Solid pivoted pendulum attached to a spring - oscillation period?

In summary: Since the moment of inertia for a solid thin rod pivoting about its edge is ##I = \frac{1}{3}ML^2##, the equation for the period becomes:##T = 2\pi\sqrt{\frac{\frac{1}{3}ML^2}{(kL^2+\frac{MgL}{2})}} = 2\pi\sqrt{\frac{ML}{3(kL+\frac{Mg}{2})}}##
  • #1
goraemon
67
4

Homework Statement


The figure shows a 200 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. What is the rod’s oscillation period? You can assume that the rod’s angle from vertical is always small.

Homework Equations



τ=2∏/ω
ω=√(k/M) for spring
ω=√(Mgl/I) for solid pendulum

The Attempt at a Solution



If not for the spring, this problem would simply be a solid pendulum, and the solution would simply be:
τ = [itex]\frac{2∏}{√(Mgl/I)}[/itex]

And if this were just a particle mass attached to the spring, the solution would be:
τ = [itex]\frac{2∏}{√(k/M)}[/itex]

I'm having trouble figuring out what happens to the equation in this case, though. One guess is as follows:

τ = [itex]\frac{2∏}{√(k/M) + √(Mgl/I)}[/itex]

But I'm clueless as to whether this is right or how to derive a correct equation. It seems like I might have to figure out the restoring force and/or torque in this case, but isn't that just -kx and -MglΘ? If anyone could help me toward the right direction, I'd appreciate it.
 

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  • #2
Thinking about torques is a good idea.

When the rod has a small angular displacement θ from the vertical, what is the net torque acting on the rod (about the axle) expressed in terms of θ?
 
  • #3
TSny said:
Thinking about torques is a good idea.

When the rod has a small angular displacement θ from the vertical, what is the net torque acting on the rod (about the axle) expressed in terms of θ?

OK, the gravitational force is being imposed downwards on the rod's center of mass, so the torque is:

-M*g*l*sin θ (where l = the distance between the pivot and the rod's center of mass, and θ = the angle of the rod from vertical)

I think? And when θ is small, then the above can be approximated to:

-M*g*l*θ

Is that right? And where to go from here? I feel like I should combine the above somehow with ω = √(k/m) for the spring. But how?
 
  • #4
What does ##l## mean? Is this the rod's length, is it its half-length?

What about the torque from the spring?
 
  • #5
voko said:
What does ##l## mean? Is this the rod's length, is it its half-length?

What about the torque from the spring?

##l## = the distance between the pivot and the rod's center of mass, so it would be the half-length...right?

OK, about the torque from the spring...since the angle is small, isn't it just -kx*L, where x = the distance by which the spring is stretched or compressed, and L = the full length of the rod?

If the above is correct, how do I combine them into a single equation to find the oscillation period?
 
  • #6
You should express ##x## in terms of ##\theta##. Then, having the sum of torques equals, what can you do?
 
  • #7
voko said:
You should express ##x## in terms of ##\theta##. Then, having the sum of torques equals, what can you do?

OK, so then the torque imposed by the spring = ##-kL\theta##?
And the torque from gravitational force = ##-Mg(L/2)\theta##?

Adding the torques together...

Net torque = ##(-kL\theta)## + ##(-Mg(L/2)\theta)## = ##-(kL+MgL/2)\theta##

Recalling Newton's 2nd law for rotational motion...

##\alpha = τ/I##

Replacing net torque for ##τ##...

##\alpha## = [itex]\frac{-(kL+\frac{MgL}{2})\theta}{I}[/itex]Am I on the right track or way off base or somewhere in between?
 
  • #8
You are on the right track. But: does ##kL\theta## have the dimension of torque? Compare this with the earlier earlier expression.
 
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  • #9
voko said:
You are on the right track. But: does ##kL\theta## have the dimension of torque? Compare this with the earlier earlier expression.

Oh I think I see what you're saying. ##kL\theta## has units of N, which isn't right for a torque. Since ##x = Lsin \theta ≈ L\theta##, does it follow that ##kL\theta## should be replaced by ##kL^{2}\theta##?
 
  • #10
And if so, does the equation for ##\alpha## (above in post #7) change to the following?

##\alpha = \frac{-(kL^2+\frac{MgL}{2})\theta}{I}##
 
  • #11
goraemon said:
And if so, does the equation for ##\alpha## (above in post #7) change to the following?

##\alpha = \frac{-(kL^2+\frac{MgL}{2})\theta}{I}##

Yes, I think that's right.
 
  • #12
TSny said:
Yes, I think that's right.

OK, then what to do next...comparing ##\alpha## with linear acceleration ##a##...
Since ##a = \frac{-k}{m}x##, and ##\omega = \sqrt{\frac{k}{m}}##

analogously with torque...
##\alpha = \frac{-(kL^2+\frac{MgL}{2})}{I}\theta##, so ##\omega = \sqrt{\frac{(kL^2+\frac{MgL}{2})}{I}}##

meaning that the period = ##T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{I}{(kL^2+\frac{MgL}{2})}}##

And since the moment of inertia for a solid thin rod pivoting about its edge is ##I = \frac{1}{3}ML^2##, the equation for the period becomes:

##T = 2\pi\sqrt{\frac{\frac{1}{3}ML^2}{(kL^2+\frac{MgL}{2})}} = 2\pi\sqrt{\frac{ML}{3(kL+\frac{Mg}{2})}}##


Is that right? Now I just need to plug in the numbers?
 
  • #13
goraemon said:
##T = 2\pi\sqrt{\frac{\frac{1}{3}ML^2}{(kL^2+\frac{MgL}{2})}} = 2\pi\sqrt{\frac{ML}{3(kL+\frac{Mg}{2})}}##

Is that right? Now I just need to plug in the numbers?

Looks good!
 
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Related to Solid pivoted pendulum attached to a spring - oscillation period?

1. What is a solid pivoted pendulum attached to a spring?

A solid pivoted pendulum attached to a spring is a physical system consisting of a mass (the pendulum) attached to a fixed point (the pivot) and connected to a spring. When the pendulum is pulled away from its resting position and released, it will oscillate back and forth around the pivot due to the force of gravity and the restoring force of the spring.

2. What is the oscillation period of a solid pivoted pendulum attached to a spring?

The oscillation period of a solid pivoted pendulum attached to a spring is the time it takes for one full cycle or one complete back-and-forth motion. It is affected by the length of the pendulum, the strength of the spring, and the local gravitational acceleration.

3. How is the oscillation period of a solid pivoted pendulum attached to a spring calculated?

The oscillation period of a solid pivoted pendulum attached to a spring can be calculated using the equation T = 2π√(L/g), where T is the period in seconds, L is the length of the pendulum in meters, and g is the local gravitational acceleration in meters per second squared. This equation assumes small oscillations and neglects air resistance.

4. How does the amplitude affect the oscillation period of a solid pivoted pendulum attached to a spring?

The amplitude (maximum displacement from the resting position) does not affect the oscillation period of a solid pivoted pendulum attached to a spring. This is known as isochronism, meaning the period remains constant regardless of the amplitude. However, larger amplitudes will result in a larger range of motion and a longer distance traveled by the pendulum over a single cycle.

5. How does changing the mass or spring constant affect the oscillation period of a solid pivoted pendulum attached to a spring?

Changing the mass or spring constant will affect the oscillation period of a solid pivoted pendulum attached to a spring. A heavier mass or a stronger spring will result in a longer period, while a lighter mass or a weaker spring will result in a shorter period. This can be seen in the equation T = 2π√(m/k), where m is the mass in kilograms and k is the spring constant in newtons per meter.

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