Chapter 15, Problem 15.30P

Review. A 65.0-kg bungee jumper steps off a bridge with a light bungee cord tied to her body and to the bridge. The outstretched length of the cord is 11.0 m. The jumper reaches the bottom of her motion 36.0 m below the bridge before bouncing back. We wish to find the time interval between her leaving the bridge and her arriving at the bottom of her motion. Her overall motion can be separated into an 11.0-m tree fall and a 25.0-m section of simple harmonic oscillation. (a) For the free-fall part, what is the appropriate analysis model to describe her motion? (b) For what time interval is she in free fall? (c) For the simple harmonic oscillation part of the plunge, is the system of the bungee jumper, the spring, and the Earth isolated or nonisolated? (d) From your response in part (c) find the spring constant of the bungee cord. (c) What is the location of the equilibrium point where the spring force balances the gravitational force exerted on the jumper? (f) What is the angular frequency of the oscillation? (g) What time interval is required for the cord to stretch by 25.0 m? (h) What is the total time interval for the entire 36.0-m drop?

To determine

The appropriate analysis model of jumper’s motion for the free fall part.

Answer to Problem 15.30P

The jumper’s motion has the constant acceleration.

Explanation of Solution

A free falling object is an object that is falling under the sole influence of gravity.

Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. The free fall phase follows the parabolic behavior. Since the only gravity acted on the free fall, the acceleration is constant.

Conclusion:

Therefore, the jumper’s motion has the constant acceleration.

To determine

The time required for free fall.

Answer to Problem 15.30P

The time required for free fall is $1.49\text{s}$.

Explanation of Solution

The mass of bungee jumper is $65.0\text{kg}$, the stretched length of the cord and free fall height is $11.0\text{m}$, the distance where the jumper reaches before bouncing back is $36.0\text{m}$ and the height of simple harmonic oscillation is $25.0\text{m}$.

The equation for the kinematic is,

    $y_{\text{f}}=y_{\text{i}}+v_{\text{y}}t+\frac{1}{2}{a}_{\text{y}}{t}^2$

Here, $y_{\text{f}}$ is the free fall height, $t$ is the time, $y_{\text{i}}$ is the initial height, $v_{\text{y}}$ is the velocity and $a_{\text{y}}$ is the acceleration.

Substitute $-11.0\text{m}$ for $y_{\text{f}}$, $0$ for $y_{\text{i}}$, $0$ for $v_{\text{y}}$ and $-9.8\text{m}/{\text{s}}^2$ for $a_{\text{y}}$ in above equation to find $t$.

  $\begin{array}{c}-11.0\text{m}=0+\left(0\right)t+\frac{1}{2}\left(-9.8\text{m}/{\text{s}}^2\right)t^2\\ =\frac{1}{2}\left(-9.8\text{m}/{\text{s}}^2\right)t^2\\ t=\sqrt{\frac{\left(2\times 11.0\text{m}\right)}{9.8\text{m}/{\text{s}}^2}}\\ =1.49\text{s}\end{array}$

Conclusion:

Therefore, the time required for free fall is $1.49\text{s}$.

To determine

Weather the system of the bungee jumper, the spring and the earth is isolated or non-isolated for simple harmonic oscillation.

Explanation of Solution

When a system is isolated, it means that it is separated from its environment in such a way that no energy flows on or out of the system. The non-isolated system interacts with its environment and exchanges the energy.

The energy of the system of the bungee jumper, the spring and the earth is exchanged only with each other not outside from the system. Since the earth and spring act on the jumper, the system is isolated.

Conclusion:

Therefore, the system the bungee jumper, the spring and the earth is isolated.

To determine

The spring constant of the bungee cord.

Answer to Problem 15.30P

The spring constant of the bungee cord is $73.38\text{N}/{\text{m}}^2$.

Explanation of Solution

Write an expression of the law of conservation of the energy for the bungee jumper

    $mgh=\frac{1}{2}k{h}_1^2$

Here, $m$ is the mass of jumper, $g$ is the acceleration due to gravity, $h$ is the total height, $k$ is the spring constant and $h_{\text{1}}$ is the height of simple harmonic oscillation.

Substitute $65.0\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^2$ for $g$, $36.0\text{m}$ for $h$ and $25.0\text{m}$ for $h_1$ in above equation to find $k$.

    $\begin{array}{c}\left(65.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)\left(36.0\text{m}\right)=\frac{1}{2}k{\left(25.0\text{m}\right)}^2\\ k=\frac{2\left(65.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)\left(36.0\text{m}\right)}{{\left(25.0\text{m}\right)}^2}\\ =73.38\text{N}/{\text{m}}^2\end{array}$

Conclusion:

Therefore, the spring constant of the bungee cord is $73.38\text{N}/{\text{m}}^2$.

To determine

The location of the equilibrium point where the spring force balances the gravitational force exerted on the jumper.

Answer to Problem 15.30P

The location of the equilibrium point where the spring force balances the gravitational force exerted on the jumper is $16.32\text{m}$.

Explanation of Solution

The equation for the equilibrium point from the lorded is,

    $\begin{array}{c}kx=mg\\ x=\frac{mg}{k}\end{array}$

Here, $x$ is the position of the spring.

The equilibrium point is calculated as,

    $y=L+x$

Here, $L$ is the stretched length of the chord.

Substitute $\frac{mg}{k}$ for $x$ in above expression.

    $y=L+\left(\frac{mg}{k}\right)$

Substitute $65.0\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^2$ for $g$, $73.38\text{N}/{\text{m}}^2$ for $k$ and $11.0\text{m}$ for $L$ in above equation to find $y$.

    $\begin{array}{c}y=\left(11.0\text{m}\right)+\frac{\left(65.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)}{\left(73.38\text{N}/{\text{m}}^2\right)}\\ =19.68\text{m}\end{array}$

The amplitude of the motion is,

    $A=h-y$

Substitute $36.0\text{m}$ for $h$ and $19.68\text{m}$ for $y$ in above equation to find $A$.

    $\begin{array}{c}A=\left(36.0\text{m}\right)-\left(19.68\text{m}\right)\\ =16.32\text{m}\end{array}$

Conclusion:

Therefore, the location of the equilibrium point where the spring force balances the gravitational force exerted on the jumper is $16.32\text{m}$.

To determine

The angular frequency of the oscillation.

Answer to Problem 15.30P

The angular frequency of the oscillation is $1.06\text{rad}/\text{s}$.

Explanation of Solution

The formula to calculate angular frequency of the oscillation is,

    $\omega =\sqrt{\frac{k}{m}}$

Substitute $65.0\text{kg}$ for $m$ and $73.38\text{N}/{\text{m}}^2$ for $k$ in above equation to find $\omega$.

    $\begin{array}{c}\omega =\sqrt{\frac{73.38\text{N}/{\text{m}}^2}{65.0\text{kg}}}\\ =1.06\text{rad}/\text{s}\end{array}$

Conclusion:

Therefore, the angular frequency of the oscillation is $1.06\text{rad}/\text{s}$.

To determine

The time interval required for the cord to stretched by $25.0\text{m}$.

Answer to Problem 15.30P

The time interval required for the cord to stretched by $25.0\text{m}$ is $2.0\text{s}$.

Explanation of Solution

The expression for the position of a particle in simple harmonic motion is,

    $x=A\cos \left(\omega t_{\text{1}}\right)$

Substitute $-\frac{mg}{k}$ for $x$ in above expression.

    $-\frac{mg}{k}=A\cos \left(\omega t_{\text{1}}\right)$

Substitute $65.0\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^2$ for $g$, $73.38\text{N}/{\text{m}}^2$ for $k$, $16.32\text{m}$ for $A$ and $1.06\text{rad}/\text{s}$ for $\omega$ in above equation to find $t_1$.

    $\begin{array}{c}-\frac{\left(65.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)}{\left(73.38\text{N}/{\text{m}}^2\right)}=\left(16.32\text{m}\right)\cos \left(\left(1.06\text{rad}/\text{s}\right)t_{\text{1}}\right)\\ \frac{-8.68\text{m}}{\left(16.32\text{m}\right)}=\cos \left(\left(1.06\text{rad}/\text{s}\right)t_{\text{1}}\right)\\ -0.53=\cos \left(\left(1.06\text{rad}/\text{s}\right)t_{\text{1}}\right)\\ t_1=2.0\text{s}\end{array}$

Conclusion:

Therefore, the time interval required for the cord to stretched by $25.0\text{m}$ is $2.0\text{s}$.

To determine

The total time interval for the entire $36.0\text{m}$ drop.

Answer to Problem 15.30P

The total time interval for the entire $36.0\text{m}$ drop is $\text{3}.49\text{s}$.

Explanation of Solution

The total time interval for the entire $36.0\text{m}$ drop is,

    $T=t+t_1$

Substitute $1.49\text{s}$ for $t$ and $2.0\text{s}$ for $t_1$ in above equation to find $T$.

    $\begin{array}{c}T=1.49\text{s}+2.0\text{s}\\ =\text{3}.49\text{s}\end{array}$

Conclusion:

Therefore, the total time interval for the entire $36.0\text{m}$ drop is $\text{3}.49\text{s}$.