Chapter 8, Problem 32PQ

The Flybar high-tech pogo stick is advertised as being capable of launching jumpers up to 6 ft. The ad says that the minimum weight of a jumper is 120 lb and the maximum weight is 250 lb. It also says that the pogo stick uses “a patented system of elastometric rubber springs that provides up to 1200 lbs of thrust, something common helical spring sticks simply cannot achieve (rubber has 10 times the energy storing capability of steel).”

1. a. Use Figure P8.32 to estimate the maximum compression of the pogo stick’s spring. Include the uncertainty in your estimate.
2. b. What is the effective spring constant of the elastometric rubber springs? Comment on the claim that rubber has 10 times the energy-storing capability of steel.
3. c. Check the ad’s claim that the maximum height a jumper can achieve is 6 ft.

To determine

The maximum compression of the pogo sticks spring with uncertainty.

Answer to Problem 32PQ

The maximum compression of the pogo sticks spring is $\underset{\_}{0.3\text{m}\pm 0.1\text{m}}$.

Explanation of Solution

The maximum compression of the spring will be always less than the length of the stick below the person’s feet. It is estimated that the maximum compression $l$ to be 1 foot or about $0.3\text{m}$.

The uncertainty of the estimation will be about $\pm 0.1\text{m}$.

Conclusion:

Therefore, the maximum compression of the pogo sticks spring is $\underset{\_}{0.3\text{m}\pm 0.1\text{m}}$.

To determine

The effective spring constant of the elastometric rubber springs and comment on the claim that rubber has 10 times the energy storing capability of steel.

Answer to Problem 32PQ

The effective spring constant of the elastometric rubber springs is $\underset{\_}{1800\text{N}/\text{m}}$ and it does not make sense to say from this example that rubber spring stores more energy than a steel spring because it depends on the geometric parameters of the spring.

Explanation of Solution

Write the expression force exerted by the spring when it is fully compressed according to Hooke’s law.

    $F_{\text{H}}=kl$                                                                                                                   (I)

Here, $F_{{}_{\text{H}}}$ is the force exerted by the spring, $k$ is the spring constant, $l$ is the maximum compression.

The effective spring constant must be derived from the claim that the spring provides $1,200\text{lbs}$ of thrust which we assume is the force exerted by the spring when it is fully compressed.

Solve equation (I) for $k$,

    $k=\frac{F_{\text{H}}}{l}$                                                                                                                    (II)

Conclusion:

Substitute $1200\text{lbs}$ for $F_{\text{H}}$, $0.3\text{m}$ for $l$ in equation (II) to find $k$.

    $\begin{array}{c}k=\frac{F_{\text{H}}}{l}\\ =\frac{1200\text{lbs}}{0.3\text{m}}\\ =\frac{1200\text{lbs}\times \frac{4.45\text{N}}{1\text{lbs}}}{0.3\text{m}}\\ \approx 1800\text{N}/\text{m}\end{array}$

Since this is a pretty stiff spring, it doesn’t make sense to say that the rubber spring stores more energy than steel spring because it depends on the geometric c parameters of the spring.

Therefore, the effective spring constant of the elastometric rubber springs is $\underset{\_}{1800\text{N}/\text{m}}$ and it is not possible to say from this example that rubber spring stores more energy than a steel spring because it depends on the geometric parameters of the spring

To determine

Show that the maximum height a jumper can achieve is $6\text{ft}$.

Answer to Problem 32PQ

The result is far below the claimed maximum height of $6\text{ft}$.

Explanation of Solution

The total energy at the two points in the trajectory (1) the jumper is momentarily at rest on the ground before the jump with zero gravitational potential energy and the spring is fully compressed (II) is the jumper is momentarily at rest at peak height when all of the energy is in gravitational potential energy.

Write the expression for the conservation of the total mechanical energy.

    $\Delta K+\Delta U=0$                                                                                                       (III)

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

The change in kinetic energy between the two points is zero, the change in potential energy is,

    $\Delta U=\Delta U_g+\Delta U_e$                                                                                               (IV)

Here, $\Delta U_g$ is the change in gravitational potential energy, $\Delta U_e$ is the change in elastic potential energy.

Write the expression for the change in gravitational potential energy.

    $\Delta U_g=mgh$                                                                                                           (V)

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

Write the expression for the change in elastic potential energy.

    $\Delta U_e=\frac{1}{2}k{l}^2$                                                                                                          (VI)

Use equation (IV), (V) and (VI) in equation (III) and solve for $h$.

    $\begin{array}{c}0+mgh-\frac{1}{2}k{l}^2=0\\ h=\frac{k{l}^2}{2mg}\end{array}$                                                                                    (VII)

Conclusion:

Substitute $1800\text{N}/\text{m}$ for $k$, $0.3\text{m}$ for $l$, $60\text{kg}$ for $m$, $9.81\text{m}/{\text{s}}^2$ for $g$ in equation (VII) to find $h$.

    $\begin{array}{c}h=\frac{\left(1800\text{N}/\text{m}\right){\left(0.3\text{m}\right)}^2}{2\left(60\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2\right)}\\ =0.14\text{m}\times \frac{3.280\text{ft}}{1\text{m}}\\ =0.5\text{ft}\\ \end{array}$

The obtained result is far below the claimed maximum height of 6 ft. The biggest uncertainty comes from the statement about “thrust then the spring constant must be about ten times larger.

Therefore, the result is far below the claimed maximum height of $6\text{ft}$.