"Hard Oscillations: Finding the Period

[darksyesider]

Homework Statement

A rod of length L is attached to a ball of radius L/6 A spring is attached to a wall, and to the vertical rod 1/3 down the top (opposite the ball). THe rod is displaced a small theta, and released. Find period of oscillation.

Homework Equations

T = 2pi sqrt(I/mgd)

The Attempt at a Solution

I found the moment of inertia of the system. (Im confident on this).
I have no idea where to go from here. I don't see how you can factor in the spring.

 

[Simon Bridge]

You need to relate the position of the pendulum in it's swing with the compression and extension of the spring.
This will involve making some decisions about how you want to model the spring ...

If the bob moves to the right - the spring drawn can be expected to sag ... unless it is always under tension somehow. But I suspect you are intended to assume the spring just give a torque to the pendulum that depends on angular displacement.

 

[darksyesider]

That's the part I don't know how to calculate. I know that F=-kx.
So if the pendulum moves slightly to the right (a small theta), the distance it stretches the spring is the length times the sin of theta. Length I can calculate. But how do you approximate theta??

 

[Simon Bridge]

You also know how x varies with the angle of the swing, and you know the oscillations are "small" ... and you know how to relate a linear force to a torque.

The lynchpin is that the oscillations are small ...
when ##\theta## is very small, ##\sin\theta \approx \theta,\; \cos\theta \approx 1##

 

[rude man]

That's the part I don't know how to calculate. I know that F=-kx.
So if the pendulum moves slightly to the right (a small theta), the distance it stretches the spring is the length times the sin of theta. Length I can calculate. But how do you approximate theta??

Use torque = I d^2theta/dt^2. Solve for theta.

 

[Simon Bridge]

That too - rotation version of Newton's law: $$\sum \tau = I\ddot{\theta}$$
Merry Xmas folks :)

 

[rude man]

That too - rotation version of Newton's law: $$\sum \tau = I\ddot{\theta}$$
Merry Xmas folks :)

To you too!