Chapter 8, Problem 85CP

(a)

To determine

The length of the cord.

Answer to Problem 85CP

The length of the cord is $\underset{\_}{25.8\text{m}}$.

Explanation of Solution

Consider the body-cord-baloon as a isolated system under gravitation force.

Since the potential energy of the daredavil is transformed to spring energy of the cord.

Write the equation for conservation of energy

  $\Delta U+\Delta K=0$

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

Since there is no initial and final motion and hence the change in kintetic energy is zero.

Subsitute $0$ for $\Delta K$ in above equation.

  $\Delta U=0$                                                                                                          (I)

Write the expression for total potential energy

  $U=U_s+U_g$

Here, $U_s$ is potential energy of the cord under spring force and $U_g$ is potential energy of system due to gravity.

Write the expression for potential energy of the cord of length $L\text{m}$ under spring force.

  $U_s=\frac{1}{2}{k}_L{x}^2$

Write the expression for potential energy due to gravity

  $U_g=mgh$

Here, $h$ is the height of the system.

Write the expression for change in total potential energy

  $\begin{array}{l}\Delta U=U_f-U_i\\ =\left(U_{sf}+U_{gf}\right)-\left(U_{si}+U_{gi}\right)\end{array}$

Here, $U_{sf}$ is final potential energy of cord of length $L\text{m}$, $U_{si}$ is initial potential energy of the cord of length $L\text{m}$, $U_{gf}$ is the final potential energy due to gravity and $U_{gi}$ is the initial potential energy due to gravity.

Substitute $\frac{1}{2}{k}_L{x}^2$ for $U_{sf}$, $mg{h}_f$ for $U_{g,f}$, $0$ for $U_{s,i}$ and $mg{h}_i$ for $U_{g,i}$ in above equation.

  $\Delta U=\left(\frac{1}{2}{k}_L{x}^2+mg{h}_f\right)-\left(0+mg{h}_i\right)$

Here, $h_f$ is the final height of the system of body-cord-balloon system and $h_i$ is the initial height of the system.

Simplify the above equation.

  $\Delta U=mg(h_f-h_i)+\frac{1}{2}{k}_L{x}^2$                                                                          (II)

Since the daredavil finds out the length of stretched cord when hanging at rest by it.

Write the expression for equation for force from Figure 1.

  $F_g=F_s$

Here, $F_g$ is the force due to gravity or body weigh and $F_s$ is the spring force due to elastic cord.

Write the expression for body weight

  $F_g=mg$

Here, $m$ is mass and $g$ is acceleration due to gravity.

Write the expression for spring force

  $F_s=k_{L}x$                                                                                                       (III)

Here, $k_L$ is the spring constant for elastic cord of $L\text{m}$ length and $x$ is the stretched length of cord.

Substitute $k_{L}x$ for $F_s$ and $mg$ for $F_g$ in above equation.

  $mg=k_{L}x$

Since the cord is uniform and elastic.

Hence, spring constant of cord is proportional to the length of the cord.

Rearange the above equation.

  $k_L=\frac{mg}{x}$                                                                                                     (IV)

When the daredavil test the cord he finds out that his body weight stretches it.

The rate of stretch of the spring or cord is depends on the length and stiffness of the spring or cord. The rate of stretches is the ratio of change in length or stretches to the orginal length of the spring or cord. If the the length of the spring or cord is $L$ than the stretches in length will becomes the product of rate of stretches to the length of the spring or cord.

Substitute $\frac{1.5}{5}L$ for $x$ in equation (III).

  $k_L=\frac{mg}{\left(\frac{1.5}{5}L\right)}$

Simplify the above equation.

  $k_L=\frac{10}{3}\left(\frac{mg}{L}\right)$                                                                                               (V)

Substitute $\frac{10}{3}\left(\frac{mg}{L}\right)$ for $k_L$ in equation (II).

  $\Delta U=mg(h_f-h_i)+\frac{1}{2}\left(\frac{10}{3}\left(\frac{mg}{L}\right)\right)x^2$

Since the length of stretched cord should be the equal to the length of the jump.

Write the equation for streched length of the cord which has $\left(h_i-h_f\right)$ stretched length

  $x=\left(h_i-h_f\right)-L$

Sustitute $\left(h_i-h_f\right)-L$ for $x$ in above equation.

  $\Delta U=mg(h_f-h_i)+\frac{1}{2}\left(\frac{10}{3}\frac{mg}{L}\right){\left(\left(h_i-h_f\right)-L\right)}^2$                                                 (VI)

Conclusion:

Substitute $0$ for $\Delta U$ in equation (IV).

  $0=mg(h_f-h_i)+\frac{1}{2}\left(\frac{10}{3}\frac{mg}{L}\right){\left(\left(h_i-h_i\right)-L\right)}^2$

Simplify the above expression in terms of $L$.

  $\begin{array}{l}0=mg(h_f-h_i)+\frac{1}{2}\left(\frac{10}{3}\frac{mg}{L}\right){\left(\left(h_i-h_f\right)-L\right)}^2\\ 0=(h_f-h_i)+\left(\frac{5}{3L}\right){\left(\left(h_i-h_f\right)-L\right)}^2\end{array}$

Substitute $65\text{m}$ for $h_i$ and $10\text{m}$ for $h_f$ in above equation.

  $\begin{array}{c}(10-65)+\left(\frac{5}{3L}\right){\left(\left(65-10\right)-L\right)}^2=0\\ -55+\frac{5}{3L}{\left(55-L\right)}^2=0\\ -11+\frac{{55}^2+L^2-110L}{3L}=0\\ L^2-143L+3225=0\end{array}$

Solve the above quadratic equation.

  $\begin{array}{c}L=\frac{-\left(-143\right)\pm \sqrt{{\left(-143\right)}^2-4\left(1\right)\left(3025\right)}}{2}\\ =\frac{143\pm \sqrt{8349}}{2}\\ =\frac{143\pm 91.373}{2}\\ =25.813\text{or}117.186\end{array}$

Since the length of the cord cannot be more than $55\text{m}$ which is the total jump height.

Thus, the length of the cord is $\underset{\_}{25.8\text{m}}$.

(b)

To determine

The maximum acceleration which daredevil will experience.

Answer to Problem 85CP

The maximum acceleration which daredevil will experience is $\underset{\_}{27.1{\text{m/s}}^{\text{2}}}$.

Explanation of Solution

Write the equation for motion of daredevil

  $F_s-mg=ma$

Here, $F_s$ is the force acting on daredevil by the cord to which he is hanging, $m$ is the mass and $a$ is acceleration due to gravity.

Substitute $k_{l}x$ for $F_s$ in equation above from equation (III).

  $k_{l}x-mg=ma$

Solve the above equation to find the expression for $a$.

  $a=\frac{k_{l}x-mg}{m}$

Substitute $\frac{10}{3}\frac{mg}{L}$ for $k_l$ from equation (V) in above equation.

  $a=\frac{\left(\frac{10}{3}\frac{mg}{L}\right)x-mg}{m}$

Simplify the above equation.

  $a=g\left(\frac{10x}{3L}-1\right)$                                                                                          (VII)

Write the expression for stretched cord length

  $X=L+x$

Rearrange the above equation.

  $x=X-L$                                                                                                (VIII)

Here, $X$ is the length of cord after stretching it, $x$ is the stretched length of cord and $L$ is the actual length of cord.

When the daredevil uses the cord of $25.8\text{m}$ length as concluded in part (a). the daredevil experiences maximum acceleration when all of the spring force is being applied to the body or when the cord is fully stretched.

Conclusion:

Substitute $55\text{m}$ for $X$ and $25.8\text{m}$ for $L$ in equation (VIII).

  $\begin{array}{c}x=\left(55\text{m}\right)-\left(25.8\text{m}\right)\\ =29.2\text{m}\end{array}$

Substitute $29.2\text{m}$ for $x$, $9.8{\text{m/s}}^{\text{2}}$ for $g$ and $25.8\text{m}$ for $L$ in equation (VII).

  $\begin{array}{c}a=\left(9.8{\text{m/s}}^{\text{2}}\right)\left(\frac{10\left(29.2\text{m}\right)}{3\left(25.8\text{m}\right)}-1\right)\\ =27.1{\text{m/s}}^2\end{array}$

Thus, the maximum acceleration which daredevil will experience is $\underset{\_}{27.1{\text{m/s}}^{\text{2}}}$.