Chapter 8, Problem 81CP

(a)

To determine

The minimum speed with that Jane begins her swing to just make it to another side.

Answer to Problem 81CP

The minimum speed with that Jane begin her swing to just make it to other side is $\underset{\_}{6.15\text{m/s}}$.

Explanation of Solution

Consider the horizontal motion of Jane, as shown in figure (I) Jane moves from initial point to final point on the other side.

Write the expression for the total horizontal distance

  $D=\left(L\sin \varphi \right)+\left(L\sin \theta \right)$

Rearrange the above expression.

  $\sin \varphi =\frac{D-L\sin \theta }{L}$

Simplify the above expression.

  $\varphi ={\sin }^{-1}\left(\frac{D-L\sin \theta }{L}\right)$                                                                                   (I)

Here, $D$ is the width of the river, $L$ is the length of the vine and $\varphi$ is the angle between an initial position and an equilibrium position vector and $\theta$ is the angle between final position and an equilibrium position vector.

Consider the body-string-earth system as a non-isolated system. Since there is a force acting on the vine in a positive horizontal direction.

Write the equation for conservation of energy.

  $\Delta U+\Delta K=W$                                                                                              (II)

Here, $\Delta U$ is the change in potential energy and $\Delta K$ is the change in kinetic energy.

Write the equation for the change in potential energy

  $\begin{array}{l}\Delta U=U_f-U_i\\ =mg{h}_f-mg{h}_i\end{array}$

Simplify the above expression as.

  $\Delta U=mg\left(h_f-h_i\right)$                                                                                       (III)

Here, $m$ is the mass of the system, $g$ is the acceleration due to gravity, $h_f$ is the final height of the system and $h_i$ is the initial height of the system from equilibrium point.

The height can be calculated as shown in figure (I) from the equilibrium point.

Substitute $L-L\cos \varphi$ for $h_f$ and $L-L\cos \theta$ for $h_i$ in equation (III).

  $\begin{array}{l}\Delta U=mg\left(\left(L-L\cos \varphi \right)-\left(L-L\cos \theta \right)\right)\\ =mgL\left(\cos \theta -\cos \varphi \right)\end{array}$

As the body moves with an initial speed and final speed of the body is zero.

Write the expression for the change in kinetic energy as.

  $\Delta K=K_f-K_i$                                                                                             (IV)

Since the body moves with an initial speed and final speed of the body is zero. As the final speed of the body is zero than final the kinetic energy becomes zero.

Substitute $0$ for $k_f$ in equation (IV).

  $\begin{array}{l}\Delta K=0-\frac{1}{2}m{v}^2\\ =-\frac{1}{2}m{v}^2\end{array}$

Here, $v$ is the initial speed of the system.

Write the expression for work done by wind as.

  $W=-FD$

Here, $F$ is the force exerted by the wind on the system and $D$ is the distance traveled by the system under the force in a negative direction.

Substitute $mgL\left(\cos \theta -\cos \varphi \right)$ for $\Delta U$, $-\frac{1}{2}m{v}^2$ for $\Delta K$ and $-FD$ for $W$ in equation (II).

  $mgL\left(\cos \theta -\cos \varphi \right)+\left(-\frac{1}{2}m{v}^2\right)=-FD$

Rearrange the above equation.

  $\begin{array}{c}\frac{1}{2}m{v}^2=FD+mgL\left(\cos \theta -\cos \varphi \right)\\ v^2=\frac{2FD}{m}+2gL\left(\cos \theta -\cos \varphi \right)\\ v^2=2gL\left(\cos \theta -\cos \varphi \right)+\frac{2FD}{m}\end{array}$

Simplify the above equation.

  $v=\sqrt{2gL\left(\cos \theta -\cos \varphi \right)+\frac{2FD}{m}}$                                                                  (V)

Conclusion:

Substitute $50.0\text{m}$ for $D$, $40.0\text{m}$ for $L$ and $50.0°$ for $\theta$ in equation (I).

  $\begin{array}{c}\varphi ={\sin }^{-1}\left(\frac{\left(50.0\text{m}\right)-\left(40.0\text{m}\right)\sin \left(50.0°\right)}{\left(40.0\text{m}\right)}\right)\\ ={\sin }^{-1}\left(\frac{\left(50.0\text{m}\right)-\left(40.0\text{m}\right)0.766}{\left(40.0\text{m}\right)}\right)\\ ={\sin }^{-1}\left(0.4839\right)\\ =28.9°\end{array}$

Substitute $9.8{\text{m/s}}^{\text{2}}$ for $g$, $40.0\text{m}$ for $L$, $50\text{kg}$ for $m$, $50.0\text{m}$ for $D$, $110\text{N}$ for $F$, $50.0°$ for $\theta$ and  $28.9°$ for $\varphi$ in equation (IV).

  $\begin{array}{c}v=\sqrt{2\left(9.8{\text{m/s}}^{\text{2}}\right)\left(40.0\text{m}\right)\left(\cos \left(50.0°\right)-\cos \left(28.9°\right)\right)+\frac{2\left(110\text{N}\right)\left(50\text{m}\right)}{\left(50\text{kg}\right)}}\\ =\sqrt{\frac{\left(\left(784{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right)\left(0.6428-0.8755\right)\left(50\text{kg}\right)\right)+\left(11000\text{N}\cdot \text{m}\right)}{\left(50\text{kg}\right)}}\\ =\sqrt{37.8{\text{m}}^2{\text{/s}}^2}\\ \approx 6.15\text{m/s}\end{array}$

Thus, the minimum speed with that Jane begins her swing to just make it to the other side is $\underset{\_}{6.15\text{m/s}}$.

(b)

To determine

The minimum speed with that Jane and Tarzan begin their swing to just make it to the other side.

Answer to Problem 81CP

The minimum speed with that Jane and Tarzan begin their swing to just make it to other side is $\underset{\_}{9.87\text{m/s}}$.

Explanation of Solution

Consider the body-string-earth system as a non-isolated system. Since there is a force acting on the vine in a positive horizontal direction.

The height can be calculated as shown in figure (I) from the equilibrium point.

Substitute $L-L\cos \theta$ for $h_f$ and $L-L\cos \varphi$ for $h_i$ in equation (III).

  $\begin{array}{l}\Delta U=mg\left(\left(L-L\cos \theta \right)-\left(L-L\cos \varphi \right)\right)\\ =mgL\left(\cos \varphi -\cos \theta \right)\end{array}$

Since the body moves with an initial speed and final speed of the body is zero. As the final speed of the body is zero than final the kinetic energy becomes zero.

Substitute $0$ for $k_f$ in equation (IV).

  $\begin{array}{l}\Delta K=0-\frac{1}{2}m{v}^2\\ =-\frac{1}{2}m{v}^2\end{array}$

Here, $v$ is the initial speed of the system.

Write the expression for work done by wind as.

  $W=FD$

Here, $F$ is the force exerted by the wind on the system and $D$ is the distance traveled by the system under the force in a positive direction.

Substitute $mgL\left(\cos \varphi -\cos \theta \right)$ for $\Delta U$, $-\frac{1}{2}m{v}^2$ for $\Delta K$ and $FD$ for $W$ in equation (III).

  $mgL\left(\cos \varphi -\cos \theta \right)+\left(-\frac{1}{2}m{v}^2\right)=FD$

Rearrange the above equation.

  $v=\sqrt{\frac{2}{m}\left(mgL\left(\cos \varphi -\cos \theta \right)-FD\right)}$

Simplify the above equation.

  $v=\sqrt{2gL\left(\cos \varphi -\cos \theta \right)-\frac{2FD}{m}}$                                                                (VI)

Since the total mass of the system is the mass of Jane and Tarzan.

Conclusion:

Substitute $9.8{\text{m/s}}^{\text{2}}$ for $g$, $40.0\text{m}$ for $L$, $130\text{kg}$ for $m$, $50.0\text{m}$ for $D$, $110\text{N}$ for $F$, $50.0°$ for $\theta$ and $28.9°$ for $\varphi$ in equation (VI).

  $\begin{array}{l}v=\sqrt{2\left(9.8{\text{m/s}}^{\text{2}}\right)\left(40.0\text{m}\right)\left(\cos \left(28.9°\right)-\cos \left(50.0°\right)\right)-\frac{2\left(110\text{N}\right)\left(50\text{m}\right)}{\left(130\text{kg}\right)}}\\ =\sqrt{\frac{\left(\left(784{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right)\left(0.8755-0.6428\right)\left(130\text{kg}\right)\right)-\left(11000\text{N}\cdot \text{m}\right)}{\left(130\text{kg}\right)}}\\ =\sqrt{97.4{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\\ =9.87\text{m/s}\end{array}$

Thus, the minimum speed with that Jane and Tarzan begin their swing to just make it to the other side is $\underset{\_}{9.87\text{m/s}}$.