# 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

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.
• 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