Rolling Ball & Pendulum Project

The motion of a pendulum and a ball rolling in a circular track are compared with a DIY apparatus to better understand their unique use of energy.

Oscillation due to Gravity

The passage of time can be marked using a pendulum, consisting of a mass m suspended by a length R of string. The mass traces out part of a circle of radius R (measured from the pivot to the ball’s center), oscillating with a period $T = 2\pi \sqrt{\frac{R}{g}}$, where $g =  9.8 \frac{m}{s}$ is the Earth’s gravitational acceleration. The period is independent of the mass and can be made longer by increasing the length of string. Essentially the mass is falling and rising while constrained to move in a circle.

Another way to force the same circular motion is to have a spherical mass on a valley shaped track of radius R. But does this result in the same oscillation period?

Oscillation of a Sliding Ball on a Track vs. Pendulum

Assume a ball of radius r slides on a frictionless valley track of radius R, where the ball does not rotate. A line drawn from the center of the ball to the center of the track makes an angle $\gamma$ from the vertical that oscillates as the ball slides back and forth. The forces acting on the ball as it slides are gravity (mg) pointing down and a Normal force N pointing radially upward. The forces along the radial direction are balanced: $N = m g \cos{\gamma}$. Newton’s second law (F = ma) gives for the x and y components of the forces:

$m \frac{d^2x}{dt^2} = -N \sin{\gamma}$

$m \frac{d^2y}{dt^2} = -mg + N \cos{\gamma}$

We can relate x and y to $\gamma$ using $x = (R-r) \sin{\gamma}, y = (R-r)(1-\cos{\gamma})$. Some calculus and algebra then gives:

$\frac{d^2x}{dt^2} = (R – r)[\cos{\gamma} \frac{d^2\gamma}{dt^2} - \sin{\gamma} (\frac{d\gamma}{dt})^2]$

$\frac{d^2y}{dt^2} = (R – r)[\sin{\gamma} \frac{d^2\gamma}{dt^2} + \cos{\gamma} (\frac{d\gamma}{dt})^2]$

which simplifies to: $\frac{d^2\gamma}{dt^2} = -\sin{\gamma}\frac{g}{R – r}$

Using $\gamma = \gamma_0 \cos{\omega t}$ and $\sin{\gamma}\approx \gamma$ gives us $\omega^2 = \frac{g}{R-r}$, or period $T = 2\pi \sqrt{\frac{R – r}{g}}$. This is exactly what we would expect for a pendulum of length R – r. This derivation is identical for the pendulum, where the string tension S takes the place of N.

Apparatus to Measure Period T

A circular valley track was made using two ½” thick wooden forms cut with a circle-jig and router, topped with 1/8” thick steel bars bent to conform and epoxied in place. The two rails were bolted together with a 3.8 mm spacer to form a curved track with an angular range of +/- 19°. A pendulum was made just behind the track using a 1” steel ball encased in a thin metal shell. The pendulum was tied to a pivot so that its center of mass follows the same path as a 1” ball on the track. 

Observations:

• The pendulum had a measured period T = 1.923 s. This is very close to the calculated T = 1.925 s, based on $T = 2\pi \sqrt{\frac{R}{g}}$.
• A 1” steel ball placed on the track had an oscillation period of 2.315 s, 1.2 times longer than the swinging pendulum, even though the circular paths were essentially identical.
• This indicates that additional physics is at work. 

In the videos below, the pendulum completes about 21 cycles after 40 seconds while the ball completes only 17.

Oscillation of a Ball Rolling on a Track

We now assume the ball is rolling about its center without sliding. The rotation angle is $\theta$. Rolling without sliding requires a frictional force f between the ball and track. Newton’s second law for the center of mass motion is now:

$m \frac{d^2x}{dt^2} = -N \sin{\gamma} – f \cos{\gamma}$

$m \frac{d^2y}{dt^2} = -mg + N \cos{\gamma} – f \sin{\gamma}$

Some calculus and algebra yields:

$(R – r) \frac{d^2\gamma}{dt^2} = -g \sin{\gamma} – \frac{f}{m}$. This is the same result as the sliding ball with an additional term of $–\frac{f}{m}$.

To account for the ball’s rotation, we also need another form of Newton’s second law: $I \frac{d^2\theta}{dt^2} = r f$, where $I$ is the ball’s moment of inertia. We can relate $\theta$ to $\gamma$ noting that if the ball center traces an arc of length $(R-r)\gamma$ without slipping, then the ball must roll an equal distance about its center $r\theta$, so $(R-r)\gamma = r\theta$. Then $f = \frac{I}{r} \frac{d^2\theta}{dt^2} = \frac{I}{r}\frac{R-r}{r} \frac{d^2\gamma}{dt^2}$. Using this in the above equation of motion gives 

$\frac{d^2\gamma}{dt^2} = -g \frac{\sin{\gamma}}{1 + \frac{I}{mr^2}}$

This tells us the period of oscillation of a rolling object $T_r = 2\pi \sqrt{(R-r)(1 + \frac{I}{mr^2})}/g$. Comparing to the period of a pendulum $T_p = 2\pi \sqrt{\frac{R-r}{g}}$, we can write this as $T_r = T_p \sqrt{1 + \frac{I}{mr^2}}$. 

For a sphere, $I = \frac{2}{5} mr^2$, so $T_r = T_p \sqrt{7/5} = 1.18 T_p$. This is close to the value 1.2 measured in the experiment. Physically, the energy of a rotating ball on the track is divided between the potential and kinetic energy of the center of mass and the rotational energy of the ball, reducing its oscillation speed.

The Effect of the Gap Width and Ball Radius

Measurements were also made as a function of the gap width d between the rails and the ball radius r. For a narrow gap (d = 3.8mm), there is essentially no difference in T between ball diameters 7/8” and 1”. However, for a wide gap of 19.6 mm, the period was much longer and increased as r decreased.

Below is a video of ball diameter = 7/8” for a narrow gap (d = 3.8mm) and wide gap (d = 19.6mm), showing slower motion (larger T) for the wider gap.

 

Below is a video of ball diameter = 7/8” and 1” on a wide track (d = 19.6mm), showing slower motion (larger $T_t$) for the smaller diameter.

7/8” diameter $\rightarrow \frac{T_t}{T_p} = 1.79$

1” diameter $\rightarrow \frac{T_t}{T_p} = 1.45$

Oscillation of a Ball Rolling in a Wide Gap Track

With a wide gap, the center of the ball of radius r is a distance h from the contact edge of the rails. If the gap between the rails is d, we can calculate h using $r^2 = h^2 + (\frac{d}{2})^2$.  The previous analysis is modified by changing r to h (except in the expression for the ball’s moment of inertia). Then the period of oscillation of a ball rolling within the gap is $T_t = 2\pi \sqrt{(R-h)(1 + \frac{I}{mh^2})/g}$ or $T_t = T_p \sqrt{1 + \frac{I}{mh^2}}$. For a sphere, $I = \frac{2}{5} mr^2$, so $T_t = T_p \sqrt{1 + \frac{(2/5)r^2}{h^2}}$. As the gap widens or the radius increases, h/r decreases, resulting in an increase in $T_t$. Physically, smaller h requires the ball to roll more to cover the same path length, increasing its rotational energy, reducing the speed traveling along the track.

The plot below shows the measured oscillation period relative to the pendulum for two ball diameters as a function of $\frac{d}{2r}$. The grey curve is the prediction from the above analysis, showing reasonable agreement. Slower oscillation at high gaps than predicted may be due to more frictional energy loss at higher rotation rates.