91Ó°ÊÓ

When the Tarzan of mass m moves in a circular arc of radius rhaving a constant speed v, it will need a force that will keep the Tarzan moving in the circular motion. This type of force acts on the Tarzan towards the arc center, which can be termed the centripetal force.

Mathematically, it is written as:

$F_{\mathrm{R}}=\frac{m{v}^2}{r}$

Here, Tarzan is moving in a circle with the centripetal force pointing upwards at the lowest point of his swing.

• The maximum tension that can be created in the vine is, $T_{\max .}=1150\mathrm{N}$.
• The length of the vine is, l = 4.7 m.
• The mass of the Tarzan is, m = 78 kg.

Let the maximum speed that Tarzan can tolerate at the lowest point of the swing be v.

The free-body diagram of the Tarzan at the lowest point of the swing is:

Here, the forces acting on the Tarzan are. his weight (mg) acting downwards and tension in the vine acting upwards (T).

As it is clear from the figure, the net force acting on the Tarzan in the vertical direction is, $\left(T-mg\right)$. This force must provide the centripetal force required by the Tarzan to swing in an arc. Thus, the centripetal force is:

$\begin{array}{c}{F}_R=T-mg\\ m\frac{v^2}{r}=T-mg\end{array}$

Thus, the expression for the speed of the Tarzan can be written as:

$v=\sqrt{\frac{\left(T-mg\right)r}{m}}$

At the lowest point of the swing, the speed will be the maximum if tension in the vine is maximum. Thus, the maximum speed is:

$v_{\max }=\sqrt{\frac{\left(T_{\max }-mg\right)r}{m}}$

Substitute all the values in the above equation.

$\begin{array}{c}{v}_{\max }=\sqrt{\frac{\left[1150\mathrm{N}-\left(78\mathrm{kg}×9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(\frac{1\mathrm{N}}{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}\right)\right]×4.7\mathrm{m}}{78\mathrm{kg}}}\\ =\sqrt{23.235}{\mathrm{m}}^2/{\mathrm{s}}^2\\ =4.82\mathrm{m}/\mathrm{s}\end{array}$

Thus, the maximum speed that Tarzan can tolerate is 4.82 m/s.