Thick cylindrical ring find inertia help

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To find the moment of inertia of a thick cylindrical ring, the correct formula involves the outer and inner radii, where the outer radius is the sum of the inner radius and the thickness. The moment of inertia is calculated using I = (1/2)(M)(R^2) for the outer radius and subtracting the moment of inertia of the inner cylinder. The confusion arose from incorrectly applying the formula for a hollow cylinder instead of considering the solid cylinder's relationship to the thick ring. Clarifications were made regarding the removal of the inner cylinder to form the ring. Understanding the relationship between the dimensions and the mass is crucial for accurate calculations.
blackbyron
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Homework Statement



A thick cylindrical ring of inner radius 29.0cm and thickness 2.8cm has a mass of 10.0kg. What is the moment of inertia of this cylinder about its central axis?

Homework Equations



I = (.5)(m)(ri^2+ro^2)

The Attempt at a Solution


I tried to use hollow cylinder formula, but the answer says it's wrong.
I don't know if I use the right inertia formula. I don't know if thickness is the outer radius. I confused.

Thanks
 
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The thickness is the difference between the outer radius and the inner one. Use the formula of the solid cylinder. The thick cylindrical ring is made from a solid cylinder by removing a thinner cylinder from its inside.

ehild
 
Ah okay, yeah cause I thought the formula I mention is the one I have to use. But thanks for your help.
 
Sorry, but I think I'm lost here. How does solid cylinder relate to the thick cylindrical ring? Because when I use solid cylinder formula, it didn't work on the answer tho.

And also

"The thick cylindrical ring is made from a solid cylinder by removing a thinner cylinder from its inside."
So you're saying that the inner solid cylinder is removed right? And the thick cylindrical ring is the only thing left. Is that correct.


outer radius .29meters, thickness is .028m mass is 10 kg.

I = (1/2)(M)(R)^2
 
blackbyron said:
So you're saying that the inner solid cylinder is removed right? And the thick cylindrical ring is the only thing left. Is that correct.


outer radius .29meters, thickness is .028m mass is 10 kg.

I = (1/2)(M)(R)^2

The outer radius is R=(0.29 + 0.028) m, the inner radius is r=0.29 m. The mass of the ring is 10 kg. If the mass of the thinner cylinder is m, that of the thick one is

M=10+m.

The density of both cylinders is the same as that of the ring.

The moment of inertia is

I = 1/2 (MR2-mr2)

ehild
 
Oh I see, I forgot to add the thickness by radius to get the outer radius.

Sorry about that

Thanks for your help

Now I'm getting it.
 

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