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I have to find the total energy $V$ of an inverted pendulum (rod).

The following parameters and their values are given:

Mass $m$ and length $l$ of the pendulum

$\theta$ as the angle of the pendulum from the vertical axis

Moment of inertia (rotation about the center of mass) $I = \frac{1}{3}ml^{2}$

Is this given moment of inertia really the rotation about the center of mass? I thought that $I = \frac{1}{3}ml^{2}$ is the rotation about the end of the rod.

There is a hint given that confuses me:

Remember that the mass of the pendulum is distributed along its length (not located at the center of mass).

Without the given hint I assumed a pendulum with mass $m$ at the center. This pendulum is fixed at one end via a bearing to a carriage that will balance the rod.

inverted pendulum with center of mass at l/2

Then the total energy $V$ denpending on the angle $\theta$ should be: $$ V_{Total} = V_{Potential} + V_{Kinetic} $$ with $$ V_{Pot} = mgy(\theta) = mg\frac{l}{2}\cos{\theta} $$ $$ V_{Kin} = \frac{1}{2}I\dot\theta^2 $$

I assume that this solution is not correct regarding the hint. I think $V_{Kin}$ should be correct anyway because of using the moment of inertia. But how do I calculate $V_{Pot}$ regarding the hint? Is there some sort of integration along the pendulum necessary?

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  • $\begingroup$ There seems to be a lot of confusion here. First - the kinetic energy is $\frac12 I \omega^2$ not $\frac12 I \theta^2$. Second, I agree that the moment of inertia given is that of a uniform rod rotating about its end (not its center of mass). Third, unless you know something about the initial conditions, you cannot express your answer in terms of $\theta$ - the best you can do is write it in terms of $\dot\theta$, the derivative of $\theta$ with respect to time. I don't think the "hints" are particularly helpful in this case. Any chance you can talk to your teacher about it? $\endgroup$ – Floris Apr 18 '15 at 14:58
  • $\begingroup$ Your right I forgot the derivative: $\frac{1}{2}I\dot\theta^2$. I don't need to express it in terms of $\theta$. $\dot\theta$ is totally fine. Is my solution of $V_{Pot}$ and $V_{Kin} = \frac{1}{2}I\dot\theta^2$ without the hint correct? Thanks for your help and I will ask the instructor if I won't find out how to use the hint. $\endgroup$ – evolving Apr 18 '15 at 15:19

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