Could Someone Check This Answer?

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Homework Help Overview

The problem involves determining the radius of a thin hoop that oscillates as a physical pendulum with a specified period of 2.0 seconds. The original poster attempts to apply the formula for the period of a physical pendulum, incorporating the moment of inertia and the distance from the pivot point to the center of gravity.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the moment of inertia about the pivot point and the application of the parallel-axis theorem. There are questions about the correctness of the moment of inertia used and adjustments needed for the pivot location.

Discussion Status

Participants are actively engaging with the problem, exploring the implications of the moment of inertia and its adjustment for the pivot point. Some guidance has been offered regarding the correct moment of inertia, but no consensus has been reached on the original poster's calculations.

Contextual Notes

There is a mention of uncertainty regarding the application of the moment of inertia formula and the original poster's calculations, indicating potential gaps in understanding or assumptions made in the setup.

kitz
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Hello!

I did this problem and have gotten it wrong-- but my math is right. Maybe I'm missing something? Here's the question:

We want to support a thin hoop by a horizontal nail and have the hoop make one complete small-angle oscillation each 2.0 s. What must the hoop's radius be?


So this is a physical pendulum, and in order to find the radius, I need to know the period. The period is found by:

[tex]T=2\pi\sqrt{\displaystyle{\frac{I}{mgd}}}[/tex]
Where I is the moment of inertia, g is the acceleration due to gravity, and d is the distance from the pivot point to the center of gravity.

For a hoop, the moment of inertia is: [tex]I=MR^2[/tex]
And the center of gravity is in the center of the hoop, so d is the radius.

So with that information, I set up the following:

[tex]T=2\pi\sqrt{\displaystyle{\frac{MR^2}{MgR}}}[/tex]
[tex]T=2\pi\sqrt{\displaystyle{\frac{R}{g}}}[/tex]

And in plugging in the values, I get:

[tex]2=2\pi\sqrt{\displaystyle{\frac{R}{9.8}}}[/tex]
Solving for R, I get .9929475

Which, when I plug into my equation, I get very close to 2 for my period. However, this is incorrect...

Any ideas as to why?


Thanks!
 
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That makes sense. How would I go about finding the moment of inertia about the nail?

...This is a longshot, but would the moment of inertia be 2MR^2??
(using the parallel-axis theorem...)
 
...There was a reply here a minute ago... ...That suddenly disappeared...
 
Sorry, that was me. I wasn't sure about my answer because I haven't done this stuff in a while, but you can try it if you want. And yes, that is the correct moment of inertia around the nail.
 
Yes, you're right, you have to adjust the MI by adding the appropriate amount because you're spinning the hoop around a spot on the rim instead of its center.

If I recall, the formula for adjusting the MI is to take the formula you look up, and add to it [tex]md^2[/tex], where "m" is the mass of the object, and "d" is the distance away from the center of mass. So you get [tex]MR^2+MR^2 = 2MR^2[/tex], like you suspected.

Carl
 
Last edited:

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