Physical Pendulum and moment of inertia

In summary, the physics student can find the acceleration due to gravity without measuring the moment of inertia of the pendulum by finding the location of the center of gravity, which can be done experimentally. This is because for a physical pendulum, the ratio of the moment of inertia to the distance from the center of gravity is constant.
  • #1
kudoushinichi88
129
2
A physics student measures the period of a physical pendulum about one pivot point to be T. Then he finds another pivot point on the opposite side of the center of mass that gives the same period. The two points are separated by a distance L. Can he find the acceleration due to gravity, g, without measuring the moment of inertia of the pendulum? Why?

My answer:
For a physical pendulum, the angular speed is

[tex]\omega=\sqrt{\frac{mgd}{I}}[/tex]

Where m is the mass of the pendulum, I is the moment of inertia at the axis of rotation and d is the distance to the center of gravity of the pendulum. So, the period of the pendulum is

[tex]T=2\pi\sqrt{\frac{I}{mgd}}[/tex]

Since the two points have the same period, then the ratio of the moment of inertia about those points to the distance between that point and the cg should be the same. ie

[tex]\frac{I_1}{d_1}=\frac{I_2}{d_2}[/tex]

Using the parallel axis theorem,

[tex]
\frac{I_{CM}+md_1^2}{d_1}=\frac{I_{CM}+md_2^2}{d_2}[/tex]

simplifying,

[tex]I_{CM}=md_1d_2[/tex]

Therefore, the student could find g without measuring the moment of inertia of the pendulum. He just needs to find the location of the center of gravity.But how do I relate L to my answer?
 
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  • #2
anyone?
 
  • #3
L is your d in parallel axis theorem.

You need to understand how the period of physical pendulum was derived in the first place.
 
  • #4
Uh. No? The L is the distance between the two points which have the same period, not the d.
 
  • #5
kudoushinichi88 said:
Since the two points have the same period, then the ratio of the moment of inertia about those points to the distance between that point and the cg should be the same. ie

[tex]\frac{I_1}{d_1}=\frac{I_2}{d_2}[/tex]

Using the parallel axis theorem,

[tex]
\frac{I_{CM}+md_1^2}{d_1}=\frac{I_{CM}+md_2^2}{d_2}[/tex]

simplifying,

[tex]I_{CM}=md_1d_2[/tex]

Therefore, the student could find g without measuring the moment of inertia of the pendulum. He just needs to find the location of the center of gravity.

My mistake. I didn't read properly. However, starting at this part i got lost. Your final result for Icm is not correct. And the best way to prove this is to assume that the physical pendulum now i question is a rod or ruler, with centre of mass at the centre.

It is now pivoted at its end. Base on your final equation, will it give me back I_cm=1/12ml^2?

And even if the last equation is correct, how are you going to calculate g?

I don't think there is enough information for you to determine g.
 
  • #6
A physical pendulum is not a rod or a ruler. It could be a potato or even a truck for that matter. My reasoning is that instead of measuring the I, he just need to find the cg. That could be done experimentally.
 
  • #7
A physical pendulum is CAN not only be a rod or a ruler. But a rod or a ruler is a physical pendulum.

A normal bob pendulum is also a physical pendulum. If you were to take any of the examples i gave you and use the physical pendulum period equation, you can find their period. So unless your final result work for a ruler, You don't even have to think about a potato.

And after you measured the position of centre of mass, what are you going to do with it?
 
  • #8
Oh yeah, sorry. My point is everything is a physical pendulum. XD once we know the position of the cg, we can measure d1 or d2 and just plug into the formula.
 
  • #9
But is your final formula correct? The Icm = m d1 d2
 
  • #10
that is what I hope to know from the other users in the forum. 8D
 
  • #11
Well, i already told you to use this on a ruler. Does it give you back a Icm of 1/12ml^2?
 
  • #12
You have a point. It does not. So I consulted my tutor. And he said that the equation from my omega for the physical pendulum is only valid for oscillations of small angles.

There's nothing wrong with my math, but I guess my derivation of I here cannot be used?
 
  • #13
[tex]T = 2\pi\sqrt{\frac{I}{mgd}}[/tex]

[tex]I = I_cm + md^2 = mk^2 + md^2[/tex] where k is the radius of gyration. Hence

[tex]T = 2\pi\sqrt{\frac{mk^2 + md^2}{mgd}} = 2\pi\sqrt{\frac{k^2 + d^2}{gd}}
[/tex]

Squaring both the sides you get

[tex]T^2 = 4\pi^2{\frac{k^2 + d^2}{gd}}[/tex]

[tex]d^2 - {\frac{gT^2}{4\pi^2}}d + k^2[/tex]

If the roots of the above quadratic are d1 and d2 then the product of the roots are d1*d2 = k^2. So L = d1 + d2 and

[tex]T = 2\pi\sqrt{d_1 + \frac{d_2}{g}}[/tex]
 
  • #14
[tex]
I = I_{cm} + md^2 = mk^2 + md^2
[/tex]

I am lost here. What is radius of gyration?
 
  • #15
Radius of gyration is defined as the distance from the axis of rotation at which if whole mass of the body were supposed to be concentrated, the moment of inertia would be the same as with the actual distribution of the mass of body into small particles.
 

1. What is a physical pendulum?

A physical pendulum is a rigid body that is able to rotate about a fixed axis. It is similar to a simple pendulum in that it consists of a mass suspended from a pivot point, but a physical pendulum has a more complex shape and can exhibit different types of motion.

2. How is the moment of inertia related to a physical pendulum?

The moment of inertia is a measure of an object's resistance to rotational motion, and it plays a crucial role in the motion of a physical pendulum. The moment of inertia depends on the mass distribution and shape of the pendulum, and it determines how easily the pendulum can rotate about its axis.

3. What factors affect the period of a physical pendulum?

The period of a physical pendulum is affected by several factors, including the length of the pendulum, the mass distribution, and the moment of inertia. Other factors that can influence the period include the amplitude of the pendulum's motion and the strength of the gravitational field.

4. How can the moment of inertia of a physical pendulum be calculated?

The moment of inertia can be calculated using the equation I = MR^2, where I is the moment of inertia, M is the mass of the pendulum, and R is the distance between the pivot point and the center of mass. For more complex shapes, the moment of inertia can be calculated using integrals or by using tables of known moments of inertia for common shapes.

5. Can the moment of inertia of a physical pendulum be changed?

Yes, the moment of inertia can be changed by altering the mass distribution or shape of the pendulum. For example, if the mass is distributed further from the pivot point, the moment of inertia will increase, making it more difficult for the pendulum to rotate. Additionally, changing the shape of the pendulum can also affect the moment of inertia.

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