# Need help with hw problem (torque question)

bigbuck
The question is:

A solid disk with m=60.0kg and a diameter of 48.0cm is to be turned from rest through 12.0 revolutions in 6.00s. Calculate the torque required to accomplish this.

What ive got is this:

Tau=I*al(angular acceleration)

I=.5*m*r^2 .5*60.0kg*.240m^2=1.728kg*m^2

Al = om/t (convert to om and then to al)12revs/6sec=2.00revs/sec = 12.56rads/s / 6

sec = 2.09 rads/s^2=al

1.728kg*m^2 * 2.09rads/s^2= 3.61N*m ...is this right?

This is an online physics course and I am getting a little lost at this point. The course notes are vague at best and I am just not sure if i have the irght answers or not.
ANy help would be appreciated.

## Answers and Replies

Homework Helper
The question is:

A solid disk with m=60.0kg and a diameter of 48.0cm is to be turned from rest through 12.0 revolutions in 6.00s. Calculate the torque required to accomplish this.

What ive got is this:

Tau=I*al(angular acceleration)

I=.5*m*r^2 .5*60.0kg*.240m^2=1.728kg*m^2

Al = om/t (convert to om and then to al)12revs/6sec=2.00revs/sec = 12.56rads/s / 6

sec = 2.09 rads/s^2=al
If a disk has a constant angular acceleration, a rad/s2, then in t seconds, it will have reached angular speed at rad/s and will have moved (1/2)at2 rad or (1/2)[\pi]at2 revolutions. Another way of looking at that is that if it has constant acceleration, a, it will have average speed (1/2)at over time t. Moving through 12.0 revolutions in 6 seconds means it had an angular acceleration of 12/72= 1/6 rev/s2= $\pi$/6= 0.523 rad/sec2, not 2.09.

1.728kg*m^2 * 2.09rads/s^2= 3.61N*m ...is this right?

This is an online physics course and I am getting a little lost at this point. The course notes are vague at best and I am just not sure if i have the irght answers or not.
ANy help would be appreciated.

bigbuck
still confused

I am sorry.
One of the problems is that this proffesor, while a good and helpful person, has used his own characters for the variables..ie "om" instead of w for omega, and "al" for angular acceleration. While i know he thinks hes helping, it is causing me great problems when i refer to the text or get help from others, I am getting "lost in translation" because I dont understand the language. I am very frustrated because i know this is simple.

Is "at" angular acceleration?

I dont see how to get the acceleration from the given info.

bigbuck
anyone?

robertm
It is unfortunate that your professor does that. I would suggest buying a used textbook and create a key or something so you can translate. Anyway, you are looking for angular acceleration. Heres the equation in the proper language: