Big Bang Theory Season 2, Episode 5: “The Euclid Alternative”

In the opening scene of the “The Euclid Alternative”, we see Sheldon (Jim Parsons) demanding that Leonard (Johnny Galecki)needs to drive him around to run various errands. Leonard, after spending a night in the lab using the new Free Electron Laser to perform X-ray diffraction experiments. In the background, we can see equations that describe a rolling ball problem on the whiteboard in the background.

Rolling motion plays an important role in many familiar situations so this type of motion is paid considerable attention in many introductory mechanics courses in physics and engineering. One of the more challenging aspects to grasp is that rolling (without slipping) is a combination of both translation and rotation where the point of contact is instantaneously at rest.The equations on the white board describe the velocity at the point of contact on the ground, the center of the object and at the top of the object.

Pure Translational Motion When an object undergoes pure translational motion, all of its points move with the same velocity as the center of mass– it moves in the same speed and direction or \(v_{\textrm{cm}}\).

Pure Rotational Motion In the case of a rotating body, the speed of any point on the object depends on how far away it is from the axis of rotation; in this case, the center. We know that the body’s speed is \(v_{\textrm{cm}}\) and that the speed at the edge must be the same. We may think that all these points moving at different speeds poses a problem but we know something else — the object’s angular velocity.

The angular speed tells us how fast an object rotates. In this case, we know that all points along the object’s surface completes a revolution in the same time. In physics, we define this by the equation:

\begin{equation}

\omega=\frac{v}{r}

\end{equation}

where \(\omega\) is the angular speed. We can use this to rewrite this equation to tell us the speed of any point from the center:

\begin{equation}

v(r)=\omega r

\end{equation}

If we look at the center, where \(r=0\), we expect the speed to be zero. When we plug zero into the above equation that is exactly what we get:

\begin{equation}

v(0)= \omega \times 0 = 0

\label{eq:zero}

\end{equation}

If we know the object’s speed, \(v_{\textrm{cm}}\) and the object’s radius, \(R\), using a little algebra we can define \(\omega\) as:

\[\omega=\frac{v_{\textrm{cm}}}{R}\]

or the speed at the edge, \(v_{\textrm{cm}}\) to be \(v(R)\) to be:

\begin{equation}

v_{\textrm{cm}}=v(R) = \omega R

\label{eq:R}

\end{equation} Putting it all Together To determine the absolute speed of any point of a rolling object we must add both the translational and rotational speeds together. We see that some of the rotational velocities point in the opposite direction from the translational velocity and must be subtracted. As horrifying as this looks to do, we can reduce the problem somewhat to what we see on the whiteboard. Here we see the boys reduce the problem and look at three key areas, the point of contact with the ground (\(P)\), the center of the object, (\(C\)) and the top of the object (\(Q\)). We have done most of the legwork at this point and now the rolling ball problem is easier to solve. At point \(Q\) At point \(Q\), we know the translational speed to be \(v_{\textrm{cm}}\) and the rotational speed to be \(v(R)\). So the total speed at that point is

\begin{equation}

v = v_{\textrm{cm}} + v(R)

\label{eq:Q1}

\end{equation}

Looking at equation \eqref{eq:R}, we can write \(v(R)\) as

\begin{equation}

v(R) = \omega R

\end{equation}

Putting this into \eqref{eq:Q1} and we get,

\begin{aligned}

v & = v_{\textrm{cm}} + v(R) \\

& = v_{\textrm{cm}} + \omega R \\

& = v_{\textrm{cm}} + \frac{v_{\textrm{cm}}}{R}\cdot R \\

& = v_{\textrm{cm}} + v_{\textrm{cm}} = 2v_{\textrm{cm}}

\end{aligned}

which looks almost exactly like Leonard’s board, so we must be doing something right. At point \(C\) At point \(C\) we know the rotational speed to be zero (see equation \eqref{eq:zero}).

Putting this back into equation \eqref{eq:Q1}, we get

\begin{aligned}

v & = v_{\textrm{cm}} + v(r) \\

& = v_{\textrm{cm}} + v(0) \\

& = v_{\textrm{cm}} + \omega \cdot 0 \\

& = v_{\textrm{cm}} + 0 \\

& = v_{\textrm{cm}}

\end{aligned}

Again we get the same result as the board. At point \(P\) At the point of contact with the ground, \(P\), we don’t expect a wheel to be moving (unless it skids or slips). If we look at our diagrams, we see that the rotational speed is in the opposite direction to the translational speed and its magnitude is

\begin{aligned}

v(R) & = -\omega R \\

& = -\frac{v_{\textrm{cm}}}{R}\cdot R \\

& = -v_{\textrm{cm}}

\end{aligned}

It is negative because the speed is in the opposite direction. Equation \eqref{eq:Q1}, becomes

\begin{aligned}

v & = v_{\textrm{cm}} + v(r) \\

& = v_{\textrm{cm}} – \omega R \\

& = v_{\textrm{cm}} – \frac{v_{\textrm{cm}}}{R}\cdot R \\

& = v_{\textrm{cm}} – v_{\textrm{cm}} = 0

\end{aligned}

Not only do we get the same result for the rolling ball problem we see on the whiteboard but it is what we expect. When a rolling ball, wheel or object doesn’t slip or skid, the point of contact is stationary.

Cycloid and the Rolling ball If we were to trace the path drawn by a point on the ball we get something known as a cycloid. The rolling ball problem is an interesting one and the reason it is studied is because the body undergoes two types of motion at the same time — pure translation and pure rotation. This means that the point that touches the ground, the contact point, is stationary while the top of the ball moves twice as fast as the center. It seems somewhat counter-intuitive which is why we don’t often think about it but imagine if at the point of contact our car’s tires wasn’t stationary but moved. We’d slip and slide and not go anywhere fast. But that is another problem entirely.