Chapter 7, Problem 81P

Jane, whose mass is 50.0 kg, needs to swing across a river (having width D) filled with person-eating crocodiles to save Tarzan from danger. She must swing into a wind exerting constant horizontal force $\stackrel{\to }{\text{F}}$, on a vine having length L and initially making an angle θ with the vertical (Fig. P7.81). Take D = 50.0 m, F = 110 N, L = 40.0 m, and θ = 50.0°. (a) With what minimum speed must Jane begin her swing to just make it to the other side? (b) Once the rescue is complete, Tarzan and Jane must swing back across the river. With what minimum speed must they begin their swing? Assume Tarzan has a mass of 80.0 kg.

To determine

Minimum speed to reach the opposite side.

Answer to Problem 81P

The speed is $v_i=\underset{\_}{6.15\text{m}/\text{s}}$.

Explanation of Solution

This system has to be solved using spherical polar coordinates with $L$ as the radius.

Considering the geometry, write the equation to determine $\varphi$

  $\begin{array}{c}D=L\sin \theta +L\sin \varphi \\ \varphi ={\sin }^{-1}\left(\frac{D-L\sin \theta }{L}\right)\end{array}$        (I)

Here $D$ is the displacement of the person, $\theta$ is the angle between radius and origin and $\varphi$ is the angle made with the z-axis

Write the equation for energy conservation of the system

    $\Delta K+\Delta U+\Delta E_{\mathrm{int}}=0$        (II)

Here $\Delta K$ is the change in kinetic energy, $\Delta U$ is the gravitational potential energy and $\Delta E_{\mathrm{int}}$ is the change in internal energy.

Write the equation for initial and final kinetic energy.

  $\begin{array}{c}{K}_i=\frac{1}{2}m{v}_i{}^2\\ K_f=0\end{array}$        (III)

Here $m$ is the mass of the block, $v_i$ is the initial velocity. Final velocity is zero

Write the expression for gravitational potential energy at A, B and C.

  $\begin{array}{c}{U}_i=mg\left(-L\cos \theta \right)\\ U_f=mg\left(-L\sin \varphi \right)\end{array}$        (IV)

Here $U_i$ is the initial gravitational potential energy, $U_f$ is the final gravitational potential energy $m$ is the mass of the block, $g$ is the acceleration due to gravity, $L$ is the radius, $\theta$ is the angle between radius and the origin and $\varphi$ is the angle made with the z-axis

Write down the equation for change in internal energy.

    $\Delta E_{\mathrm{int}}=FD$        (V)

Here $F$ is the force and $D$ is the displacement.

Substitute (III), (IV) and (V) in (II)

    $\left[0-\left(\frac{1}{2}m{v}_i{}^2\right)\right]+\left[mg\left(-L\sin \varphi \right)-mg\left(-L\cos \theta \right)\right]+FD=0$        (VI)

Rearrange (VI) in terms of $v_i$

    $\begin{array}{c}\left(\frac{1}{2}m{v}_i{}^2\right)=\left[mg\left(-L\sin \varphi \right)-mg\left(-L\cos \theta \right)\right]+FD\\ v_i=\frac{\sqrt{2\left[mg\left(-L\sin \varphi \right)-mg\left(-L\cos \theta \right)\right]+FD}}{m}\end{array}$        (VII)

Conclusion:

Substitute $40\text{m}$ for $L$, $50\text{m}$ for $D$ and $50°$ for $\theta$ in (I)

    $\begin{array}{c}\varphi ={\sin }^{-1}\left(\frac{50\text{m}-40\text{m}\sin 50°}{40\text{m}}\right)\\ =28.9°\end{array}$

Substitute $50\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $40\text{m}$ for $L$, $50\text{m}$ for $D$, $28.9°$ for $\varphi$, $110\text{N}$ for $F$ and $50°$ for $\theta$ in (VII)

    $\begin{array}{c}{v}_i=\frac{\sqrt{2\left[\left(50\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(-\left(40\text{m}\right)\sin 28.9°\right)-\left(50\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(-\left(40\text{m}\right)\cos 50°\right)\right]+\left(110\text{N}\right)\left(50\text{m}\right)}}{50\text{kg}}\\ =6.15\text{m}/\text{s}\end{array}$

The speed is $v_i=\underset{\_}{6.15\text{m}/\text{s}}$.

To determine

Minimum speed for combined swing.

Answer to Problem 81P

The speed is $\underset{\_}{9.87\text{m}/\text{s}}$.

Explanation of Solution

Write down the equation for energy conservation

    $\Delta K+\Delta U=\Delta E_{\text{mech}}$        (VIII)

Here $\Delta E_{\text{mech}}$ is the mechanical energy.

Write down the equation for mechanical energy

        $\Delta E_{\text{mech}}=FD$        (IX)

Here $F$ is the force and $D$ is the displacement.

Substitute (III), (IV) and (IX) in (VIII)

    $\left[0-\left(\frac{1}{2}m{v}_i{}^2\right)\right]+\left[mg\left(-L\sin \varphi \right)-mg\left(-L\cos \theta \right)\right]=FD$        (X)

Here $m$ is the combined mass of two persons.

Rearrange (X) in terms of $v_i$ .

    $\begin{array}{c}\left(\frac{1}{2}m{v}_i{}^2\right)=FD-\left[mg\left(-L\sin \varphi \right)-mg\left(-L\cos \theta \right)\right]\\ v_i=\frac{\sqrt{2\left(FD-\left[mg\left(-L\sin \varphi \right)-mg\left(-L\cos \theta \right)\right]\right)}}{m}\end{array}$        (XI)

Conclusion:

Substitute $50\text{kg+80}\text{kg=130}\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $40\text{m}$ for $L$, $50\text{m}$ for $D$, $28.9°$ for $\varphi$, $110\text{N}$ for $F$ and $50°$ for $\theta$ in (XI)

    $\begin{array}{c}{v}_i=\frac{\sqrt{2\left(\left(110\text{N}\right)\left(50\text{m}\right)-\left[\left(\text{130}\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(-\left(40\text{m}\right)\sin 28.9°\right)-\left(\text{130}\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(-\left(40\text{m}\right)\cos 28.9°\right)\right]\right)}}{\text{130}\text{kg}}\\ =9.87\text{m}/\text{s}\end{array}$

The speed is $\underset{\_}{9.87\text{m}/\text{s}}$.