# Analyzing the motion of a ball rolling without slipping inside a hemispherical bowl

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). Now, I assume the oscillations are small and so the small angle approximation holds. I wish to find the period of oscillation and I analyze the motion in two ways, first using conservation of energy and secondly using dynamics. However, I receive two inconsistent answers. One or both of the solutions must be wrong, but I cannot figure out which one and more importantly, I cannot figure out why.

Method 1: We write the energy conservation equation for the ball

$mgh + \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = Constant$

from the center of mass, we take the height as $h = R-(R-r)cos\theta$ where $\theta$ is the angle from the vertical. Applying the no slip condition $v = r\omega$ and taking the moment of inertia for a solid sphere $I = \frac{2}{5}mr^2$ we can write the energy equation as

$mg(R-(R-r)cos\theta) + \frac{7}{10}mr^2\omega^2 = Constant$

Differentiating with respect to time:

$mg(R-r)sin\theta\cdot\omega + \frac{7}{5}mr^2\omega\cdot\alpha = 0$

taking the small angle approximation $sin\theta = \theta$ we get

$g(R-r)\theta + \frac{7}{5}r^2\alpha=0$

$-\frac{5g(R-r)}{7r^2}\theta = \alpha$

from which we can get $T = 2\pi\sqrt{\frac{7r^2}{5g(R-r)}}$

Method 2: 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

$F = ma = mgsin\theta - f$

$\frac{7}{5}a = gsin\theta$

taking the small angle approximation and converting $a$ to $\alpha$ we get

$\alpha = \frac{5g}{7r}\theta$

and a corresponding period of $T = 2\pi\sqrt{\frac{7r}{5g}}$

Now the solutions are very different and I would appreciate it if someone would point out where I went wrong.

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There are a lot of things wrong with this. The most glaring is that you take energy to be constant in the first approach, but you introduce friction into the second, which is an outside force that will reduce energy over time. So the two approaches are necessarily going to give different results because you're making different assumptions. – Mitchell Jun 18 '11 at 5:25
@Mitchell, no, that is not a problem. The friction force does no work. I think Yuqing explained this pretty clearly. – Mark Eichenlaub Jun 18 '11 at 5:33
Closely related: physics.stackexchange.com/q/10798 – qftme Jun 18 '11 at 12:23

## 3 Answers

Your first derivation, using energy, uses two different meanings for the same symbol $\omega$. In one place, you interpret it as

$$\omega = \dot{\theta}$$

the time derivative of the angle of the line from the center of the ball to the center of the bowl with the vertical.

In another place, you interpret $\omega$ as the time derivative of the unnamed angle through which the ball itself has rotated.

These two angles are related to each other by the $r/(R-r)$ factor by which you are off.

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A problem I see in your first solution is that $v$ is the velocity of the center of mass. You assume $v=r\omega$ but that is the velocity of the bottom of the ball. The center of mass is moving parallel to this in a circular motion, the radius of which is $R-r$, $v=((R-r)/R)r\omega$.

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Both answers are incorrect. The correct answer taking into account both angles as mentioned by others, has R-r in the numerator. The limiting case, in which the ball gets really small, produces a period that cannot go to zero, and this consideration alone can be used to eliminate both of your answers.

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