Rotational Mechanics: Calculating Torque as a Function of Time

AI Thread Summary
The discussion revolves around calculating the torque exerted by a hamster on an exercise wheel, given a specific angular velocity function. The user initially derived the torque using the relationship τ = Iα and differentiated the angular velocity to find angular acceleration. The resulting torque function was τ(t) = 2250t² + 4000/r. However, a correction was noted regarding the application of Newton's second law for rotation versus translation, emphasizing the importance of using τ = Iα correctly. The conversation highlights the need for clarity in applying rotational dynamics principles.
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:
\omega(t)= 3.0 rads/s + (8.0 rad/s^{}2)t + (1.5 rad/s^{}4)t^{}3. Calculate the torque on the wheel as a function of time. Assume that the moment of inertia is 500 kg*m^{}2 and is constant."

\tau=Fr F=m\alpha and I=mr^{}2

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

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

Is this correctly done?
 
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PoofyHair said:
F=m\alpha
One error is mixing up Newton's 2nd law for rotation and translation.
For translation:
F = m a
For rotation:
\tau = I \alpha
 
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:
\omega(t)= 3.0 rads/s + (8.0 rad/s^{}2)t + (1.5 rad/s^{}4)t^{}3. Calculate the torque on the wheel as a function of time. Assume that the moment of inertia is 500 kg*m^{}2 and is constant."

\tau=Fr F=m\alpha and I=mr^{}2

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

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