# Homework Help: Physical pendulum: frequency of rotated square

1. Jul 5, 2010

### emr13

1. A square object of mass m is constructed of four identical uniform thin sticks, each of length L, attached together. This object is hung on a hook at its upper corner. If it is rotated slightly and then released, at what frequency will it swing back and forth?

2. Relevant equations

frequency of a physical pendulum:

parallel-axis theorem:

I=Icm+MD2

3. The attempt at a solution

Each thin stick has an I of (1/12)ML2. There are four of them, and the distance from the center of mass of each rod to the center of the square is L/2 so the I for the square (using parallel-axis theorem) is I=(4)((1/12)mL2)+m(L/2)2

Then plug it into the frequency equation...d here is from the axis of rotation to the center of mass, so L/sqrt(2)

Simplified a little:

Is this right at all? Thanks.

2. Jul 5, 2010

### kuruman

In the expression

$$f=\frac{1}{2 \pi}\sqrt{\frac{mgd}{I}}$$

about what axis is I, the center of mass or the point of support?

3. Jul 5, 2010

### emr13

I guess it should be the point of support...

4. Jul 5, 2010

### kuruman

That's what is. You need to recalculate. Since you already have the moment of inertia about the CM, all you need to do is apply the parallel axis theorem once more.

5. Jul 5, 2010

### emr13

so Icm=(4)((1/12)mL2)+m(L/2)2

which simplifies to (7/12)mL2.

Then the "new" I would be

I=(7/12)mL2 + m(L/sqrt(2))2 ?

6. Jul 6, 2010

### kuruman

Almost. Read the problem. If the square object has mass m what is the mass of one of the arms?

7. Jul 6, 2010

### emr13

m/4...which m is that though? My first thought is that it's the first one in the "old" inertia equation...is that the only one I would need to switch? Because the others reflect the mass of the entire object, and I'm pretty sure they're correct.

8. Jul 6, 2010

### kuruman

Pretend that each arm has mass m,i.e. symbol "m" stands for the "mass of one side", then find the moment of inertia of the square about its CM correctly. (It is not (7/12)mL2)
Use the parallel axis theorem consistently (i.e. pretend that the square's mass is 4m) to find Isupport. Having done that, divide all occurrences of m by 4 to take into account the fact that each side does not have mass m but m/4.

9. Jul 6, 2010

### ehild

It would be a bit simpler to derive the moment of inertia directly from its definition.

$$I = \int{\rho s^2 dl}$$

where q is the density of the rods, and s is the distance of a small piece of length dl from the axis and the integration goes around the rectengular frame.

ehild

10. Jul 6, 2010

### emr13

I am really confused now. I thought in the parallel axis theorem, m was supposed to be the entire mass of the object...I understand that each stick is m/4, but after that, you've lost me.

And my professor isn't big on calculus, so I don't think I'm supposed to integrate anything...

11. Jul 6, 2010

### kuruman

Then follow my suggestion. It completes what you started, but correctly, so it should be easy for you to finish.

12. Jul 6, 2010

### emr13

Okay, trying again:

Icm=4(1/12)(m/4)(L2)+(m/4)(L/2)2
Icm=(7/48)mL2

then

Isupport=(7/48)mL2+4m(L/sqrt(2))2
Isupport=(103/48)mL2

then divide m by 4 so

Isupport=(103/192)mL2

But that seems weird...so another way:

Icm=4(1/12)(m/4)(L2)+(4m)(L/2)2
Icm=(13/12)mL2

then

Isupport=(13/12)mL2+4m(L/sqrt(2))2
Isupport=(37/12)mL2

then divide m by 4 so

Isupport=(37/48)mL2

Are either of those correct?

13. Jul 6, 2010

### kuruman

Where do you get the 7? It seems you think that (L/2)2=L2/2 instead of L2/4.

14. Jul 6, 2010

### emr13

Icm=(4)(1/12)(m/4)(L2)+(m/4)(L/2)2
=(1/3)(m/4)(L2)+(m/4)(L2)(1/4)
=(1/12)mL2+(1/16)mL2
=(16/192)mL2+(12/192)mL2
=(28/192)mL2
=(7/48)mL2

15. Jul 6, 2010

### kuruman

The factor of 4 must multiply both terms in the sum and it should be
ICM=(4)*[(1/12)(m/4)(L2)+(m/4)(L/2)2]
because there are 4 sides each of moment of inertia
(1/12)(m/4)(L2)+(m/4)(L/2)2

16. Jul 6, 2010

### emr13

So I=(1/3)mL^2

Then Isupport is (partially simplified)

(1/3)mL^2 + 2mL^2
=(7/3)mL^2

Then if I divide that by 4, it's (7/12)mL^2...

17. Jul 6, 2010

### kuruman

That is correct.

I don't think so. Please show me exactly how you got this result.
Why would you do that? One more time, when you write

ICM=4*[(1/12)(m/4)(L2)+(m/4)(L/2)2],

the symbol "m" stands for the mass of the four sides combined. Each side has mass m/4, that's why m/4 appears in the above expression. The overall factor of 4 is there because you have four sides to the square. In other words,

ICM = 4*(Iof one side about the center of the square).

Last edited: Jul 6, 2010
18. Jul 6, 2010

### emr13

I don't know why I divided by four...thanks for being so patient with me.

The Isupport I did before is wrong (found a mistake in my work) so

Trying again:
Isupport= (1/3mL^2) + m(L/sqrt(2))^2
= 5/6mL^2

19. Jul 6, 2010

### kuruman

20. Jul 6, 2010

### emr13

Okay. and then I just plug that into the frequency equation:

21. Jul 7, 2010

### kuruman

I believe you are done.

22. Jul 7, 2010

### emr13

Okay. Thank you so much for all your help!

23. Jan 9, 2011

### sayebms

I={(1/12)mL^2}+{m(L^2)/4}=(mL^2)/3
then you should multiply it by 4 so:

I=(4/3)*(mL^2) this is for the center of mass of the square and for the support you should just add 4m{(L/2)sqrt(2)}^2