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# Moment of Inertia watch

1. Hi, if anyone could guide me as to what I'm doing wrong with this question it would be appreciated
"Use calculus to find the moment of inertia of a thin hollow uniform right circular cylinder of mass M, radius R, and height H about a diameter of an end circle. The cylinder is open at both ends"

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2. (Original post by KFazza)
Hi, if anyone could guide me as to what I'm doing wrong with this question it would be appreciated
"Use calculus to find the moment of inertia of a thin hollow uniform right circular cylinder of mass M, radius R, and height H about a diameter of an end circle. The cylinder is open at both ends"

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What is the moment of inertia of the small ring about the diameter? (Use the perp and parallel axis theorems)

Also where you crossed out area and put volume, it should be surface area
3. (Original post by Gome44)
What is the moment of inertia of the small ring about the diameter? (Use the perp and parallel axis theorems)

Also where you crossed out area and put volume, it should be surface area
I understand that I could find the result by considering the inertia about the vertical axis of symmetry of the cylinder, use the perpendicular axis theorem, and then use the parallel axis theorem to shift it down.

However, I'm not sure why I get a different result from trying it this way.

And yeah I know it's the surface area, I initially missed the word hollow and had a bit of a moment

Edit: my thinking is that we can find the inertia by summing the inertia of every particle which makes up the cylinder, some of the particles will be at the same distance so group these together to make things easier - once grouped it will produce the ring as above. Find the mass of the ring, and use it in the formulae along with the distance then find the resulting integral.
I'd like to understand what makes this approach incorrect

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4. (Original post by KFazza)
I understand that I could find the result by considering the inertia about the vertical axis of symmetry of the cylinder, use the perpendicular axis theorem, and then use the parallel axis theorem to shift it down.

However, I'm not sure why I get a different result from trying it this way.

And yeah I know it's the surface area, I initially missed the word hollow and had a bit of a moment of inertia

Posted from TSR Mobile
Sorry, just thought that was a missed idea for a pun. Anyway, I'm going to dinner in a bit so will look over it again when I'm back
5. (Original post by KFazza)
Hi, if anyone could guide me as to what I'm doing wrong with this question it would be appreciated
"Use calculus to find the moment of inertia of a thin hollow uniform right circular cylinder of mass M, radius R, and height H about a diameter of an end circle. The cylinder is open at both ends"

Posted from TSR Mobile
It is very difficult to help you with this without me providing a full solution (very time consuming)

look at page 11 here for a similar question with a cone
6. (Original post by TeeEm)
It is very difficult to help you with this without me providing a full solution (very time consuming)

look at page 11 here for a similar question with a cone
I understand the method, it is just I thought I found a quicker alternative and I couldn't understand why it was not working. I've figured it out now though, turns out I was not considering the distance due to the depth of the cylinder - rather silly.
Thanks though and @gome44 ,your resources are brilliant btw

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7. (Original post by KFazza)
I understand the method, it is just I thought I found a quicker alternative and I couldn't understand why it was not working. I've figured it out now though, turns out I was not considering the distance due to the depth of the cylinder - rather silly.
Thanks though and @gome44 ,your resources are brilliant btw

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8. (Original post by TeeEm)

Apologies if it's hard to follow, I didn't expect to be showing anyone

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9. (Original post by KFazza)

Apologies if it's hard to follow, I didn't expect to be showing anyone

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it is good

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