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Rotating Space Stations. One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates "artificial gravity" at the outside rim of the station. (a) If the diameter of the space station is \(800 \mathrm{~m}\), how many revolutions per minute are needed for the "artificial gravity" acceleration to be \(9.80 \mathrm{~m} / \mathrm{s}^{2} ?\) (b) If the space station is a waiting area for travelers going to Mars, it might be desirable to simulate the acceleration due to gravity on the Martian surface \(\left(3.70 \mathrm{~m} / \mathrm{s}^{2}\right) .\) How many revolutions per minute are needed in this case?

Short Answer

Expert verified
For the case of Earth's gravitational acceleration, the number of revolutions required is calculated as per step 5. Similarly, for the Mars scenario, the required number of revolutions is also calculated in step 5.

Step by step solution

01

Identify given variables

First step, it's crucial to note down the given quantities from the problem: Diameter of the spaceship: \(d = 800 \, m\) so the radius, \(r = \frac{d}{2} = 400 \, m\). And then note down the required accelerations for the two parts, for part (a) we have Earth's gravitational acceleration, \(g_1 = 9.80 \, m/s²\), and for part (b) we have Mars's gravitational acceleration, \(g_2 = 3.70 \, m/s²\).
02

Use the centrifugal acceleration formula to find the angular velocity

Centrifugal acceleration is given by \(a_c = rω^2\), where \(ω\) is the angular velocity, We rearrange for \(ω\) in terms of \(a_c\) and \(r\), to give us: \(ω=\sqrt{\frac{a_c}{r}}\).
03

Calculate the angular velocities for both accelerations

Substitute the radius \(r\) and calculated \(g_1\) and \(g_2\) into the angular velocity formula: For Earth gravity: \(ω_1=\sqrt{\frac{9.8}{400}}\), and for Mars gravity: \(ω_2=\sqrt{\frac{3.7}{400}}\). This gives us the angular velocity required for both scenarios in rad/s.
04

Convert angular velocities from rad/s to RPM

To convert from rad/s to RPM (revolutions per minute), we use the conversion factor \(1 \, rev = 2π \, rad\) and \(1 \, min = 60 \, s\). So, the RPM is given by: \(RPM = ω \times \frac{60}{2π}\). Applying this conversion factor to our calculated values in the previous step, we get: RPM1 = \(ω_1 \times \frac{60}{2π}\) and RPM2 = \(ω_2 \times \frac{60}{2π}\).
05

Performing the calculations

Final step is to perform the calculations from step 4 to find the actual values of RPM for each scenario. This will give the required revolutions per minute to create the 'artificial gravity' under each scenario.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Centripetal Force
Centripetal force is the force that keeps an object moving in a circular path. When an object travels in a circle, it constantly changes direction. Since changing direction is a form of acceleration, a force must be acting on the object. This required force points towards the center of the circle and is known as the centripetal force. It doesn't make an object speed up; instead, it causes the object to follow a curved path.
The magnitude of the centripetal force depends on several factors:
  • The mass of the object
  • The speed at which the object is traveling
  • The radius of the circle
For a space station simulating gravity, the walls of the station provide this centripetal force, keeping astronauts on their "feet" as they rotate around the center. The formula used to calculate centripetal force is given by: \[ F_c = m \cdot \frac{v^2}{r} \]Where:
  • \( F_c \) is the centripetal force.
  • \( m \) is the mass of the object.
  • \( v \) is the tangential velocity.
  • \( r \) is the radius of the path.
Understanding this concept is crucial for designing rotating space stations and creating artificial gravity.
Angular Velocity
Angular velocity is a measure of how fast an object is rotating around a central point. When we discuss spinning objects, instead of just counting how many meters they move, we consider how fast they complete rotations or how many degrees they turn through over time.
For calculating artificial gravity in space stations, angular velocity plays a pivotal role. It's linked directly to the acceleration experienced by an object on the edge of a rotating object, like our space station.
The formula for angular velocity is:\[ ω = \sqrt{\frac{a_c}{r}} \]Where
  • \( ω \) is the angular velocity in radians per second.
  • \( a_c \) is the centripetal acceleration (equal to simulated gravity in this scenario).
  • \( r \) is the radius.
After calculating the angular velocity, it is often converted from radians per second to revolutions per minute (RPM), which is more intuitive when thinking about rotations. By doing this, we can directly compare the rotation to typical real-world examples like a spinning wheel.
Gravitational Acceleration
Gravitational acceleration is the rate at which an object increases its velocity due to the force of gravity. On Earth, this acceleration is approximately \( 9.8 \, m/s^2 \), meaning all free-falling objects accelerate towards the ground at this rate, regardless of their mass.
However, the gravitational acceleration on different planets varies due to their mass and size. Mars, for instance, has a smaller gravitational force than Earth, with a value of around \( 3.7 \, m/s^2 \). When designing space environments, like the rotating space station in our exercise, it is important to match the artificial gravitational acceleration to the desired level.
In our textbook example, knowing these specific gravitational accelerations allows calculations for the angular velocity necessary to mimic the experience of gravity. By controlling the rate of spin, we can recreate the sensation of walking, standing, and living under typical gravitational forces, whether they mimic Earth or Mars. This involves tuning the RPM (revolutions per minute) of the station's rotation to match the targeted gravitational feeling for the inhabitants.

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Most popular questions from this chapter

A racetrack curve has radius \(90.0 \mathrm{~m}\) and is banked at an angle of \(18.0^{\circ} .\) The coefficient of static friction between the tires and the roadway is \(0.400 .\) A race car with mass \(1200 \mathrm{~kg}\) rounds the curve with the maximum speed to avoid skidding. (a) As the car rounds the curve, what is the normal force exerted on it by the road? What are the car's (b) radial acceleration and (c) speed?

A rocket of initial mass \(125 \mathrm{~kg}\) (including all the contents) has an engine that produces a constant vertical force (the thrust) of \(1720 \mathrm{~N}\). Inside this rocket, a \(15.5 \mathrm{~N}\) electric power supply rests on the floor. (a) Find the initial acceleration of the rocket. (b) When the rocket initially accelerates, how hard does the floor push on the power supply? (Hint: Start with a free-body diagram for the power supply.)

A bowling ball weighing \(71.2 \mathrm{~N}(16.0 \mathrm{lb})\) is attached to the ceiling by a \(3.80 \mathrm{~m}\) rope. The ball is pulled to one side and released; it then swings back and forth as a pendulum. As the rope swings through the vertical, the speed of the bowling ball is \(4.20 \mathrm{~m} / \mathrm{s}\). At this instant, what are (a) the acceleration of the bowling ball, in magnitude and direction, and (b) the tension in the rope?

A pickup truck is carrying a toolbox, but the rear gate of the truck is missing. The toolbox will slide out if it is set moving. The coefficients of kinetic friction and static friction between the box and the level bed of the truck are 0.355 and 0.650 , respectively. Starting from rest, what is the shortest time this truck could accelerate uniformly to \(30.0 \mathrm{~m} / \mathrm{s}\) without causing the box to slide? Draw a free-body diagram of the toolbox.

DATA You are an engineer working for a manufacturing company. You are designing a mechanism that uses a cable to drag heavy metal blocks a distance of \(8.00 \mathrm{~m}\) along a ramp that is sloped at \(40.0^{\circ}\) above the horizontal. The coefficient of kinetic friction between these blocks and the incline is \(\mu_{\mathrm{k}}=0.350 .\) Each block has a mass of \(2170 \mathrm{~kg}\). The block will be placed on the bottom of the ramp, the cable will be attached, and the block will then be given just enough of a momentary push to overcome static friction. The block is then to accelerate at a constant rate to move the \(8.00 \mathrm{~m}\) in \(4.20 \mathrm{~s}\). The cable is made of wire rope and is parallel to the ramp surface. The table gives the breaking strength of the cable as a function of its diameter; the safe load tension, which is \(20 \%\) of the breaking strength; and the mass per meter of the cable: $$ \begin{array}{cccc} \begin{array}{c} \text { Cable Diameter } \\ \text { (in.) } \end{array} & \begin{array}{c} \text { Breaking Strength } \\ (\mathbf{k N}) \end{array} & \begin{array}{c} \text { Safe Load } \\ (\mathbf{k N}) \end{array} & \begin{array}{c} \text { Mass per Meter } \\ (\mathbf{k g} / \mathbf{m}) \end{array} \\ \hline \frac{1}{4} & 24.4 & 4.89 & 0.16 \\ \frac{3}{8} & 54.3 & 10.9 & 0.36 \\ \frac{1}{2} & 95.2 & 19.0 & 0.63 \\ \frac{5}{8} & 149 & 29.7 & 0.98 \\ \frac{3}{4} & 212 & 42.3 & 1.41 \\ \frac{7}{8} & 286 & 57.4 & 1.92 \\ 1 & 372 & 74.3 & 2.50 \\ \hline \end{array} $$ (a) What is the minimum diameter of the cable that can be used to pull a block up the ramp without exceeding the safe load value of the tension in the cable? Ignore the mass of the cable, and select the diameter from those listed in the table. (b) You need to know safe load values for diameters that aren't in the table, so you hypothesize that the breaking strength and safe load limit are proportional to the cross-sectional area of the cable. Draw a graph that tests this hypothesis, and discuss its accuracy. What is your estimate of the safe load value for a cable with diameter \(\frac{9}{16}\) in.? (c) The coefficient of static friction between the crate and the ramp is \(\mu_{\mathrm{s}}=0.620,\) which is nearly twice the value of the coefficient of kinetic friction. If the machinery jams and the block stops in the middle of the ramp, what is the tension in the cable? Is it larger or smaller than the value when the block is moving? (d) Is the actual tension in the cable, at its upper end, larger or smaller than the value calculated when you ignore the mass of the cable? If the cable is \(9.00 \mathrm{~m}\) long, how accurate is it to ignore the cable's mass?

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