- #1

- 2,261

- 996

Given the pendulum setup below:

Details:

##m## is the mass of the bob

## r ## is the instantaneous length of the spring

## \theta ## is the angle the bob makes w.r.t vertical

## I ## is the bobs mass moment of inertia about pivot

## l_o## is the free length of the spring.

##k## is the spring constant

The spring is assumed not to deflect laterally and is ideally massless. It is elevated to the angle ## \theta## at its free length without stretching and released. Summing torques about the piviot ## \circlearrowright ^+## I arrive at:

$$ r m g \sin \theta = I \ddot \theta \tag{1}$$

Next shift frame of reference to the mass and apply Newtons Second ( non-inertial frame) in the direction of ##r## with ##\searrow ^+##

$$ m \ddot r = mg \cos \theta -k\left( r - l_o \right) + mr {\dot \theta}^2 \tag{2}$$

I've never really gone further with solving a system like this. What I was thinking of doing was solving ##(2)## first treating ## \theta ## and ## \dot \theta ## as parameters such that ##(2)## becomes:

$$ \ddot r + \left( \frac{k}{m} - {\dot \theta}^2 \right) r = g\cos \theta +\frac{k l_o}{m}$$

where I solve for ##r## as a second order ODE with constant coefficients, in other words solve##(2)## as:

$$ \ddot r + \beta r = \kappa $$

Then use that result to numerically solve ##(1)## for ##\ddot \theta## with a small time step where I assume it is constant over that small interval.

Is that nonsense?

EDIT:

##(1)## should be:

$$ r^2 \ddot \theta + 2 r \dot r \dot \theta + rg \sin \theta = 0 \tag{1}$$

Details:

##m## is the mass of the bob

## r ## is the instantaneous length of the spring

## \theta ## is the angle the bob makes w.r.t vertical

## I ## is the bobs mass moment of inertia about pivot

## l_o## is the free length of the spring.

##k## is the spring constant

The spring is assumed not to deflect laterally and is ideally massless. It is elevated to the angle ## \theta## at its free length without stretching and released. Summing torques about the piviot ## \circlearrowright ^+## I arrive at:

$$ r m g \sin \theta = I \ddot \theta \tag{1}$$

Next shift frame of reference to the mass and apply Newtons Second ( non-inertial frame) in the direction of ##r## with ##\searrow ^+##

$$ m \ddot r = mg \cos \theta -k\left( r - l_o \right) + mr {\dot \theta}^2 \tag{2}$$

I've never really gone further with solving a system like this. What I was thinking of doing was solving ##(2)## first treating ## \theta ## and ## \dot \theta ## as parameters such that ##(2)## becomes:

$$ \ddot r + \left( \frac{k}{m} - {\dot \theta}^2 \right) r = g\cos \theta +\frac{k l_o}{m}$$

where I solve for ##r## as a second order ODE with constant coefficients, in other words solve##(2)## as:

$$ \ddot r + \beta r = \kappa $$

Then use that result to numerically solve ##(1)## for ##\ddot \theta## with a small time step where I assume it is constant over that small interval.

Is that nonsense?

EDIT:

##(1)## should be:

$$ r^2 \ddot \theta + 2 r \dot r \dot \theta + rg \sin \theta = 0 \tag{1}$$

Last edited: