# Equations of motion for elastic pendulum

Having some trouble deriving the equation of motion in $y$-direction ($x(t)=0$). It said in the problem description that the equation should be:

$$my'' = -\frac{dV(r)}{dy} -mg.$$

$y''$ is the second derivative of $y$

$\frac{dV(r)}{dy}$ is the partial derivative of $V(r)$ on $y$

So I got pendulum hanging on an elastic but rigid massless rod that can swing in the $xy$-plane. Pivot point is the origin of the coordinate system. The forces acting on the pendulum is: Sum of elastic force directed towards the origin and gravity which goes in positive $y$-direction (so $y$ is positive pointing towards the ground). (friction of any kind (air, in pivot point) can be disregarded)

From the problem description I get: Gravity - $g = 9.81 m/s^2$,

Potential energy of the rod $V(r) = 0.5k(r-L)^2$,

$r = \sqrt{x^2+y^2}$,

Length of rod with no external force $L = 1$m.

So I tought I could just do Sum of forces = $ma$. The two forces acting on the pendulum would be $F_e = -dV(r)/dy$ (the negative derivative of potential energy is force right?) and $mg.$ $F_e$ working in negative y-direction and $mg$ in positive. And I can set $a=y''$.

$$my'' = -dV(r)/dy + mg$$ But like stated above it should be $-mg$. Why? Am I missing something?

Sorry for long post and for not using correct math font (don't know how). But hope I made it readable. All help is appreciated!

• Welcome to Physics SE :) See here for using the correct math font. Also, think about what would happen if you write $m y'' = m g$ (i.e., in the absence of a potential), and consider whether that makes sense ... – Sanya Oct 29 '16 at 16:30
• Thanks, and I will check ou the font later. Hm sorry doesn't make more sense for me. Can you explain more specific please? – ToyMan Oct 29 '16 at 16:56
• "elastic but rigid massless rod" - it can be one or the other, but not both. It is either elastic (can deflect) or rigid (cannot deflect). – ja72 Oct 9 '18 at 16:55

## 2 Answers

Mg is directed downwards, while the elastic force is directed upwards. The partial differential tells you only about the force arising due to the potential, and not gravity. So, you just subtract the gravitational force from the elastic force to get the net force. Note that this net force is regarding upwards as positive.

I guess you have to switch you y-axis so that up is positive. Then, as you said, $$my''=F_e-F_g=-\frac{dV(r)}{dy}-mg.$$ You pointed out that the two forces have opposite direction, thus the formula you wrote is wrong even with your choice of coordinates.

• Well thats what I'm thinking too. But the problem is that the formula I wrote is from the problem description same with the choice of coordinates. So not sure if it's something I have missed or the teacher have written the formula wrong or something. – ToyMan Oct 30 '16 at 7:38
• As @Sanya said, think about the problem without the potential. With your choice of coordinates, since $F_g$ is directed downwards, you have $my''=mg$. Then the sign of the formula you are trying to reproduce is wrong. Maybe there's just some misunderstanding about the direction of the y-axis. – Charlie Oct 30 '16 at 7:50