Tag Archives: Leonard Hofstadter

The Big Bang Theory of the Inclined Plane

Season 01, Episode 02: “The Big Bran Hypothesis”
Sheldon and Leonard solve a problem using an inclined plane

Sheldon and Leonard contemplate using the stairs as an inclined plane to move Penny’s furniture to her apartment.

In Season 1 Episode 2 of The Big Bang Theory, “The Big Bran Hypothesis”, Penny (Kaley Cuoco) asks Leonard (Johnny Galecki) to sign for a furniture delivery if she isn’t home. Unfortunately for Leonard and Sheldon, they are left with the task of getting a huge (and heavy) box up to Penny’s apartment.

To solve this problem, Leonard suggest using the stairs as an inclined plane, one of the six classical simple machines defined by Renaissance scientists. Both Leonard and Sheldon have the right idea here. Not only are inclined planes used to raise heavy loads but they require less effort to do so. Though this may make moving a heavy load easier the tradeoff is that the load must now be moved over a greater distance. So while, as Leonard correctly calculates, the effort required to move Penny’s furniture is reduced by half, the distance he and Sheldon must move Penny’s furniture twice the distance to raise it directly.

Mathematics of the Inclined Plane

Effort to lift block on Inclined Plane

Now we got an inclined plane. Force required to lift is reduced by the sine of the angle of the stairs… call it 30 degrees, so about half.

Steps being used as an inclined plane to raise a block

Free-body Diagram of a block on an inclined plane. It shows the forces acting on a block and the force needed to keep it stationary and not let it slip down.


To analyze the forces acting on a body, physicists and engineers use rough sketches or free body diagrams. This diagram can help physicists model a problem on paper and to determine how forces act on an object. We can resolve the forces to see the effort needed to move the block up the stairs.

If the weight of Penny’s furniture is \(W\) and the angle of the stairs is \(\theta\) then
\[\angle_{\mathrm{stairs}}\equiv\theta \approx 30^\circ\]
and
\[\Rightarrow\sin 30^\circ = \frac{1}{2}\]
So the effort needed to keep the box in place is about half the weight of the furniture box or \(\frac{1}{2}W\), just as Leonard says.

Distance moved along Inclined Plane

Inclined plane and steps

The relationship between the height \(h\) the block is raised and the distance it moves \(d\) is the sine of the angle \(\theta\).

While the inclined plane allows Leonard and Sheldon to push the box with less effort, the tradeoff is that the distance they move along the incline is twice the height to raise the box vertically. Geometry shows us that
\[\sin \theta = \frac{h}{d}\]
We again assume that the angle of the stairs is approximately \(30^\circ\) and \(\sin 30^{\circ} = 1/2\) then we have \(d=2h\).

Uses of the Inclined Plane

We see inclined planes daily without realizing it. They are used as loading ramps to load and unload goods. Wheelchair ramps also allow wheelchair users, as well as users of strollers and carts, to access buildings easily. Roads sometimes have inclined planes to form a gradual slope to allow vehicles to move over hills without losing traction. Inclined planes have also played an important part in history and were used to build the Egyptian pyramids and possibly used to move the heavy stones to build Stonehenge.

Lombard Street (San Francisco)

Lombard Street, San Francisco Inclined Plane

Lombard Street is one of the most visited street in San Francisco as seen from Coit Tower. It is best known for the one-way section on Russian Hill between Hyde and Leavenworth Streets, in which the roadway has eight sharp turns (or switchbacks) that have earned the street the distinction of being the crookedest “most winding “street in the world, though this title is contested. (Photo by David Yu).

Lombard Street in San Francisco is famous for its eight tight hairpin turns (or switchbacks) that have earned it the distinction of being the crookedest street in the world (though this title is contested). These eight switchbacks are crucial to the street’s design as the reduce the hills natural 27° grade which is too steep for most vehicles. It is also a hazard to pedestrians, who are more accustomed to a more reasonable 4.86° incline due to wheel chair navigability concerns.

Technically speaking, the “zigzag” path doesn’t make climbing or coming down the hill any easier. As we have seen, all it does is change how various forces are applied. It just requires less effort to move up or down but the tradeoff is that you travel a longer distance. This has several advantages. Car engines have to be less powerful to climb the hill and in the case of descent, less force needs to be applied on the brakes. There are also safety considerations. A car will not accelerate down the switch back path as fast than if it was driven straight down, making speeds safer and more manageable for motorists.

This idea of using zigzagging paths to climb steep hills and mountains is also used by hikers and rock climbers for very much the same reason Lombard Street zigszags. The tradeoff is that the distance traveled along the path is greater than if a climber goes straight up.

The Descendants of Archimedes

We don’t need strength, we’re physicists. We are the intellectual descendants of Archimedes. Give me a fulcrum and a lever and I can move the Earth. It’s just a matter of… I don’t have this, I don’t have this!

We see that Leonard had the right idea. If we were to assume are to assume — based on the size of the box — that the furniture is approximately 150 lbs (65kg) and the effort is reduced by half, then they need to push with at least 75 lbs of force. This is equivalent to moving a 34kg mass. If they both push equally, they are each left pushing a very manageable 37.5 lbs, the equivalent of pushing a 17kg mass.

Penny’s apartment is on the fourth floor and we if we assume a standard US building design of ten feet per floor, this means a 30 foot vertical rise. The boys are left with the choice of lifting 150 lbs vertically 30 feet or moving 75lbs a distance of 60 feet. The latter is more manageable but then again, neither of our heroes have any upper body strength.

The Big Bang Theory of the Rolling Ball Problem

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.

Equations of a rolling ball

Scene with Sheldon Cooper. The background shows a whiteboard with the physics equations of a rolling ball.

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

Pure Translatoinal Motion

An object undergoing 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

Pure Rotation

An object undergoing pure rotational motion. The velocity of each point of the object depends on the distance away from the center. The further away the faster it moves as can be seen by the length of the arrows.

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

Rotating Object

An object that rolls undergoes both translational and rotational motion. To determine the speed at any point we must add both of these speeds. We see the translational speed show in red and the rotational speed show in in blue.

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 (\(Q\)).

Whiteboard of the Rolling Ball Problem

Rendering copy of the whiteboard showing the rolling ball problem as seen on the Big Bang Theory.

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

Cycloid drawn by a Rolling Ball

The path a point on a rolling ball draws is called a cycloid.

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.