# Spring Pendulum - Lagrangian Mechanics

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1. May 22, 2016

### bigguccisosa

1. The problem statement, all variables and given/known data

2. Relevant equations
Euler-Lagrange Equation
$\frac{\partial{L}}{\partial{q}} - \frac{d}{dt}\frac{\partial{L}}{\partial{\dot{q}}} = 0$
$L = T - V$

3. The attempt at a solution
a. The potential energy V is the potential energy from the spring and the gravitational potential energy. The kinetic energy is the energy in the radial direction and in the theta direction.
$$L = \frac{1}{2}m(\dot{r}^2+\dot{\theta}^2(r_0+r)^2) + mg(r_0+r)cos\theta - \frac{1}{2}kr^2$$

b. Use one Euler-Lagrange equation for r and one for theta and I get: $$m\dot{\theta}^2 + mgcos\theta - kr = \frac{d}{dt}(m\dot{r})$$ $$-mg(r_0+r)sin\theta = \frac{d}{dt}(m\dot{\theta}(r_0+r)^2)$$
Which are the F=ma equation for the radial direction and the Torque = (d/dt) Angular momentum equation.

c. Confused here, if angular position and velocity are to be fixed, I assume they mean $\theta$ and $\dot{\theta}$, do I consider them both to be zero, or just a constant number? And if the position is fixed doesn't that mean that the velocity should be zero? Is this a differential equation I have to solve?

d. and e. Need c to continue.

Any help is appreciated, thanks !

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2. May 23, 2016

### vela

Staff Emeritus
"Fixed" means "constant." The position and velocity aren't actually fixed (so $\dot{\theta} \ne 0$), but because the timescale over which the radial motion occurs is so short, you can analyze the motion as if $\theta$ and $\dot{\theta}$ were constant. It's an approximation.

By the way, the first term in the radial differential equation is incorrect.

3. May 23, 2016

### bigguccisosa

Thanks I must have missed a (r_0 + r) factor in the first term when I was putting it into tex, I'll give it a shot with theta and theta dot as constants.