Final Angular Speed: 0.5m, 0.25kg, 23rad/s, 0.33Nm

[wholf09]

I am trying to do a pratice test and I am stuck on this problem.
a disk of radius r=0.5m,mass m=0.25kg has an initial angular speed of w=23rad/s.assume a torque of 0.33Nm is applied against the rotation for an angular distance of 20 rad. what is the final angular speed?
thanks for any help.

 

[Tide]

What have you tried so far? Do you know how torque relates to angular acceleration?

 

[wholf09]

distance?

I am getting thrown off by the 20 rad distance. I thought you just find the two torques and add them ,one being neg and the other pos . I don't know what to do with the distance? I am useing the formula t=mr^2*tnet/I thanks

 

[Tide]

Two torques? I only see one torque in your statement of the problem.

You should recognize that the torque is related to angular acceleration:

$$\tau = - I \alpha$$

where $\alpha$ is the angular acceleration and I use the negative sign to indicate the torque "is applied against the rotation."

It follows that $\omega = \omega_0 - (\tau/I) t$ and

$$\theta = \theta_0 + \omega_0 t + \frac {1}{2} \frac {\tau}{I} t^2$$

and you should be able to take it from there.