Moment of Inertia and anemometer

AI Thread Summary
The discussion centers on calculating the moment of inertia for an anemometer consisting of four metal cups mounted on rods. Participants clarify that the moment of inertia is calculated using the formula I = mR^2, where R is the distance from the axis of rotation. Confusion arises regarding the correct value of R, with some suggesting it could be L or L/2 depending on the configuration of the rods. Ultimately, it is confirmed that using R = L/2 yields the correct moment of inertia calculation. The exchange highlights the importance of accurately defining the geometry of the system in physics problems.
masamune
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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
 

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masamune said:
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.
 
Doc Al said:
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.
 
masamune said:
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.
 
Doc Al said:
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.
 
Last edited:
masamune said:
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.
 
Yup! That worked. Thanks a lot for your patience
 
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