Chapter 8, Problem 82CP

(a)

To determine

The maximum height the ball attains when it swings up.

Answer to Problem 82CP

The maximum height of the ball when it swings up is $\underset{\_}{\frac{1.6}{1+\frac{8.64{\text{N}}^2}{F^2}}}$.

Explanation of Solution

Consider the ball-string-wind as a system under gravitational force.

Since the work done by the wind on the system is transformed into potential energy of the system.

Write the equation for conservation of energy

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

Here, $\Delta U$ is the change in potential energy, $\Delta K$ is the change in kinetic energy and $W$ is the work done by the force exerted by the wind.

Write the expression for the change in potential energy

  $\Delta U=U_f-U_i$                                                                                               (II)

Here, $U_f$ is the final potential energy of the system and $U_i$ is the initial potential energy of the system.

Since the kinetic energy of the systems remains zero because in both initial and final cases the system is not in rest.

Write the expression for potential energy

    $U=mgh$

Here, $m$ is the mass of the system, $g$ is the acceleration due to gravity and $h$ is the height of the system from the ground.

.

Substitute $mg{h}_f$ for $U_f$ and $mg{h}_i$ for $U_i$ in equation (II).

    $\begin{array}{l}\Delta U=\left(mg{h}_f\right)-\left(mg{h}_i\right)\\ =mg\left(h_f-h_i\right)\end{array}$

Here, $h_f$ is the final height of the system from the ground and $h_i$ is the initial height of the system from the ground.

Since the difference in initial and final height is given by $H$.

    $H=h_f-h_i$

Substitute $h_f-h_i$ for $H$ in above expression as.

  $\Delta U=mgH$

Substitute $mgH$ for $\Delta U$ and $0$ for $\Delta K$ in equation (I).

  $mgH+0=W$

Simplify the above equation.

  $W=mgH$

Write the expression for work done by the force

  $W=Fd$

Here, $F$ is the force exerted by wind and $d$ is the distance in the direction of force.

Substitute $Fd$ for $W$ in above equation.

  $Fd=mgH$                                                                                                 (III)

Write the expression for distance in the direction of the force exerted from the figure (I).

    $d=\sqrt{L^2-{\left(L-H\right)}^2}$

Here, $d$ is the horizontal distance traveled by the ball, $L$ is the height of the string and $H$ is the vertical distance traveled by the ball.

Simplify the above equation.

    $d=\sqrt{2LH-H^2}$                                                                                         (IV)

Substitute $\sqrt{2LH-H^2}$ for $d$ in equation (III).

    $F\sqrt{2LH-H^2}=mgH$

Simplify the above equation.

    $\begin{array}{l}{F}^2\left(2LH-H^2\right)=m^2{g}^2{H}^2\\ 2H{F}^{2}L-H^2{F}^2=m^2{g}^2{H}^2\\ \left(F^2+m^2{g}^2\right)H^2-2L{F}^{2}H=0\\ H\left(\left(F^2+m^2{g}^2\right)H-2L{F}^2\right)=0\end{array}$

Since one root of the above equation is zero, so take non-zero value of $H$.

Solve the above equation.

    $\begin{array}{l}H\left(F^2+m^2{g}^2\right)-2L{F}^2=0\\ H=\frac{2L{F}^2}{F^2+m^2{g}^2}\end{array}$

Simplify the above equation to find the expression for $H$

    $H=2L\left(\frac{1}{1+{\left(\frac{mg}{F}\right)}^2}\right)$                                                                                   (V)

Conclusion:

Substitute $300\text{g}$ for $m$, $9.8{\text{m/s}}^{\text{2}}$ for $g$ and $80.0\text{cm}$ for $L$ in equation (V).

    $\begin{array}{l}H=2\left(80.0\text{cm}\right)\left(\frac{1}{1+{\left(\frac{\left(300\text{g}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)}{F}\right)}^2}\right)\\ =160.0\left(\frac{1\text{m}}{100\text{cm}}\right)\left(\frac{1}{1+\left(\frac{{\left(9.8{\text{m/s}}^{\text{2}}\right)}^2{\left(300\left(\frac{1\text{kg}}{1000\text{g}}\right)\right)}^2}{F^2}\right)}\right)\\ =\frac{1.6\text{m}}{1+\left(\frac{\left(96.04{\text{m}}^2{\text{/s}}^4\right)\left(0.09{\text{kg}}^{\text{2}}\right)}{F^2}\right)}\end{array}$

Simplify the above equation.

    $H=\frac{1.6}{1+\frac{8.64{\text{N}}^{\text{2}}}{F^2}}$                                                                                        (VI)

Thus, the maximum height of the ball when it swings up is $\underset{\_}{\frac{1.6}{1+\frac{8.64{\text{N}}^2}{F^2}}}$.

(b)

To determine

The maximum height the ball attains when the force exerted by wind is $1\text{N}$.

Answer to Problem 82CP

The maximum height attained by the ball, when the force exerted by wind is $1\text{N}$. is $\underset{\_}{0.166\text{m}}$.

Explanation of Solution

Conclusion:

Substitute $1.00\text{N}$ for $F$ in equation (V).

  $\begin{array}{c}H=\frac{1.6\text{m}}{1+\frac{8.64{\text{N}}^{\text{2}}}{{\left(1.00\text{N}\right)}^2}}\\ =\frac{1.6\text{m}}{9.64}\\ =0.16597\text{m}\\ \approx 0.166\text{m}\end{array}$

Thus, the maximum height attained by the ball, when the force exerted by wind is $1\text{N}$ is $\underset{\_}{0.166\text{m}}$.

(c)

To determine

The maximum height the ball attains when the force exerted by wind is $10\text{N}$.

Answer to Problem 82CP

The maximum height attained by the ball, when the force exerted by wind is $10\text{N}$. is $\underset{\_}{1.47\text{m}}$.

Explanation of Solution

Consider the equation (V) for the maximum height attained due to the force exerted by the wind.

The ball swings in the upward direction by the wind force exerted on the ball.

Conclusion:

Substitute $10.0\text{N}$ for $F$ in equation (VI).

  $\begin{array}{c}H=\frac{1.6\text{m}}{1+\frac{8.64{\text{N}}^{\text{2}}}{{\left(10.0\text{N}\right)}^2}}\\ =\frac{1.6\text{m}}{1.0864}\\ =1.47275\text{m}\\ \approx 1.47\text{m}\end{array}$

Thus, The maximum height attained by the ball, when the force exerted by wind is $10\text{N}$. is $\underset{\_}{1.47\text{m}}$.

(d)

To determine

The maximum height when the force exerted by wind approaches to zero.

Answer to Problem 82CP

The maximum height approaches to zero when $F$ approaches to zero.

Explanation of Solution

Write the expression for the maximum height in the limit form as.

    $\begin{array}{c}\underset{F\to 0}{\lim }H=\underset{F\to 0}{\lim }\left(\frac{1.6\text{m}}{1+\frac{8.64{\text{N}}^2}{F^2}}\right)\\ =\frac{1.6\text{m}}{1+\frac{8.64{\text{N}}^2}{\underset{F\to 0}{\lim }{F}^2}}\end{array}$

Substitute $0$ for $\underset{F\to 0}{\lim }{F}^2$ in the above equation.

    $\begin{array}{c}H=\frac{1.6\text{m}}{1+\frac{8.64{\text{N}}^2}{0}}\\ =\frac{1.6\text{m}}{1+\infty }\\ =\frac{1.6\text{m}}{\infty }\\ H=0\end{array}$

Conclusion:

Thus, the maximum height approaches to zero when $F$ approaches to zero.

(e)

To determine

The behavior of maximum height when the force exerted by wind approaches to infinity.

Answer to Problem 82CP

The maximum height approaches to $\underset{\_}{1.6\text{m}}$ when $F$ approaches to infinity.

Explanation of Solution

Write the expression for the maximum height in the limit form as.

    $\begin{array}{c}\underset{F\to \infty }{\lim }H=\underset{F\to \infty }{\lim }\left(\frac{1.6\text{m}}{1+\frac{8.64{\text{N}}^2}{F^2}}\right)\\ =\frac{1.6\text{m}}{1+\frac{8.64{\text{N}}^2}{\underset{F\to \infty }{\lim }{F}^2}}\end{array}$

Substitute $0$ for $\left(\underset{F\to \infty }{\lim }\frac{8.64{\text{N}}^{\text{2}}}{F^2}\right)$ in the above equation.

    $\begin{array}{l}H=\frac{1.6\text{m}}{1+0}\\ =1.6\text{m}\end{array}$

Thus, the maximum height approaches to $\underset{\_}{1.6\text{m}}$ when $F$ approaches to infinity.

(f)

To determine

The equivalent height of the ball with the wind blowing.

Answer to Problem 82CP

The equivalent height of the ball with the wind blowing is $\underset{\_}{\left(0.8\text{m}\right)\left(1-\sqrt{\frac{1}{1+\frac{F^2}{8.64{\text{N}}^{\text{2}}}}}\right)}$.

Explanation of Solution

Consider the system in an equilibrium state with net force zero as shown in figure (II).

Write the equation for net force in a vertical direction from the figure (II)

    $\sum F_y=T\cos \theta -mg$

Here, $\sum F_y$ is the net force in a positive y-axis direction. $T$ is the force due to tension in the string, $\theta$ is the angle between the initial and final position of the ball-string as shown in the figure and $m$ is the mass of the ball and $g$ is the acceleration due to gravity.

When the ball-string system is in equilibrium, the net force will be zero in all directions.

  $\sum F_{axis}=0$

Here, $F_{axis}$ is algebric sum of required direction or axis.

Substitute $0$ for $\sum F_y$ in above expression as.

    $T\cos \theta -mg=0$

Simplify the above equation.

    $\cos \theta =\frac{mg}{T}$                                                                                                (VII)

Write the equation for net force in a horizontal direction from the figure (II)

    $\sum F_x=F-T\sin \theta$

Here, $\sum F_x$ is the net force in the positive x-axis direction and $F$ is the force exerted by the wind.

Substitute $0$ for $\sum F_x$ in above expression as.

    $F-T\sin \theta =0$

Simplify the above equation.

    $\sin \theta =\frac{F}{T}$                                                                                                  (VIII)

Divide equaiton (VIII) by equation (VII).

  $\begin{array}{c}\frac{\cos \theta }{\sin \theta }=\frac{\left(\frac{mg}{T}\right)}{\left(\frac{F}{T}\right)}\\ \tan \theta =\frac{F}{mg}\end{array}$

Write the expression for $\cos \theta$ in terms of $\tan \theta$

  $\cos \theta =\sqrt{\frac{1}{1+{\tan }^2\theta }}$

Substitute $\frac{F}{mg}$ for $\tan \theta$ in the above equation.

    $\cos \theta =\sqrt{\frac{1}{1+\frac{F^2}{m^2{g}^2}}}$                                                                                    (IX)

Write the expression for equivalent height from the figure (II).

    $h_{eq}=L-L\cos \theta$                                                                                           (X)

Here, $h_{eq}$ is the equivalent height.

    $\frac{mg}{F}=\frac{1}{\tan \theta }$

Substitute $\frac{1}{\tan \theta }$ for $\frac{mg}{F}$ in equation (VI).

    $H=2L\left(\frac{1}{1+{\left(\frac{1}{\tan \theta }\right)}^2}\right)$                                                                              (XI)

Conclusion:

Substitute $300\text{g}$ for $m$ and $9.8{\text{m/s}}^{\text{2}}$ for $g$ in equation (IX).

    $\begin{array}{c}\cos \theta =\sqrt{\frac{1}{1+\frac{F^2}{{\left(300\text{g}\right)}^2{\left(9.8{\text{m/s}}^{\text{2}}\right)}^2}}}\\ =\sqrt{\frac{1}{1+\frac{F^2}{{\left(300\text{g}\right)}^2{\left(\frac{1\text{kg}}{1000\text{g}}\right)}^2{\left(9.8{\text{m/s}}^{\text{2}}\right)}^2}}}\\ =\sqrt{\frac{1}{1+\frac{F^2}{\left(0.09{\text{kg}}^{\text{2}}\right)\left(96.04{\text{m}}^2{\text{/s}}^4\right)}}}\\ =\sqrt{\frac{1}{1+\frac{F^2}{8.64{\text{N}}^{\text{2}}}}}\end{array}$

Substitute $\sqrt{\frac{1}{1+\frac{F^2}{8.64{\text{N}}^{\text{2}}}}}$ for $\cos \theta$ and $80\text{cm}$ for $L$ in equation (X).

    $\begin{array}{l}{h}_{eq}=\left(80\text{cm}\right)\left(1-\sqrt{\frac{1}{1+\frac{F^2}{8.64{\text{N}}^{\text{2}}}}}\right)\\ =\left(80\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\left(1-\sqrt{\frac{1}{1+\frac{F^2}{8.64{\text{N}}^{\text{2}}}}}\right)\end{array}$

Simplify the above expression as.

    $h_{eq}=\left(0.8\text{m}\right)\left(1-\sqrt{\frac{1}{1+\frac{F^2}{8.64{\text{N}}^{\text{2}}}}}\right)$                                                              (XII)

Thus, the equivalent height of the ball with the wind blowing is $\underset{\_}{\left(0.8\text{m}\right)\left(1-\sqrt{\frac{1}{1+\frac{F^2}{8.64{\text{N}}^{\text{2}}}}}\right)}$.

(g)

To determine

The equivalent height when $10\text{N}$ of force is exerted by the wind.

Answer to Problem 82CP

The equivalent height when $10\text{N}$ of force is exerted by wind is $\underset{\_}{0.574\text{m}}$.

Explanation of Solution

Conclusion:

Substitute $10\text{N}$ for $F$ in equation (XII).

    $\begin{array}{c}{h}_{eq}=\left(0.8\text{m}\right)\left(1-\sqrt{\frac{1}{1+\frac{{\left(10\text{N}\right)}^2}{8.64{\text{N}}^{\text{2}}}}}\right)\\ =\left(0.8\text{m}\right)\left(1-\sqrt{\frac{1}{1+11.57}}\right)\\ =0.5746\\ \approx 0.574\text{m}\end{array}$

Thus, the equivalent height when $10\text{N}$ of force is exerted by wind is $\underset{\_}{0.574\text{m}}$.

(h)

To determine

The equivalent height when the force exerted by wind approaches to infinity.

Answer to Problem 82CP

The equivalent height when the force exerted by wind approaches to infinity is $\underset{\_}{0.8\text{m}}$.

Explanation of Solution

Consider the equivalent height from the equation (XII).

Write the expression for equivalent height from the equation (XII).

    $h_{eq}=0.8\left(1-\sqrt{\frac{1}{1+\frac{F^2}{8.64}}}\right)$

Write the equation for $\underset{F\to \infty }{\lim }{h}_{eq}$ from the above equation.

    $\begin{array}{l}\underset{F\to \infty }{\lim }{h}_{eq}=\underset{F\to \infty }{\lim }\left(0.8\text{m}\right)\left(1-\sqrt{\frac{1}{1+\frac{F^2}{8.64{\text{N}}^{\text{2}}}}}\right)\\ =\left(0.8\text{m}\right)\left(1-\sqrt{\underset{F\to \infty }{\lim }\frac{1}{1+\frac{F^2}{8.64{\text{N}}^{\text{2}}}}}\right)\end{array}$

Substitute $0$ for $\left(\sqrt{\underset{F\to \infty }{\lim }\frac{1}{1+\frac{F^2}{8.64}}}\right)$ in the above equation.

    $\begin{array}{l}\underset{F\to \infty }{\lim }{h}_{eq}=0.8\text{ m}\left(1-0\right)\\ =0.8\text{ m}\end{array}$

Conclusion:

Thus, the equivalent height when the force exerted by wind approaches to infinity is $\underset{\_}{0.8\text{m}}$.