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Rotational inertia Problem

  • Thread starter sskk221
  • Start date
  • #1
10
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Homework Statement



Three identical balls, with masses of M, 2M, and 3M, are fastened to a massless rod of length
L as shown. The rotational inertia about the left end of the rod is:

http://img13.imageshack.us/img13/1424/1234sjy.jpg [Broken]

Homework Equations



I = MR^2

The Attempt at a Solution



I total = (3M)(0)^2 + (2M)(L/2)^2 + (M)(L)^2

I total = 3ML^2/2

It says the answer is 3ML^2/4 though. Thanks for the help
 
Last edited by a moderator:

Answers and Replies

  • #2
33
0
Hmm...

[tex]
I = mr^{2}
[/tex]

[tex]
I = (3M)0^{2} + 2M(\frac{L}{2})^{2}) + ML^{2}
[/tex]

Therefore;

[tex]
I = \frac{1}{2}ML^{2} + ML^{2}
[/tex]

[tex]
I = \frac{3}{2} ML^{2}
[/tex]

I get the same as you.

The definition of the moment of inertia is;

SUM( miri2 ) So I'm preaty certian that we are following the correct method. So perhaps it is some arithmetical mistake we are both making.

2/4 is 1/2 yes and 1/2mr2 + mr2 is mr2(1/2 + 1)
(1/2 + 1) = 3/2. Nope. I think your answer book might be flawed.

UNLESS you are meant to use the distribution of mass in a sphere for the first mass (3M). Mindyou, because they havn't given you a radius for the sphere I would assume not.

Haths
 

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