# Find the period of a small oscillation

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1. Jan 20, 2016

### diredragon

1. The problem statement, all variables and given/known data
A rod of lenght $2L$ is bent at point of its middle so that the rods now created are in a upside V shape and the angle between them is $120°$. The system oscilates. Find the expression of the period of oscilation.

2. Relevant equations
3. The attempt at a solution

I thought of one thing. Since the angle is $120°$ the system oscilates the same way as a normal pendulum rod of lenght $L/2$. If that is true all that follows is the center of mass which for a rod would be $L/2$ i think. Therefore we have imagined a pendulum of lenght $L/6$. Its period is $T=2*\pi * \sqrt{\frac{L}{6g}}$. Is this correct?

2. Jan 20, 2016

### Staff: Mentor

A pendulum that is not a simple bob on a massless string does not behave in the same way. You need to look at the moment of inertia around the pivot point.

Look up: "Physical Pendulum". The Hyperphysics web site would be a good place to start.

3. Jan 21, 2016

### diredragon

Well it seems i need to find the center of mass. But that means that the center of mass for a poll of lenght $L$ is $L/2$. I can think of the center of mass being located at point $L/4$ below the point $O$. Is that correct?

4. Jan 21, 2016

### Staff: Mentor

Could be. Show how you worked it out.

5. Jan 21, 2016

### diredragon

The poll of lenght $L$ has a center of mass located at $L/2$. $(L/2)*cos60=L/4$ but that gives the period of $\pi*\sqrt{\frac{L}{g}}$. The solution in the book read $T=4*\pi* \sqrt{\frac{L}{3g}}$ How can this be? That means that the center of mass is at $4L/3$ No idea how...

6. Jan 21, 2016

### Staff: Mentor

What did you use for the moment of inertia of the pendulum about its pivot point?

7. Jan 22, 2016

### diredragon

I didnt. I should have used $I=mL^2/3$ but i dont know how to use it to get the period.

8. Jan 22, 2016

### haruspex

Do you understand how to write down and solve the differential equation for a pendulum? It involves angular acceleration and moment of inertia.
The formula for period is obtained from that.

9. Jan 22, 2016

### Staff: Mentor

Did you investigate the term "physical pendulum" as I had suggested previously? The period of such a pendulum depends upon the moment of inertia about the pivot point and the distance of the pivot point from the center of mass.

Hint: when you formulate your moment of inertia, remember that your pendulum was originally a single rod of mass m.

10. Jan 23, 2016

### diredragon

I see now that the period of a physical pendulum is $2*\pi*\sqrt{\frac{I}{mgL_{cm}}}$
I know that the moment of inertia of a rod is $mL^2/3$ but that inputed doesnt give correct answer. I need the center of mass. Im not sure what to do

11. Jan 23, 2016

### diredragon

I came up with something. The center of mass of these two rods is a rod of lenght $L/2$ staying vertical and oscilating. With above equation and inputing values of $I=mL^2/3$ $L_{cm}= L/4$ i get $T= 4*\pi*\sqrt{\frac{L}{3g}}$ correct?

12. Jan 23, 2016

### haruspex

Looks right to me.