What is the Equilibrium Position of a Pivoting Meter Stick?

[Hughng]

Homework Statement

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) .

Homework Equations

The Attempt at a Solution

I attached my note.

 

[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.

 

[Hughng]

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.

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

 

[TSny]

I tried that approach but I failed.

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 $\omega$.

Am I annoying?

Not at all.

 

[Hughng]

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.

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

 

[TSny]

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

Yes.

 

[Hughng]

Yes.

Can you help me check out my steps please?

 

[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.

 

[Hughng]

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.

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

 

[TSny]

OK. Good luck with it.