Angular Momentum of a thin rod

  • Thread starter roeb
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  • #1
roeb
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Homework Statement



A thin rod (of width zero, but not necessarily uniform) is pivoted freely at one end about the horizontal z axis, being free to swing in the xy plane (x horizontal, y vertically down). Its mass is m, its CM (center of mass) is a distance a from the pivot, and its moment of inertia (about the z axis) is I. Write down the equation of motion d/dt(Lz) = Torque_z and assuming the motion is confined to small angles, find the period of this compound pendulum.

Homework Equations



T = 2pi/w
Lz = mvr
Torque = r cross F = I*(angular accel)


The Attempt at a Solution



Hmm, I think I am close to getting the right answer, but just can't quite get it.
I know Torque = I*alpha = I*a*d^2/dt^2(phi)
I also know that Lz = m*(a*d/dt(phi))*a = m*a^2*d/dt(phi)
Taking the deriv. of Lz seems to yield an equation where the phi's cancel out.
I also know that I = 1/3*MR^2 for a rod, but in this problem, I don't think that I need to use it.

Hmm, I need to somehow work gravity into the equation... but I can't quite figure out how.

Does anyone have any hints?
 

Answers and Replies

  • #2
millitiz
79
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Hi roeb,
similar to any classical mechanic problem, you probably want to draw a force diagram, things will be clearer from this point.
Using torque as I[tex]\alpha[/tex] is going to be a bit obscure.
If I am you, I'll use its definition. What is torque by definition?
After you figure out what torque is, then you'll figure out that it is related to gravitational force (yes, it comes in here).
After you find out the torque, try to write out the whole equation, torque = d(L)/dt.
And remember to use small angle approximation.
After doing some algebra, you'll probably see something very familiar to you, or if not, something like a periodic oscillation equation with respect to theta.
Then it shouldn't be hard from there.

Good luck
 
  • #3
roeb
107
1
Thanks for your reply.

r = (asin(phi), acos(phi))
F = (0, mg, 0)

Torque = r cross F
Torque_z = m*g*a*sin(phi)

momentum = m(acos(phi) d/dt(phi), -asin(phi) d/dt(phi)
L = r cross momentum
Lz = -m*a^2*d/dt(phi)

Using the small angle approximation I get, the simple harmonic oscillator equation:

ma^2*d^2/dt^2(phi) + mga*phi = 0

so omega^2 = mga/ma^2.

I think everything up to this point is correct. My point problem now is that according to my book omega^2 (which is the angular frequency) is mga/I.
--> Can I say that I = ma^2? I'm not sure if that is correct to say?
 
  • #4
millitiz
79
0
Almost, but not quite. I think the problem is at your momentum.
It might be easier to look it from the other way around
L = I*d(theta)/dt
Now you can see that you yourself assume that I = ma^2
Otherwise, I don't see anything incorrect
 

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