The Oscillation Period of A Rod Pivoted at One End

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Homework Help Overview

The problem involves a uniform rod pivoted at one end, connected to a horizontal spring. The task is to determine the oscillation period of the rod, assuming small angles from the vertical. The parameters include the mass of the rod, the spring constant, and the length of the rod.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the relationship between the spring force and the moment of inertia, with one participant attempting to derive the oscillation period using angular frequency. Others question the formulation and suggest considering the system as a simple pendulum with added stiffness from the spring.

Discussion Status

There is an ongoing exploration of the problem, with some participants providing guidance on re-evaluating the approach. The discussion reflects differing interpretations of the system dynamics, particularly regarding the role of the spring and the nature of the pendulum.

Contextual Notes

Some participants express uncertainty about the correctness of calculations and the terminology used to describe the system, indicating potential confusion about the physical principles involved.

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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?
 

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