# A rolling can of soup!

1. Nov 21, 2005

### Impathy

Disregard this post, I will reply when I can get the equations to work for me! Sorry ...

Here's a problem I'm struggling with. It says I have a can of soup with a given mass (m), height (h), and diameter (2r). It's placed at rest on the top of an incline with a given length (l) and angle (theta) to the horizontal. I need to calculate the moment of inertia (I) of the can if it takes so much time (t) to reach the bottom of the incline. Here's what I tried:
$$mgl\sin\theta = \frac{1}{2}mv^{2}+\frac{1}{2}I\omega^{2} \\ mgl\sin\theta = \frac{1}{2}m\left(\frac{l}{t}\right)^{2}+\frac{1}{2}I\omega^{2} \\ mgl\sin\theta = \frac{1}{2}m\left(\frac{l}{t}\right)^{2}+\frac{1}{2}I\left(\frac{v}{r}\right)^{2} \\ mgl\sin\theta = \frac{1}{2}m\left(\frac{l}{t}\right)^{2}+\frac{1}{2}I\left(\frac{\left(\frac{l}{t}\right)^{2}}{{r}^{2}}\right)} \\ mgl\sin\theta = \frac{1}{2}m\left(\frac{l}{t}\right)^{2}+\frac{1}{2}I\left(\frac{\left l^{2}}{{r}^{2}t^{2}}\right)} \\ mgl\sin\theta - \frac{1}{2}m\left(\frac{l}{t}\right)^{2} = \frac{1}{2}I\left(\frac{\left l^{2}}{{r}^{2}t^{2}}\right)} \\ 2\left[m\left(gl\sin\theta - \frac{1}{2}\left(\frac{l}{t}\right)^{2}\right)\right] = \frac{1}{2}I\left(\frac{\left l^{2}}{{r}^{2}t^{2}}\right)} \\ \left\frac{(2\left[m\left(gl\sin\theta - \frac{1}{2}\left(\frac{l}{t}\right)^{2}\right)\right]}{\frac{1}{2}\left(\frac{\left}\right) = I \\ \frac{2m\left(gl\sin\theta-\frac{1}{2}\frac{l^{2}}{t^{2}}\right)}{\frac{l^{2}}{r^{2}t^{2}}} = I \\$$

It looks like the units work out for the moment of inertia, but I'm getting the wrong answer. Is my algebra questionable or is there something else I'm not seeing? Thanks in advance.

Last edited: Nov 21, 2005
2. Nov 22, 2005

### Fermat

Equations (Latex source) from Impathy

$$mgl\sin\theta = \frac{1}{2}mv^{2}+\frac{1}{2}I\omega^{2}$$
$$mgl\sin\theta = \frac{1}{2}m\left(\frac{l}{t}\right)^{2}+\frac{1}{2}I\omega^{2}$$
$$mgl\sin\theta = \frac{1}{2}m\left(\frac{l}{t}\right)^{2}+\frac{1}{2}I\left(\frac{v}{r}\right)^{2}$$
$$mgl\sin\theta = \frac{1}{2}m\left(\frac{l}{t}\right)^{2}+\frac{1}{2}I\left(\frac{\left(\frac{l}{t}\right)^{2}}{{r}^{2}}\right)}$$
$$mgl\sin\theta = \frac{1}{2}m\left(\frac{l}{t}\right)^{2}+\frac{1}{2}I\left(\frac{\left l^{2}}{{r}^{2}t^{2}}\right)}$$
$$mgl\sin\theta - \frac{1}{2}m\left(\frac{l}{t}\right)^{2} = \frac{1}{2}I\left(\frac{\left l^{2}}{{r}^{2}t^{2}}\right)}$$
$$2\left[m\left(gl\sin\theta - \frac{1}{2}\left(\frac{l}{t}\right)^{2}\right)\right] = \frac{1}{2}I\left(\frac{\left l^{2}}{{r}^{2}t^{2}}\right)}$$
$$\left[\frac{(2\left[m\left(gl\sin\theta - \frac{1}{2}\left(\frac{l}{t}\right)^{2}\right)\right]}{\frac{1}{2}\left(\frac{\left}\right) = I$$
$$\frac{2m\left(gl\sin\theta-\frac{1}{2}\frac{l^{2}}{t^{2}}\right)}{\frac{l^{2}}{r^{2}t^{2}}} = I$$
It looks like the units work out for the moment of inertia, but I'm getting the wrong answer. Is my algebra questionable or is there something else I'm not seeing? Thanks in advance

Last edited: Nov 22, 2005
3. Nov 22, 2005

### Fermat

Your equations are fine, except just at the beginning!
Also a couple of arithmetic errors in the last two lines (per my first post)

You have set v = l/t, but that is the average velocity, and v, in your 1st line, is for the final velocity.

You have to put in the proper expression for the final velocity and it should all fall out.

4. Nov 22, 2005

### Impathy

Thanks for your help -- I got it. :) I just needed 2l/t instead of l/t and it worked out. :)