Rotational Mechanics

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:
[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?

Doc Al

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]

PoofyHair

Ok, thank you very much.

Marco_84

NO

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