Physical pendulum: frequency of rotated square

[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?

Homework Equations

frequency of a physical pendulum:

parallel-axis theorem:

I=I_cm+MD²

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.

 

[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?

 

[emr13]

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

 

[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.

 

[emr13]

so I_cm=(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 ?

 

[kuruman]

so I_cm=(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 ?

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

 

[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.

 

[kuruman]

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.

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 I_support. 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.

 

[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

 

[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...

 

[kuruman]

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...

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

 

[emr13]

Okay, trying again:

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

then

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

then divide m by 4 so

I_support=(103/192)mL²

But that seems weird...so another way:

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

then

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

then divide m by 4 so

I_support=(37/48)mL²

Are either of those correct?

 

[kuruman]

Okay, trying again:
I_cm=4(1/12)(m/4)(L²)+(m/4)(L/2)²
I_cm=(7/48)mL²

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

 

[emr13]

Where do you get the 7?

I_cm=(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

 

[kuruman]

I_cm=(4)(1/12)(m/4)(L²)+(m/4)(L/2)²

The factor of 4 must multiply both terms in the sum and it should be
I_CM=(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

 

[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...

 

[kuruman]

So I=(1/3)mL^2

That is correct.

Then Isupport is (partially simplified)
(1/3)mL^2 + 2mL^2
=(7/3)mL^2

I don't think so. Please show me exactly how you got this result.

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

Why would you do that? One more time, when you write

I_CM=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,

I_CM = 4*(I_{of one side about the center of the square}).

 

[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

 

[kuruman]

That looks about right.

 

[emr13]

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

 

[kuruman]

I believe you are done.

 

[emr13]

Okay. Thank you so much for all your help!

 

[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