The Oscillation Period of A Rod Pivoted at One End

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Homework Statement



The figure shows a 230 g uniform rod pivoted at one end. The other end is attached to a horizontal spring. The spring is neither stretched nor compressed when the rod hangs straight down. K=3.0N/m and the length of the rod is 0.20m.

What is the rod's oscillation period? You can assume that the rod's angle from vertical is always small.


Homework Equations


Restorative force=F=-kΔx
Torque=Fd=force * length of lever arm
Moment of inertia for a rod pivoted about one end: I=(1/3)mL2
Angular frequency(w)=2π/T

The Attempt at a Solution


-kΔx=I*-w2Θ
<=>
-kr2Θ=I*-w2Θ
<=>
-kr2Θ/I=-w2Θ
Substitute I into the equation, L=r in this case.
-3K/M*Θ=-w2Θ

Therfore w=sqt(3K/M)
<=>
2π/T=sqt(3K/M)
<=>
T=2π/sqt(M/3K)=1.00s

But, I don't really know if my calculations are correct.
 
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Sorry, I don't know how to do this problem. But check
-kΔx=I*-w2Θ
Looks like kg m/s^2 on the left and kg m^2/s^2 on the right.
 
If the spring were not there, you would have a simple pendulum. It does not look to me like you have taken this aspect of the problem into your formulation at all. What you have in this problem is a simple pendulum with additional stiffness added. I suggest that you take a second look at the problem (start over).
 
Dr.D said:
If the spring were not there, you would have a simple pendulum. It does not look to me like you have taken this aspect of the problem into your formulation at all. What you have in this problem is a simple pendulum with additional stiffness added. I suggest that you take a second look at the problem (start over).


Are you sure? It rotates around an axle..that's why I thought it was a physical pendulum. See: http://session.masteringphysics.com/problemAsset/1070632/3/14.CP80.jpg
 
OK, pardon the error in termnology. It is a physical pendulum, you are correct. That said, it is just a physical pendulum with a spring added to increase stiffness. Now, shall we start from the beginning?