Globe of Steel!

Make a free body diagram at the highest point of the loop to write one equation, and use energy conservation to write another equation. Combining these equations will allow you to solve for the answer.

Newton’s Second Law in the \({y-}\)direction at the highest point can be written as:

\begin{equation*}
N+mg = ma,
\end{equation*}

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where \(a\) is the centripetal acceleration which is equal to \(\frac{v^2}{R}\). In the limit case \(N \to 0 \), and solving for \(v^2\) we get:

\begin{equation*}
v^2 = gR.
\end{equation*}

Now, the principle of mechanical energy conservation can be stated as \(E_A=E_B \). In our case, we will choose point \(A\) to be on the ground, just before entering the loop, and \(B\) as the highest point of the loop. The equation for the conservation of energy can be written as:

\begin{equation*}
\frac{1}{2}mv_A^2=mg2R+\frac{1}{2}mv_B^2,
\end{equation*}

where \(v_B\) is the velocity which was solved previously. Substituting the velocity into the last equation, and solving for \(v_A\), we get:

\begin{equation*}
v_A=\sqrt{5gR}.
\end{equation*}

For a more detailed explanation of any of these steps, click on “Detailed Solution”.

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1. General Strategy

We need to find the minimum speed the bike rider needs when entering the loop in order to have success in the maneuver. That is, we need to find the speed illustrated in figure 1.

Figure 1: Velocity \(\vec{v}\) of the rider as it enters the loop.

To provide an answer to this problem, we’ll first need to analyze what condition must hold so that the bike has a successful maneuver. From this analysis, we will obtain the minimum speed that the bike rider must have at the most critical point in the trajectory in order to successfully make the loop. Then, we will use conservation of mechanical energy to relate this minimum speed at the critical point to the speed of the bike rider before entering the loop.

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Following the procedure previously described, let’s consider the condition required for the bike rider to make a successful loop. At all points in the bike rider’s trajectory there will be two forces exerted over him: the weight and the contact force of the bike with the track. The weight is constant, and its direction is downwards at all times. The magnitude of the weight is \(mg\) where \(m\) is the mass of the bike rider plus its bike and \(g\) the gravitational constant on Earth. The contact force \(\vec{N}\) changes its magnitude as the biker goes over the loop, and its direction is always perpendicular to the trajectory.

The condition required for the bike rider to execute a successful loop is that the contact force is always greater than zero, that is \(N>0\). This condition ensures that at all points in the trajectory the wheels of the bike make contact with the track (otherwise, the biker would be falling). The critical point at which the bike may lose contact with the track is the highest point in the trajectory, where the direction of the weight mostly contributes to this loss of contact. To better see this, let’s make a free body diagram at this critical point, using a coordinate system where Y points downwards (and in this case, towards the center of the loop):

2. Identify the forces and make a free-body diagram

Figure 2: Force diagram for the biker at the highest point in the loop. The forces shown are: the weight \(\vec{W}\) and the contact force with the loop \(\vec{N}\). The coordinate axis at this point is chosen with the Y axis pointing downwards.

Since the biker is following a circular motion, he will have a centripetal (or radial) acceleration pointing towards the center of the loop. This is illustrated in figure 3.

Figure 3: Tangent velocity \(\vec{v}\) and centripetal acceleration \(\vec{a}\) at the highest point of the biker’s trajectory in the loop. The coordinate system is oriented such that the positive Y axis points downwards.

3. Newton’s Second Law in Y

Using these figures as a guide, we can now use Newton’s second along the Y axis:

\begin{equation}
\label{newton}
N\,\hat{\textbf{j}}+mg\,\hat{\textbf{j}}=ma\,\hat{\textbf{j}},
\end{equation}

where \(a\) is the magnitude of the centripetal acceleration (the acceleration is positive in Y, according to the system used).  The centripetal acceleration has a  magnitude of

\begin{equation}
\label{ac}
a=\frac{v^2}{R},
\end{equation}

where \(v\) is the magnitude of the velocity (this is the speed). Using equation \eqref{ac} in equation \eqref{newton}, we get

\begin{equation}
\label{newton2}
N\,\hat{\textbf{j}}+mg\,\hat{\textbf{j}}=m\frac{v^2}{R}\,\hat{\textbf{j}},
\end{equation}

or dropping the vector notation and focusing on the magnitudes, the result is

\begin{equation}
\label{newton3}
N+mg=m\frac{v^2}{R}.
\end{equation}

4. Manipulate equations to find an expression for the minimum speed

Now let’s consider the most critical case, which is that, at the highest point of the trajectory, the contact force tends to zero (that is, the biker is about to fall). We can then take \(N\to 0\) as our limit case and write equation \eqref{newton3} as

\begin{equation}
\label{newton4}
mg=m\frac{v^2}{R},
\end{equation}

where we can cancel out the mass to obtain

\begin{equation}
g=\frac{v^2}{R}.
\end{equation}

Solving for \(v^2\) we get

\begin{equation}
v^2=gR,
\end{equation}

and taking the square-root on both sides, we finally obtain

\begin{equation}
\label{velmin}
v=\sqrt{gR}.
\end{equation}

We have thus obtained an expression for the minimum speed that the bike-rider must have at the highest point to be able to successfully perform the loop.

5. Conservation of energy

Now we can relate this speed with the speed before entering the loop using the principle of mechanical energy conservation. We’re able to use this principle because non-conservative forces such as friction are not present. The mechanical energy \(E\) can be written as the sum of potential energy \(U\) and kinetic energy \(K\), namely

\begin{equation}
E=U+K.
\end{equation}

In our case, the potential energy is due to gravity and can be calculated using the following expression

\begin{equation}
U=mgh,
\end{equation}

where \(h\) is the height with respect to a coordinate system. In this case, we’ll choose the origin of the coordinate system at the floor. The reader might recall that in part a) we placed the coordinate system on the top of the loop, but now it is more convenient to choose a coordinate system on the floor (we are always free to use different coordinate systems in the same problem, as long as we are careful not mixing equations used for one system with equations used for the other).

Figure 4: Point A is the lowest point in the loop, and point B is the highest point in the loop. The coordinate system in this case is located at point A. The height difference between point A and B is twice the radius of the loop \(R\).

For the kinetic energy \(K\), we have the explicit expression

\begin{equation}
K=\frac{1}{2}mv^2.
\end{equation}

Therefore, the mechanical energy is

\begin{equation}
\label{mechanicale}
E=mgh+\frac{1}{2}mv^2.
\end{equation}

Now, the principle of mechanical energy conservation can be stated as

\begin{equation}
\label{eaeb}
E_A=E_B,
\end{equation}

where \(E_A\) and \(E_B\) are the energies at points \(A\) and \(B\) respectively. In our case, we will choose point \(A\) to be floor just before entering the loop and \(B\) as the highest point in the loop, as shown in the last figure. Using expression \eqref{mechanicale} in both sides of equation \eqref{eaeb}, we obtain

\begin{equation}
\label{conservation}
mgh_A+\frac{1}{2}mv_A^2=mgh_B+\frac{1}{2}mv_B^2,
\end{equation}

where \(h_A\) and \(h_B\) are the height of points \(A\) and \(B\) respectively. The terms \(v_A\) and \(v_B\) are the velocities at points \(A\) and \(B\) respectively. From the last figure, we can see that the height at point \(A\) is zero \(h_A=0\) and the height at point \(B\) is twice the radius of the loop, \(h_B=2R\). Speed \(v_B\) is the speed at the highest point, which we will take as the minimum speed given by equation \eqref{velmin}. Using all of this into equation \eqref{conservation}, we get

\begin{equation}
\frac{1}{2}mv_A^2=mg(2R)+\frac{1}{2}m(gR).
\end{equation}

4. Manipulate equations to find an expression for the minimum speed

Cancelling out the mass in all the terms, we get

\begin{equation}
\frac{1}{2}v_A^2=2gR+\frac{1}{2}gR.
\end{equation}

Performing the fraction sum on the right of the equation, we obtain

\begin{equation}
\frac{1}{2}v_A^2=\frac{5}{2}gR.
\end{equation}

Solving for \(v_A\), we get

\begin{equation}
v_A=\sqrt{5gR}.
\end{equation}

Thus, we have found the minimum speed that the bike-rider must have at the bottom of the loop in order to make a successful maneuver.

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