Physical Pendulum


by kudoushinichi88
Tags: pendulum, physical
kudoushinichi88
kudoushinichi88 is offline
#1
Nov21-10, 07:56 AM
P: 133
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 seperated 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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kudoushinichi88
kudoushinichi88 is offline
#2
Nov21-10, 07:12 PM
P: 133
anyone?
Delzac
Delzac is offline
#3
Nov21-10, 08:53 PM
P: 389
L is your d in parallel axis theorem.

You need to understand how the period of physical pendulum was derived in the first place.

kudoushinichi88
kudoushinichi88 is offline
#4
Nov21-10, 08:57 PM
P: 133

Physical Pendulum


Uh. No? The L is the distance between the two points which have the same period, not the d.
Delzac
Delzac is offline
#5
Nov21-10, 09:17 PM
P: 389
Quote Quote by kudoushinichi88 View Post
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.
kudoushinichi88
kudoushinichi88 is offline
#6
Nov21-10, 09:43 PM
P: 133
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.
Delzac
Delzac is offline
#7
Nov21-10, 09:52 PM
P: 389
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?
kudoushinichi88
kudoushinichi88 is offline
#8
Nov21-10, 09:56 PM
P: 133
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.
Delzac
Delzac is offline
#9
Nov21-10, 09:58 PM
P: 389
But is your final formula correct? The Icm = m d1 d2
kudoushinichi88
kudoushinichi88 is offline
#10
Nov21-10, 10:11 PM
P: 133
that is what I hope to know from the other users in the forum. 8D
Delzac
Delzac is offline
#11
Nov21-10, 10:14 PM
P: 389
Well, i already told you to use this on a ruler. Does it give you back a Icm of 1/12ml^2?
kudoushinichi88
kudoushinichi88 is offline
#12
Nov22-10, 03:40 AM
P: 133
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?
rl.bhat
rl.bhat is offline
#13
Nov22-10, 05:00 AM
HW Helper
P: 4,442
[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]
kudoushinichi88
kudoushinichi88 is offline
#14
Nov22-10, 05:24 AM
P: 133
[tex]
I = I_{cm} + md^2 = mk^2 + md^2
[/tex]
I am lost here. What is radius of gyration?
rl.bhat
rl.bhat is offline
#15
Nov22-10, 07:34 AM
HW Helper
P: 4,442
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.


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