# Oscillation of a stick

1. Sep 17, 2016

### Hughng

1. The problem statement, all variables and given/known data
A meter stick is free to pivot around a position located a distance x below its top end, where 0 < x < 0.50 m(Figure 1) .

2. Relevant equations

3. The attempt at a solution
I attached my note.

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2. Sep 17, 2016

### TSny

Your expression for $I$ about the axis of rotation looks correct to me. Although I think I would have just used $I = I_{cm} + Md^2$ with $I_{cm} = \frac{1}{12} M L^2$.

It appears to me that you have a mistake in the numerator of your expression inside the square root for $\omega$. Review the general formula for $\omega$ and make sure you are interpreting the symbols correctly.

3. Sep 17, 2016

### Hughng

I tried that approach but I failed. Am I annoying?

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4. Sep 17, 2016

### TSny

It should give the correct answer. If you show your work, we can identify any mistakes. Make sure you are interpreting $d$ correctly. $d$ also occurs in the numerator of $\omega$.
Not at all.

5. Sep 17, 2016

### Hughng

I think d is the distance from the center to the pivot point which is (1/2 - x)

6. Sep 17, 2016

### TSny

Yes.

7. Sep 17, 2016

### Hughng

Can you help me check out my steps please?

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8. Sep 17, 2016

### TSny

You are setting it up correctly, but you need to be more careful with simplifying the expressions. Try it again and take your time.

Also, note that $\frac {d^2\theta}{dt^2}$ is not the correct notation for $\omega ^2$.
$\frac {d^2\theta}{dt^2}$ is the angular acceleration of the rotational motion.
But $\omega$ is the angular frequency of the simple harmonic motion; i.e., $\omega = \frac{2 \pi}{T}$, where $T$ is the period of the simple harmonic motion.

9. Sep 17, 2016

### Hughng

Yes I know that. I will take a look tomorrow again for my expression. Thanks a lot. I appreciate it. Have a good night!

10. Sep 17, 2016

### TSny

OK. Good luck with it.