# Ball Rolling in a Parabolic Bowl

I encountered a physics problem which inquired about a ball rolling inside a parabolic bowl (i.e. a bowl where any cross section through the vertex would make a parabolic shape given by $y = kx^2$). The ball was to be released from some initial height $z_0$ and allowed to roll back and forth inside the bowl.

At first, I thought that the motion would be simple harmonic for any displacement, since $U(x) = mgh = mgkx^2$, and simple harmonic motion is characterized by a potential energy which is proportional to the square of the distance from some equilibrium point. However, the answer key said that the motion was not truly simple harmonic (and the rolling ball only appeared to be so for small amplitudes of motion).

Why is motion with the a potential energy proportional to the square of displacement from equilibrium not simple harmonic in this case?

EDIT: To clarify, the "ball" should have really been a point mass.

• not sure what the answer sheet refers to. a potential of this form surely is a simple harmonic. a nitpicky argument would be that the gravitational constant is height dependent and only constant for low heights compared to the radius of the earth. – luksen Dec 19 '12 at 0:20
• @luksen No, it is not harmonic, because the displacement of the ball is the full arclength, not the horizontal projection of the displacement. – Mark Eichenlaub Dec 19 '12 at 3:04

## 2 Answers

Although the potential energy has the same form, this is not a simple harmonic motion basically because in simple harmonic oscillator the ball moves only on the x-axis, while on your problem it's forced by the bowl to move on the z-axis as well. If we write down the Lagrangian for your problem: $$L = T - U = \frac{1}{2}m(\dot{x}^2 + \dot{z}^2) - mgz = \frac{1}{2}m(\dot{x}^2 + (2kx\dot{x})^2) - mgkx^2 =$$ $$= \frac{1}{2}m\dot{x}^2(1 + 4k^2x^2) - mgkx^2$$ we can see from Euler-Lagrange equations: $$\frac{\partial L}{\partial x} = \frac{d}{dt}\frac{\partial L}{\partial \dot{x}}$$ that the equation of motion describing your problem is: $$(1+4k^2x^2)\ddot{x}+4k^2\dot{x}^2x+2gkx=0$$ which is quite different from the one for simple harmonic motion.

Edit: for simplicity I've assumed it's a point-body with no moment of inertia rather than an actual ball that rolls inside the bawl. In the second case, it requires us to take the ball's moment of inertia into account, which will result in a bit different equation of motion.

• Indeed, it is a point body. I'm not very familiar with Lagrangian mechanics (still a high school student), but I'll look into it to see if I can understand your explanation better. – Shivam Sarodia Dec 19 '12 at 2:58
• Aside from the rotation, we also must account for the ball's finite size in finding it potential energy. When the ball is on a slope, the vertical displacement from the point of contact to the center of the ball is less than when the ball is on the flat part at bottom. – Mark Eichenlaub Dec 19 '12 at 3:06
• +1 for addressing the moment of inertia point. I was just thinking that even for a bowl shape that would generate SHM (for a point mass), a physical ball would still not execute SHM. – Kyle Oman Dec 19 '12 at 3:07
• @Draksis It can, but the kinetic energy is not $1/2 m \dot{x}^2$. – Mark Eichenlaub Dec 19 '12 at 4:43
• @Draksis Think of it this way: when the ball moves from point $x_1$ to point $x_2$, it gains kinetic energy equal to $\Delta T = U(x_1)-U(x_2) = mgk(x_{1}^{2}-x_{2}^{2})$. If the movement is one dimensional, all of this change in kinetic energy will be expressed as a change in speed on the x-direction alone. If the movement is 2-dimensional, on the other hand, some of it will be expressed as a change in speed on the z-direction as well. – Andrey S Dec 19 '12 at 5:32

I would like to offer another Lagrangian approach, that tries to mirror the typical pendulum problem in lagrangian mechanics. I just thought of this, so I hope this is right...

A typical problem in lagrangian mechanics is to solve the pendulum with a constant string length, $l$. The path of the ball at the end of the string (in 2D) is a circle, this is a path of constant radius. We exemplify that we know this by first writing $$L = T-U = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2) + mgy$$ and then making the change to polar coordinates using $x = r\sin\phi$ and $y = r \cos\phi$. Now lets do this same thing, except our ball is not traveling curves of constant radius, it is traveling curves of constant parabola. Then we will use the change to parabolic coordinates, $x = \sigma\tau$ and $y = \frac{1}{2}(\tau^2 - \sigma^2)$. Here, curves of constant $\sigma$ are upward facing parabolae, so we let our $\sigma$ be a constant. With this our Lagrangian becomes $$L = \frac{1}{2}m(\sigma^2+\tau^2)\dot{\tau}^2 - \frac{1}{2}mg(\tau^2 - \sigma^2)$$ Then $$\frac{\partial L}{\partial \dot{\tau}}=m(\sigma^2 + \tau^2)\dot{\tau}\implies\frac{d}{dt}\frac{\partial L}{\partial \dot{\tau}}=m[2\tau\dot{\tau}^2+(\sigma^2+\tau^2)\ddot{\tau}]$$ and $$\frac{\partial L}{\partial \tau}=m\tau\dot{\tau}^2-mg\tau$$ so all together becomes $$m[2\tau\dot{\tau}^2+(\sigma^2+\tau^2)\ddot{\tau}]-m\tau\dot{\tau}^2+mg\tau=0$$ $$\rightarrow \ddot{\tau}+\frac{g\tau+\tau\dot{\tau}^2}{(\sigma^2+\tau^2)}=0$$ There is also a constraint equation related to constant $\sigma$, $\;f=\sigma-a=0$ which can be used with a lagrange multiplier to extract the normal force on the ball. Hope this helps/is right.

• About the units at the final expression... $g\tau$ has units of $length^{1.5} \cdot time^{-2}$ while $\tau\dot{\tau}$ has units of $length \cdot time^{-1}$, so I guess there is a mistake somewhere. – Andrey S Dec 19 '12 at 4:06
• Thanks :), I think my $\partial L /\partial \tau$ was wrong, I think I fixed it. – kηives Dec 19 '12 at 4:11