint, what equations should I look at or try to combine?
 
Last edited:
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Find the angular acceleration, then use kinematics to find the angle.
 
Well, I thought of that, but then how do I find angular velocity?
 
Seraph404 said:
Well, I thought of that, but then how do I find angular velocity?
Using kinematics as Doc Al said. You know that,

[tex]\alpha = \frac{d^2\theta}{dt} = \frac{d\omega}{dt}[/tex]

Can you take the next step?
 
Uh.. I'm still not getting 20.4 rad, for some reason.
 
Last edited:
Seraph404 said:
Uh.. I'm still not getting 20.4 rad, for some reason.
Perhaps if you showed us your working we could help you out.
 
Answers are not important if you know the correct approach.

I hope my great colleagues above would agree.

Now for the help part..

Torque is the rate of change of angular momentum.
Next, torque= M*I * alpha
alpha=omega/t or omega= alpha*t
next, omega * time= angular displacement.
finally... angular displcement is what you want .rest is easy.
 
Last edited:

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