Moment of Inertia

[masamune]

An anemometer for measuring wind speed consists of four metal cups, each of mass m = 146 g, mounted on the ends of essentially massless rods of length L = 0.3 m. The rods are at right angles to each other and the structure rigidly rotates at f = 12 rev/s. Treat the cups as point masses.

a) What is the moment of inertia of the anemometer about the axis of rotation?

I know moment of inertia is calculated by the sum of masses times distance to axis of rotation. I tried to substitute (L^2)/2 as my distance to the rotational axis, and I know it's 4 times each mass, (I also converted to kg) so I got 0.0827 but it's not correct. Any help would be appreciated

[Doc Al]

I know moment of inertia is calculated by the sum of masses times distance to axis of rotation.
Not true. The rotational inertia of a point mass is I = mR2, where R is the distance to the axis.
I tried to substitute (L^2)/2 as my distance to the rotational axis, and I know it's 4 times each mass, (I also converted to kg) so I got 0.0827 but it's not correct.
I don't know if R = L or L/2, since I don't know how "rod" is defined: are there two rods or four?

In this case, the total I = 4mR2.

[masamune]

Not true. The rotational inertia of a point mass is I = mR2, where R is the distance to the axis.

I don't know if R = L or L/2, since I don't know how "rod" is defined: are there two rods or four?

In this case, the total I = 4mR2.

The picture shows four rods. Yeah sorry I meant MR^2, basically I used pythagorem theorem and I found that R^2 equals 2(L/2)^2 which simplifies to (L^2)/2. I still can't solve the problem though because I don't know how to define the distance from each of the point mass to the axis of rotation.

[Doc Al]

The picture shows four rods.
Is it four rods of length L? Or two rods of length L attached in the middle? Yeah sorry I meant MR^2, basically I used pythagorem theorem and I found that R^2 equals 2(L/2)^2 which simplifies to (L^2)/2.
Why are you using the pythagorean theorem??

If there are four rods of length L, then the distance to the axis is L.

[masamune]

Is it four rods of length L? Or two rods of length L attached in the middle?

That's a really good question... :biggrin: unfortunately the question doesn't make it clear...can you explain to me though how I would go about trying each setup-both 2 rods connected at the center or 4 rods connected. Thank you.
Why are you using the pythagorean theorem??

If there are four rods of length L, then the distance to the axis is L.[/QUOTE]
I tried just squaring L and plugging it into our I equation (4ML^2) but it still registers as incorrect...sorry for all the grief doc.

[Doc Al]

I tried just squaring L and plugging it into our I equation (4ML^2) but it still registers as incorrect...sorry for all the grief doc.
The equation is I = 4MR^2. Try assuming that R = L/2.

[masamune]

Yup! That worked. Thanks a lot for your patience