Adventurer!

Remember to consider the centripetal acceleration for Newton’s Second Law. You are looking for the “radius”.

\(T\) points upwards and \(W\) downwards. We know the forces and accelerations involved in the motion of the adventurer, so we can use Newton’s Second Law:

\begin{equation*}
\sum \vec{F}=m\vec{a},
\end{equation*}

which is:

\begin{equation*}
-mg+T=m\frac{v^2}{L}.
\end{equation*}

Since \(T=2mg\), and solving for \(L\) we get:

\begin{equation*}
L=\frac{v^2}{g},
\end{equation*}

which, with numerical values, is given as:

\begin{equation*}
L=5\,\text{m}.
\end{equation*}

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

The problem asks us to calculate the length of the \({liana}\). Let’s first note that the length of the \({liana}\) \(L\) does not change, and so the center of mass of the person describes a circular motion of radius \(L\), meaning the person will have some centripetal acceleration. This acceleration depends on the speed, and we know the speed at the lowest point of the trajectory. Thus, using Newton’s Second Law and an expression for the centripetal acceleration (in terms of the speed), we will be able to determine the length of the \({liana}\).

In order to use Newton’s Second Law, let’s start by making the free-body diagram.

To draw this diagram, let’s identify all the forces. There are only two forces: the weight and the tension. The weight \(\vec{W}\) is directed downwards and has magnitude \(mg\), where \(m\) is the mass of the adventurer and \(g\) is the gravitational acceleration of Earth. The tension \(\vec{T}\) has magnitude \(T\) and points upwards. Hence, the force diagram is shown in Figure 1.

Figure 1: Force diagram for the adventurer at the lowest point of the trajectory. The forces exerted on her are the weight \(\vec{W}\) and the tension \(\vec{T}\). The coordinate system is chosen such that the positive Y axis points upwards.

Because the adventurer moves in a circular path, she’ll experience a centripetal acceleration \(\vec{a}_c\), directed towards the center of the circular trajectory at all points and with magnitude \(\frac{v^2}{L}\), where \(v\) is the speed and \(L\) the radius of the trajectory. This is illustrated in Figure 2.

Figure 2: Tangent velocity \(\vec{v}\) and centripetal acceleration \(\vec{a}_c\) at the lowest point in the adventurer’s trajectory.

At the lowest part of the adventurer’s trajectory, the acceleration points upwards, so we can write an expression for the acceleration at the lowest point:

\begin{equation}
\label{ac}
\vec{a}_c=\frac{v^2}{L}\,\hat{\textbf{j}}.
\end{equation}

Since we know the forces and accelerations involved in the motion of the adventurer, we can use Newton’s Second Law:

\begin{equation}
\label{newton}
\sum \vec{F}=m\vec{a},
\end{equation}

where \(\sum \vec{F}\) is the sum of all the forces exerted on the adventurer and \(\vec{a}\) their acceleration. Since we know that the only forces are the weight \(\vec{W}\) and the tension \(\vec{T}\) and the acceleration at this point is the centripetal acceleration, we can write equation \eqref{newton} as

\begin{equation}
\vec{W}+\vec{T}=m\vec{a}_c.
\end{equation}

To continue, notice that given our coordinate system, we can write \(\vec{W}=-mg\,\hat{\textbf{j}} \) and \( \vec{T}=T\,\hat{\textbf{j}}\). Using the explicit expressions for the forces and the centripetal acceleration \eqref{ac} into the equation above, we obtain

\begin{equation}
-mg\,\hat{\textbf{j}}+T\,\hat{\textbf{j}}=m\frac{v^2}{L}\,\hat{\textbf{j}}.
\end{equation}

We can now drop the vector notation noting that all quantities are along the same axis:

\begin{equation}
\label{newton2}
-mg+T=m\frac{v^2}{L}.
\end{equation}

Now, we are told in the prompt that the adventurer can barely hold on to the \({liana}\) when the force is around twice her weight, which is \(2mg\). Because at the lowest point she can barely hold onto the \({liana}\), it means the tension has reached the limit of the force she can hold. So at this point, we have \(T=2mg\). Using this result in equation \eqref{newton2}, we obtain

\begin{equation}
-mg+2mg=m\frac{v^2}{L},
\end{equation}

which simplifies to

\begin{equation}
mg=m\frac{v^2}{L}.
\end{equation}

Cancelling out the mass yields

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

We can now solve for \(L\). First, multiply both sides by \(L\)

\begin{equation}
gL=v^2.
\end{equation}

Then, divide both sides by \(mg\):

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

Using the numerical values given in the prompt, we obtain

\begin{equation}
L=\frac{(7\,\text{m/s})^2}{9.8\,\text{m/s}^2},
\end{equation}

which yields

\begin{equation}
L=5\,\text{m}.
\end{equation}

Thus, the radius of the circular trajectory, which is given by the length of the \({liana}\), is 5 meters.

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