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Rotational moments of Inertia

  • Thread starter WillP
  • Start date
  • #1
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Pivot point on a Uniform Rod and Acceleration

I'm stuck on this question, which seems like it should be fairly simple:

A uniform rod of length 1.15 m is attached to a frictionless pivot at one end. It is released from rest from an angle theta = 21.0° above the horizontal. Find the magnitude of the initial acceleration of the rod's CM.
theres a graphic of it here:
http://capaserv.physics.mun.ca/msuphysicslib/Graphics/Gtype20/prob28_1012rod.gif

ok so what I've done so far is:
Rmg(sinθ) = Iα
Rmg(sinθ) = (1/3)mL^2α

getting the expression I = (1/3)mL^2 from my text book for a uniform rod with a pivot at its end.

working that out I get:
(0.575)(9.8)(sin21) = (1/3)(1.15)^2 α
α = 4.58 rad/s

then converting α to a with:
a = Rα
a = (.575)(4.58)
a = 2.63 m/s^2

but apparently this isn't right.
can anyone help me out?
 
Last edited by a moderator:

Answers and Replies

  • #2
8
0
nevermind... my stupid mistake should have used cos instead of sin :redface:
 

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