# Rolling (without slipping) ball on a moving surface

I've been looking at examples of a ball rolling without slipping down an inclined surface. What happens if the incline angle changes as the ball is rolling?

More precisely I've been trying to find equations for programming a simulated (2D) ball rolling inside a swinging bowl/arc.

I thought I could still use the same equations for just having a ball rolling inside a still bowl (see below) and the changes in the incline angle (tangent at the contact point of the ball and the bowl) would take care of itself:

For a ball rolling inside a bowl: The only torque acting on the ball is the frictional force: $τ=Iα=fr$, using the rolling without slipping condition $a=rα$ and the moment of inertia for a solid sphere, $I = \frac{2}5 mR^2$, we get $f=\frac{2}5ma$. The net force acting on the system is gravity and the force of friction, $F=ma=mgsinθ−f$ and therefore, $a=\frac{5}7gsinθ$

I am speculating that due to the surface itself moving (swinging on a circular path), it's the relative motion that contributes to the friction? But, I don't know how to include that. Can someone please help me?

• You need diagrams: increased normal force on the ball due to upward swing of bowl affects friction. Please be specific: What's the shape of the bowl? Hemisphere? Parabolic? – raindrop Dec 20 '12 at 16:55
• Link: A circle rolls along a parabola. math.stackexchange.com/q/32629 Related: Analyzing the motion of a ball rolling without slipping inside a hemispherical bowl physics.stackexchange.com/q/11227 – raindrop Dec 20 '12 at 17:01

## 1 Answer

Here are some questions to ask before building equations: What is the shape of the bowl? What is the mathematical description of the shape of the bowl? Is the bowl massless? How does the bowl swing? Does it swing from a string? Is that string massless? Does the bowl rotate? (in addition to its swinging and having a ball roll on its surface)

From a related question, "Consider a solid ball of radius $r$ and mass $m$ rolling without slipping in a hemispherical bowl of radius $R$ (simple back and forth motion). "

"The only torque acting on the ball at any point in its motion is the friction force $f$. So we can write

$\tau = I\alpha = fr$

again using the rolling condition $a = r\alpha$ and the moment of inertia for a solid sphere,

$\frac{2}{5}ma = f$

The net force acting on the system is the tangential component of gravity and the force of friction, so

$F = ma = mgsin\theta - f$ "

Since your bowl is swinging, $\theta$ changes with time. (Imagine the bowl is like a swinging pendulum bob)

Now lets discuss the details about the swinging bowl. Consider a swinging bowl on a massless string of length $L'$ with period of oscillation $T'$ and maximum angular displacement $\theta_{max} '$

We need to form equations describing the change of the angle on inclination of the bowl with respect to the rolling bowl as the bowl swings.

Therefore, we need some initial condition. Let's say the bowl is at its maximum angular displacement to our 'left', and the hemispherical bowl always 'points' towards the 'axis' of it's swinging. Our 'total' inclination angle must not sum up to $\frac{\pi}{2} rad$, otherwise the ball would fall vertically instead of rolling without slipping.

First off, let's deal with $\theta$, the angle of inclination. Angle of inclination=angle of ball in bowl + angle of bowl in pendulum system.

Secondly, we need the equation describing the change of $\theta$ with time. Assume the bowl is massless but rigid and doesn't rotate (due to the other torque, exerted by the ball on the bowl). However, for a short while, let's imagine that the bowl does rotate. We would have some critical case (or range of cases) such that the rotation of the bowl corresponds to the swinging of the bowl in such a way that we can have large maximum angular displacements for the swinging of the bowl (perhaps even $2 \pi$, corresponding to full revolutions!). In other words, the rotation of the bowl could help increase the stability of the system.

If the bowl rotates, we need to even information about the bowl.

So in the oscillation of the bowl, we only consider the mass of the ball.

• Thanks very much @Raindrop. Can you explain what the "normal force" I am missing entail? I thought the normal force is what the surface exerts on the ball to counter the normal component of the gravitational force the ball is exerting on the surface. I thought that this component would be cancelled, leaving the mgsinθ component only. Could you explain more of how that works? Is it because the arc swinging up would give it an additional 'lift'? Thanks for your help! – Wynn Dec 20 '12 at 20:50
• when the bowl swings (accelerates) upwards the ball exerts an additional normal force $F_n=ma$ on the bowl. So, the net vertical force acting on the ball would be $\Sigma F_y=F_n-F_g=m_{ball}(a_{bowl}-g)$ (If we take upwards as positive, note that $F_n$ is upwards and $F_g$ is downwards) – raindrop Dec 20 '12 at 20:58
• It was a mistake, I think $mgsin\theta$ does hold in our case – raindrop Dec 20 '12 at 21:03
• note that the normal force causes horizontal forces too, not just vertical. As an additional note, a method to solve this problem is to divide it into 3 parts and solve each one separately. To be honest this problem isn't that hard it's just fun and long so don't give up! It's mostly finding the basic equations of motion for swinging and rolling and doing lots of substituting here and there. – raindrop Dec 21 '12 at 4:20