Chapter 5, Problem 69P

(a)

To determine

Period of rotation needed to provide an artificial gravity of $g$ at the rim.

Answer to Problem 69P

Period of rotation needed to provide an artificial gravity of $g$ at the rim $\underset{\_}{20\text{s}}$.

Explanation of Solution

The edge of the station is moving in a circle. A centripetal acceleration is needed for this motion.

Write the expression for centripetal acceleration.

    $a_c=\frac{v^2}{r}$        (I)

Here, $v$ is the speed, and $r$ is the radius.

Equate equation (I) to $g$, and sole for $v$.

    $\begin{array}{c}g=\frac{v^2}{r}\\ v=\sqrt{gr}\end{array}$        (II)

Write the expression for period.

    $T=\frac{2\pi r}{v}$        (III)

Conclusion:

Substitute, $9.80\text{m}/{\text{s}}^2$ for $g$, and $100\text{m}$ for $r$ in the equation (II).

    $\begin{array}{c}v=\sqrt{\left(9.80\text{m}/{\text{s}}^2\right)\left(100\text{m}\right)}\\ =31.3\text{m}/\text{s}\end{array}$

Substitute, $31.3\text{m}/\text{s}$ for $v$, and $100\text{m}$ for $r$ in the equation (II).

    $\begin{array}{c}T=\frac{2\pi \left(100\text{m}\right)}{31.3\text{m}/\text{s}}\\ =20\text{s}\end{array}$

Therefore, the period of rotation needed to provide an artificial gravity of $g$ at the rim $\underset{\_}{20\text{s}}$.

(b)

To determine

The speed needed to move the rim.

Answer to Problem 69P

The speed needed to move the rim is $\underset{\_}{31.3\text{m}/\text{s}}$.

Explanation of Solution

From the part (a), it is clear that the speed needed to move the rim is $31.3\text{m}/\text{s}$.

Conclusion:

Therefore, the speed needed to move the rim is $\underset{\_}{31.3\text{m}/\text{s}}$

(c)

To determine

Apparent weight of the person if he run along the rim at $4.2\text{m}/\text{s}$ opposite to the rotation direction.

Answer to Problem 69P

Apparent weight of the person if he run along the rim at $4.2\text{m}/\text{s}$ opposite to the rotation direction is $\underset{\_}{0.75\text{g}}$.

Explanation of Solution

The overall velocity along the arc of the circle if the velocity in the opposite sense of that the rim’s rotation can be expressed as,

    $v_{app}=v_{rim}-v_{run}$        (IV)

Here, $r_{rim}$ is the velocity of rim, and $r_{run}$ is the running velocity.

Conclusion:

Substitute, $31.3\text{m}/\text{s}$ for $r_{rim}$, and $4.2\text{m}/\text{s}$ for $v_{run}$ in the equation (IV).

    $\begin{array}{c}{v}_{app}=31.3\text{m}/\text{s}-4.2\text{m}/\text{s}\\ =27.1\text{m}/\text{s}\end{array}$

The apparent weight of the person decreases proportional to the reduced centripetal acceleration due to the lower velocity of the person.

Substitute, $27.1\text{m}/\text{s}$ for $v$, and $100\text{m}$ for $r$ in the equation (I).

    $\begin{array}{c}{a}_c=\frac{{\left(27.1\text{m}/\text{s}\right)}^2}{100\text{m}}\\ =7.3{\text{m}/\text{s}}^2\end{array}$

It can be express in terms of $g$ as follows,

    $\begin{array}{c}\frac{7.3{\text{m}/\text{s}}^2}{9.8{\text{m}/\text{s}}^2}=0.75\\ 7.3{\text{m}/\text{s}}^2=\left(0.75\right)\left(9.8{\text{m}/\text{s}}^2\right)\\ =0.75g\end{array}$

Therefore, apparent weight of the person if he run along the rim at $4.2\text{m}/\text{s}$ opposite to the rotation direction is $\underset{\_}{0.75\text{g}}$

(d)

To determine

The apparent weight of person if he run instead in the direction of rotation.

Answer to Problem 69P

The apparent weight of person if he run instead in the direction of rotation is $\underset{\_}{1.26g}$.

Explanation of Solution

The overall velocity along the arc of the circle if the velocity aligned with that the rim’s rotation can be expressed as,

    $v_{app}=v_{rim}+v_{run}$        (V)

Conclusion:

Substitute, $31.3\text{m}/\text{s}$ for $r_{rim}$, and $4.2\text{m}/\text{s}$ for $v_{run}$ in the equation (V).

    $\begin{array}{c}{v}_{app}=31.3\text{m}/\text{s}+4.2\text{m}/\text{s}\\ =35.5\text{m}/\text{s}\end{array}$

The apparent weight of the person increase proportional to the reduced centripetal acceleration due to the larger velocity of the person.

Substitute, $35.5\text{m}/\text{s}$ for $v$, and $100\text{m}$ for $r$ in the equation (I).

    $\begin{array}{c}{a}_c=\frac{{\left(35.5\text{m}/\text{s}\right)}^2}{100\text{m}}\\ =12.6{\text{m}/\text{s}}^2\end{array}$

It can be express in terms of $g$ as follows,

    $\begin{array}{c}\frac{12.6{\text{m}/\text{s}}^2}{9.8{\text{m}/\text{s}}^2}=1.26\\ 12.6{\text{m}/\text{s}}^2=\left(1.26\right)\left(9.8{\text{m}/\text{s}}^2\right)\\ =1.26g\end{array}$

Therefore, the apparent weight of person if he run instead in the direction of rotation is $\underset{\_}{1.26g}$.

(e)

To determine

The direction in which the person run to get the best workout.

Answer to Problem 69P

The direction in which the person run to get the best workout is $\underset{\_}{\text{in the direction of rotation}}$.

Explanation of Solution

If the person run in the direction of rotation, then the rotation speed can increase effectively. The centripetal acceleration can increase by increasing rotation speed. Thereby effective weight weighing more while working out will definitely make the workout more difficult.

Conclusion:

Therefore, the direction in which the person run to get the best workout is $\underset{\_}{\text{in the direction of rotation}}$.