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The question is:

A 90kg disc is floating in a frictionless vacuum. A 150N force is applied to the outer rim of the disc. The disc has a radius of 0.25m and a radius of gyration of 0.16m. What is the acceleration and angular acceleration of this disc?

To solve it I set up these equations:

\begin{equation} \mathbf{F}_{a} = \mathbf{m}\times\mathbf{a}\tag{1} \end{equation} \begin{equation} \mathbf{F}_{\alpha} = \frac{\mathbf{I}\times\mathbf{\alpha}}{0.25}\tag{2} \end{equation} \begin{equation} \mathbf{F}_{a}+\mathbf{F}_{\alpha} = 150 N\tag{3} \end{equation}

You can find I and plug it in along with m, but that still leaves 4 unknowns and only 3 equations. I need a 4th equation but I'm not sure what else is known about the problem.

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Did you use $I = m \rho^2$, where $\rho=0.16\,{\rm m}$ in your case? –  ja72 Dec 9 '12 at 13:21
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2 Answers

up vote 2 down vote accepted

First, it's important to properly understand the equations of rotational motion. Rather than $F = ma,$ the operative equation of motion is $\tau = I \alpha$, where $\tau$ is the torque, $I$ is the moment of inertia and $\alpha$ is the angular acceleration. This problem also requires you to know the definition of the radius of gyration in the form of $r_g \equiv \sqrt\frac{I}{m}$. With a proper understanding of the definitions of $\tau$ and $\alpha$, you then have all the information that you need to solve the problem.

EDIT: The above answer referred solely to the rotational acceleration of the disk. The translational acceleration of the center of mass of the disk must be worked out separately using $F = ma$.

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Could you be a little less vague? I believe I understand T = Iα. I rewrote the torque as a force applied to the rim of the disc, so that I could make the equation Fa + Fα = 150N. In this equation, Fα represents the force that is generating the torque. T = αI Fα = T / r Fα = αI / r –  charliehorse55 Jun 11 '12 at 21:38
    
I am focusing on the definitions of the quantities in the equation $\tau = I \alpha$ because you seem to be treating torque as a quantity with the same units as force, which is not correct. Also, since this problem is solely concerned with rotational motion, the equation $F = ma$ (which applies to translational motion) is not used at all. –  kleingordon Jun 11 '12 at 21:58
    
It is not solely concerned with rotational motion, the disc is not anchored at the center point. A portion of the force will cause angular acceleration and a portion will cause translational acceleration. –  charliehorse55 Jun 11 '12 at 22:12
    
I stand corrected regarding the presence of translational motion. However, you still need to decouple the rotational and translational motions. That is, you can solve for the rotational motion without any reference to the $F = ma$ law for the translational motion. And, as I said before, you need to make sure you are using the correct units for torques. They are not the same as forces. –  kleingordon Jun 11 '12 at 22:30
    
Thanks for the help. I didn't realize it was as simple as calculating the two values separately. For some reason I thought the force had to be 'shared' between the translation and the rotation. –  charliehorse55 Jun 12 '12 at 1:10
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There is only one force applied to the system $F = 150\,{\rm N}$ not two, $F_a$ and $F_\alpha$.

Then you have

  1. The total force applied equals the mass times the acceleration of the center of gravity.
  2. The total torque applied equals the mass moment of inertia (at the center of gravity) times the angular acceleration plus the gyroscopic forces (which are zero in your case).

So what is the torque applied when $F$ is located at the edge of the disk? What is the mass moment of inertia of disk of mass $m$ with radius of gyration $\rho$?

Once you answer the above questions you can proceed to solve your problem.

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