# Rotational Inertia of a Disc and Subsequent Acceleration

This is more of a conceptual problem as opposed to anything wrong with the maths! (I think/I hope! :) )

But essentially the problem states: imagine a circular flat disc of radius $R$ and of mass $m$ that rests on a rough horizontal table. A light string is wound onto the disc and constant tension $T$ (where $T=mg$) is applied vertically downwards to the free end. There is no slip between the disc and the table. Show that the acceleration from rest at the centre of the disc is $2g/3$. Now, I understand the equations required here are:

$Torque= I\alpha$

$a= R\alpha$

And my solution is as follows:

Inertia at Edge of disc is $3mR^2/2$ and given that the torque is $mgR$, the angular acceleration from the equation above must be $2g/3R$.

From the next equation therefore, the acceleration at the centre must be $2g/3$.

However, I'm having a lot of trouble understanding conceptually why this is the case. For example, when I'm calculating the angular acceleration, I'm at a loss as to which Torque I'm using in my calculation (the one due to the tension in the string or the one due to the mass of the disc acting through its centre). In addition, I'm not exactly sure what my angular acceleration is referring to (is it the angular acceleration of the centre or the angular acceleration of the edge?). And finally, I'm not sure as to how (if it is the angular acceleration of the centre), multiplying this value by $R$ will give the acceleration at the centre because I'd assume it would give the acceleration at the edge. If this isn't the case and the angular acceleration is that of the edge, why would the torque due to the tension on the string make any difference because surely there is no moment as the force acts through the edge (wouldn't the only moment be the one that exists due to the weight of the disc?)...

I apologise for this question because I presume the answer is exceedingly simple but I just can't get my head around understanding it! I'd very much appreciate any responses!

• How do you get the acceleration equals 2g/3? You seem to multiply both sides by R, but this R does not equal the radius of the disc, rather the radius of the point you are looking at. So for the centre of the disc, shouldn't R=0? Jan 2 '15 at 4:40
• Is the disk vertical balancing on one edge, or flat laying about. Can you include some kind of sketch please? Jan 2 '15 at 16:31
• Added a diagram to help! :) Jan 2 '15 at 17:10
• Please look at physics.stackexchange.com/help/notation in order to better format your question and comments. Jan 2 '15 at 19:16

So in dynamics only the net torque about the center of mass is important. In this case the equations of motion on the center of mass are

\begin{aligned} m \dot{v} & = F \\ 0 & =N - m g - T \\ I \dot{\omega} & = R ( F-T) \end{aligned}

With the kinematic constraint that the contact point does not move $\left. v + R \omega = 0 \right\}\left. \omega=-\frac{v}{R} \right\} \left. \dot{\omega}=-\frac{\dot{v}}{R} \right\}$

The last equation is $I \left(-\frac{\dot{v}}{R}\right) = R \left( m \dot{v}- m g \right)$ and solved for:

$$\dot{v} = \frac{g}{1+\frac{I}{m R^2}}$$

Since for a disk $I=\frac{1}{2} m R^2$ the above becomes $$\boxed{\dot{v} = \frac{2}{3} g}$$ Also from the free body diagram $F = \frac{2}{3} m g$ pointing to the right. If the results came out negative then I drew the picture with the incorrect sense for $F$.

• Many thanks @ja72 for the brilliant response. Everything is becoming a lot clearer now! I guess it turns out my maths was wrong as well haha! I have just a couple of queries just to make sure I understand everything: 1) How is N=T? If one were to take a free body of the the ball and the string, the downward force is 2mg so surely N must be 2mg upwards? 2) Why is angular acceleration acting anticlockwise? Surely a clockwise acceleration is created by T and this results in the appropriate direction of Friction F at the base (i.e. to counteract the driving force due to angular acceleration)? Jan 2 '15 at 19:05
• 2) By convention positive angles (and rotations) act CCW. Since the resulting $\dot{\omega}$ is negative the actual motion is CW. Jan 2 '15 at 19:15
• 3) Would it be possible to calculate the angular acceleration and subsequent acceleration straight off using the Inertia at the edge of the disc? If Net Torque around COM = Inertia at any Point * Angular Acceleration of COM, would TR=3mR^2/2 * alpha? Since acceleration = alpha * R, is it correct to conclude this as the acceleration of the COM? Jan 2 '15 at 19:15
• Please format your math with $...$ as it is difficult to understand what you are asking. Jan 2 '15 at 19:18
• There are cross terms to consider that couple the linear with the angular motion when torque not at the COM is considered. As a rule of thumb, always look at torque at the COM. Jan 2 '15 at 19:21

I'm finding some difficulty trying to understand your question. It seems simple enough, but I hope you'll humor me. I assume there is the one force T, since you did not mention other forces. Then, if you mean the acceleration of the center of mass when you say 'and cceleration at the center', Newton's second law gives it away. T= ma ===> a = g. Now, as for which torque to use, it comes down to your choice of axis. For this setup, assuming I did not misunderstand, a simple choice of axis is around CM. Then, the angular acceleration is simply 2*g/R. I hope this helps!

• Apologies, I seem to have got the question incorrect - I'll amend the original post in a second. But with regards to which axis to choose, I thought the centre would be the best choice as well but the answer is incorrect if you use that logic for some reason! Many thanks for your response! Jan 2 '15 at 14:55

The torque is force times perpendicular distance to the axis of rotation, $TR$, and the disk is (instantaneously) rotating not about its center but about a point $P$ on the surface (no slip): The moment of inertia for this motion is (by parallel axis theorem) $$I=\frac12 mR^2 + mR^2=\frac32 mR^2$$ And the angular acceleration follows

$$\frac{d\omega}{dt}=\frac{TR}{\frac32 m R^2}=\frac{2T}{3mR}$$

If you put $T=mg$ the above simplifies to

$$\frac{d\omega}{dt}=\frac{2g}{3R}$$

Now since we are rotating about point $P$, the velocity $v$ of the center of the disk is $\omega R$ and the acceleration of the center of the disk is $a=\frac{d\omega}{dt} R$.

The acceleration of the center of the disk is therefore

$$a = \frac23 g$$

the result you were looking for.

• Many thanks for the response! This is the way I was doing it originally, but I was taking my rotating point at the centre which was why I was getting so confused!! Of course, the pivot point must be at the point of no slip instantaneously! Doh!! Thanks - I would vote up but I can't as yet :( Jan 7 '15 at 6:56