Disregard this post, I will reply when I can get the equations to work for me! Sorry ...(adsbygoogle = window.adsbygoogle || []).push({});

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:

[tex]

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 \\

[/tex]

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.

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# Homework Help: A rolling can of soup!

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