# Could Someone Check This Answer?

1. Jun 16, 2005

### kitz

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:

$$T=2\pi\sqrt{\displaystyle{\frac{I}{mgd}}}$$
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: $$I=MR^2$$
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:

$$T=2\pi\sqrt{\displaystyle{\frac{MR^2}{MgR}}}$$
$$T=2\pi\sqrt{\displaystyle{\frac{R}{g}}}$$

And in plugging in the values, I get:

$$2=2\pi\sqrt{\displaystyle{\frac{R}{9.8}}}$$
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!!!!

2. Jun 16, 2005

### kitz

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...)

3. Jun 16, 2005

### kitz

...There was a reply here a minute ago... ...That suddenly disappeared...

4. Jun 16, 2005

### StatusX

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.

5. Jun 17, 2005

### CarlB

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 $$md^2$$, where "m" is the mass of the object, and "d" is the distance away from the center of mass. So you get $$MR^2+MR^2 = 2MR^2$$, like you suspected.

Carl

Last edited: Jun 17, 2005