Rotational Mechanics: Calculating Torque as a Function of Time

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Homework Help Overview

The problem involves calculating the torque exerted by a hamster on an exercise wheel, given the angular velocity as a function of time and a constant moment of inertia. The context is within rotational mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between torque, angular velocity, and angular acceleration, with attempts to differentiate the angular velocity to find the angular acceleration. There are questions about the correctness of the approach and the application of rotational dynamics equations.

Discussion Status

Some participants have pointed out potential errors in the application of Newton's laws for rotation versus translation. There is ongoing clarification regarding the correct formulation of the equations involved.

Contextual Notes

Participants are working under the assumption that the moment of inertia is constant, and there is a mix of initial attempts and corrections regarding the relationships between torque, force, and angular quantities.

PoofyHair
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Hello,

Hopefully this is in the correct place.
I am presented with a the following problem.
"A hamster running on an exercise wheel, exterts a torque on the wheel. If the wheel has an angular velocity that can be expressed as:
[tex]\omega[/tex](t)= 3.0 rads/s + (8.0 rad/s[tex]^{}2[/tex])t + (1.5 rad/s[tex]^{}4[/tex])t[tex]^{}3[/tex]. Calculate the torque on the wheel as a function of time. Assume that the moment of inertia is 500 kg*m[tex]^{}2[/tex] and is constant."

[tex]\tau[/tex]=Fr F=m[tex]\alpha[/tex] and I=mr[tex]^{}2[/tex]

I then said that [tex]\tau[/tex]=m[tex]\alpha[/tex]r. Next I set I=mr[tex]^{}2[/tex] equal to m and plugged it into [tex]\tau[/tex]=m[tex]\alpha[/tex]r.
I got [tex]\tau[/tex]=I[tex]\alpha[/tex]/r.

After that I differentiated the angular velocity and got [tex]\alpha[/tex](t)=8.0 + 3(1.5)t[tex]^{}2[/tex]. I plugged it in [tex]\tau[/tex]=I[tex]\alpha[/tex]/r and solved. My end result is: [tex]\tau[/tex](t)=2250t[tex]^{}2[/tex] + 4000[tex]/[/tex]r.

Is this correctly done?
 
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PoofyHair said:
F=m[tex]\alpha[/tex]
One error is mixing up Newton's 2nd law for rotation and translation.
For translation:
[tex]F = m a[/tex]
For rotation:
[tex]\tau = I \alpha[/tex]
 
Ok, thank you very much.
 
PoofyHair said:
Hello,

Hopefully this is in the correct place.
I am presented with a the following problem.
"A hamster running on an exercise wheel, exterts a torque on the wheel. If the wheel has an angular velocity that can be expressed as:
[tex]\omega[/tex](t)= 3.0 rads/s + (8.0 rad/s[tex]^{}2[/tex])t + (1.5 rad/s[tex]^{}4[/tex])t[tex]^{}3[/tex]. Calculate the torque on the wheel as a function of time. Assume that the moment of inertia is 500 kg*m[tex]^{}2[/tex] and is constant."

[tex]\tau[/tex]=Fr F=m[tex]\alpha[/tex] and I=mr[tex]^{}2[/tex]

I then said that [tex]\tau[/tex]=m[tex]\alpha[/tex]r. Next I set I=mr[tex]^{}2[/tex] equal to m and plugged it into [tex]\tau[/tex]=m[tex]\alpha[/tex]r.
I got [tex]\tau[/tex]=I[tex]\alpha[/tex]/r.

After that I differentiated the angular velocity and got [tex]\alpha[/tex](t)=8.0 + 3(1.5)t[tex]^{}2[/tex]. I plugged it in [tex]\tau[/tex]=I[tex]\alpha[/tex]/r and solved. My end result is: [tex]\tau[/tex](t)=2250t[tex]^{}2[/tex] + 4000[tex]/[/tex]r.

Is this correctly done?

NO

L=IA=Id(w)/dt=I(8+9/2t^2)
A=angular acceleration
w=angular velocity
I=momentum of inertia=mr^2
 

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