- #1

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- 47/5mR^2 (not sure how to find it)

Three identical objects, each of mass M, are fastened to a massless rod of length L as shown. The rotational inertia about one end of the rod of this array is

-5/4mL^2

- again no clue how to get this

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- Thread starter grouchy
- Start date

- #1

- 73

- 0

- 47/5mR^2 (not sure how to find it)

Three identical objects, each of mass M, are fastened to a massless rod of length L as shown. The rotational inertia about one end of the rod of this array is

-5/4mL^2

- again no clue how to get this

- #2

Doc Al

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Consider the parallel axis theorem.

- #3

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For the second one I kinda guessed and got it but I dont think its the right way.

I used 1/12ml^2 for the center mass and 1/3ml^2 for the far right mass and added it together. then since there are 3 masses I multiplied by 3 and for 5/4mL^2 but I dont think thats how i"m supposed to find it.

- #4

Doc Al

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Look at it again. For the first problem, you have the rotational inertia about the center of mass but you need it about the point of attachment. Sounds like a perfect opportunity to use the parallel axis theorem!I looked at that stuff but I dont see how to use to solve these problems.

For the second problem, please attach a diagram.

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- #6

Doc Al

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- #7

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2/5mr^2 + mL^2

2/5mr^2 + 2mL^2 and got 12/5mr^2 which is wrong...honestly I got no clue how this parallel axis thereom stuff works, if u got any tips I'd appreciate it.

- #8

Doc Al

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well for the first one I tried

2/5mr^2 + mL^2

OK, what did you use for "L"? That should be the distance between the center of mass to the point of attachment (in terms of R, of course). (Draw yourself a diagram, unless one is provided--in which case, attach it.)

- #9

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I used 2R for L since thats how long it says the string is, and no diagram with the problem :(

- #10

Doc Al

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Careful: The problem says "A light string of length 2R is attachedI used 2R for L since thats how long it says the string is, and no diagram with the problem :(

Draw your own diagram! (Just for yourself.)

- #11

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I just dont see it...I drew the diagram, but I dont see what I'm trying to find...

- #12

Doc Al

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Your diagram should show a sphere hanging from a string attached to the ceiling. The string is length 2R and attaches to theI just dont see it...I drew the diagram, but I dont see what I'm trying to find...

- #13

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the center would be another distance R. so

2/5mr^2 + 3mL^2 = 17/5mr^2 which is still wrong :(

2/5mr^2 + 3mL^2 = 17/5mr^2 which is still wrong :(

- #14

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ah, since the length is 3R it would be M(3R)^2 so it would be 9MR^2 so..

2/5mr^2 + 9mr^2 = 47/5mr^2 would that be the proper way to do it?

2/5mr^2 + 9mr^2 = 47/5mr^2 would that be the proper way to do it?

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- #15

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the middle mass mr^2 = m(L/2)^2 = 1/4mL^2

the far right mass mr^2 where r would be the length to the point of rotation so mL^2

1/4mL^2 + mL^2 = 5/4mL^2

I think I did the two problems the correct way, if there is anything you see shady about my method I'd love to know. THX for your patience and help lol. I think i'm starting to understand this some. Thx again! Peace.

- #16

Doc Al

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Your latest solutions to both problems are perfectly correct. Well done!

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