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Chapter 5: Problem 9

An orbiting space station may be made to rotate to provide an artificial gravity. Given that the radius is \(25 \mathrm{~m}\), find the rotation period required to produce an apparent gravity equal to \(0.7 g\). A man whose normal weight is \(75 \mathrm{~kg}\) weight runs around the station in one direction and then the other (i.e., on a circle on the inside of the cylindrical wall) at \(5 \mathrm{~ms}^{-1}\). Find his apparent weight in each case. What effects will he experience if he climbs a ladder to a higher level ( i.e., closer to the axis), climbing at \(1 \mathrm{~ms}^{-1} ?\)

Short Answer

Expert verified
#Answer# - The rotation period required for the space station to produce an artificial gravity of 0.7g is approximately 15.92 seconds. - When the man runs in the same direction as the rotation, his apparent weight remains the same at 514.725 N. - When the man runs in the opposite direction, his apparent weight decreases to 16.5 N. - When the man climbs the ladder, there will be no net upward or downward force. However, he will experience a lower apparent weight because the centripetal acceleration decreases as he gets closer to the axis of the space station.

Step by step solution

01

Find the rotation period

To find the rotation period required for the space station, we can use the formula for artificial gravity: \(F = m \cdot a_{c}\), where \(F\) is the force, \(m\) is the mass (75 kg), and \(a_c\) is the centripetal acceleration. We have \(a_c = 0.7g\) and we know that \(g = 9.81 \mathrm{~m/s^2}\). Moreover, we can relate the centripetal acceleration with the angular velocity \(\omega\) and the radius \(R\): \(a_c = R \cdot \omega^2 \). Finally, the angular velocity can be expressed as \(\omega = \frac{2\pi}{T}\), where \(T\) is the rotation period. Now, we will find \(T\).
02

Relate centripetal acceleration and angular velocity

First, let's relate the centripetal acceleration with the angular velocity: \(a_c = R \cdot \omega^2 \Rightarrow 0.7g = R \cdot \left(\frac{2\pi}{T}\right)^2 \Rightarrow 0.7(9.81) = 25 \cdot \frac{4 \pi^2}{T^2}\).
03

Compute the rotation period, T

Solving for \(T\): \(T^2 = \frac{25 \cdot 4 \pi^2}{0.7(9.81)} \Rightarrow T \approx 15.92 \mathrm{s}\). So, the rotation period required is approximately 15.92 seconds.
04

Find the apparent weights of the man running in each direction

Since the man runs at a speed of \(5 \mathrm{~m/s}\), we can consider two different scenarios: running in the same direction as the rotation, and running in the opposite direction. When running in the same direction, the effective speed of the man is the sum of the speeds \(v_{total} = v_{station} + v_{man}\), where \(v_{station} = R\omega\). The centripetal acceleration stays the same, so his effective weight stays the same as well. When running in the opposite direction, the effective speed is the difference between the speeds \(v_{total} = |v_{station} - v_{man}|\). The centripetal acceleration will be \(a_c' = \frac{v_{total}^2}{R}\), and his apparent weight will be: \(W_{apparent} = m \cdot a_c'\).
05

Compute the apparent weights in each case

Calculate the apparent weights in both scenarios: Running in the same direction: \(W_{same} = m \cdot a_c = 75 \cdot 0.7(9.81) \approx 514.725 \mathrm{N}\). Running in the opposite direction: First, compute \(v_{total}\): \(v_{total} = |v_{station} - v_{man}| = |25\cdot \frac{2\pi}{15.92} - 5| \approx 2.34 \mathrm{~m/s}\), Then, calculate the centripetal acceleration: \( a_c' = \frac{v_{total}^2}{R} \approx 0.22 \mathrm{~m/s^2}\), Finally, compute the apparent weight: \(W_{opposite} = m \cdot a_c' \approx 16.5 \mathrm{N}\). So, the apparent weights are approximately 514.725 N in the same direction and 16.5 N in the opposite direction.
06

Analyze the effects when climbing the ladder

When the man climbs the ladder at \(1\mathrm{~m/s}\), we need to compute the effective force acting on him in the radial direction. As he gets closer to the axis, the value of R decreases, which means the centripetal acceleration also decreases. Thus, the net upward or downward force will depend on the difference between the gravitational force and the centripetal acceleration. Since he climbs at a constant speed, there will be no net upward or downward force. However, he will experience a lower apparent weight as he gets closer to the axis, because the centripetal acceleration will decrease.

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Most popular questions from this chapter

Water in a rotating container of radius \(50 \mathrm{~mm}\) is \(30 \mathrm{~mm}\) lower in the centre than at the edge. Find the angular velocity of the container.

The co-ordinates \((x, y, z)\) of a particle with respect to a uniformly rotating frame may be related to those with respect to a fixed inertial frame, \(\left(x^{*}, y^{*}, z^{*}\right)\), by the transformation $$ \left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{ccc} \cos \omega t & \sin \omega t & 0 \\ -\sin \omega t & \cos \omega t & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{c} x^{*} \\ y^{*} \\ z^{*} \end{array}\right] $$ (Here, we use matrix notation: this stands for three separate equations, \(x=\cos \omega t \cdot x^{*}+\sin \omega t \cdot y^{*}\) etc.) Write down the inverse relation giving \(\left(x^{*}, y^{*}, z^{*}\right)\) in terms of \((x, y, z)\). By differentiating with respect to \(t\), rederive the relation (5.15) between \(\mathrm{d}^{2} \boldsymbol{r} / \mathrm{d} t^{2}\) and \(\ddot{\boldsymbol{r}}\). [Hint: Note that \(\ddot{\boldsymbol{r}}=(\ddot{x}, \ddot{y}, \ddot{z})\), while \(\mathrm{d}^{2} \boldsymbol{r} / \mathrm{d} t^{2}\) is the vector obtained by applying the above transformation \(\left.\operatorname{to}\left(\ddot{x}^{*}, \ddot{y}^{*}, \ddot{z}^{*}\right) .\right]\)

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Write down the equation of motion for a charged particle in uniform, parallel electric and magnetic fields, both in the \(z\)-direction, and solve it, given that the particle starts from the origin with velocity \((v, 0,0)\). A screen is placed at \(x=a\), where \(a \ll m v / q B .\) Show that the locus of points of arrival of particles with given \(m\) and \(q\), but different speeds \(v\), is approximately a parabola. How does this locus depend on \(m\) and \(q ?\)

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