Rotational Mechanics: Calculating Torque as a Function of Time

[PoofyHair]

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?

 

[Doc Al]

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

 

[PoofyHair]

Ok, thank you very much.

 

[Marco_84]

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