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A satellite is spinning at 6.0 rev/s. The satellite consists of a main body in the shape of a sphere of radius \(2.0 \mathrm{m}\) and mass \(10,000 \mathrm{kg}\), and two antennas projecting out from the center of mass of the main body that can be approximated with rods of length \(3.0 \mathrm{m}\) each and mass 10 kg. The antenna's lie in the plane of rotation. What is the angular momentum of the satellite?

Short Answer

Expert verified
The angular momentum of the satellite can be calculated by finding the total moment of inertia and multiplying it by the angular speed. First, find the moment of inertia of the main body, \(I_{main}= \frac{2}{5}(10000 \mathrm{kg})(2.0 \mathrm{m})^2\), and the moment of inertia of the antennas, \(I_{antenna} = 2 * \frac{1}{3}(10 \mathrm{kg})(3.0 \mathrm{m})^2\). Sum these up to get the total moment of inertia, \(I_{total} = I_{main} + I_{total_{antennas}}\). Convert the angular speed of 6.0 rev/s to radians/second, which is \(12\pi\, \mathrm{rad/s}\). Finally, find the angular momentum using the formula \(L = I_{total} * \omega\).

Step by step solution

01

Calculate the moment of inertia of the main body

The moment of inertia (I) of a sphere rotating about an axis through its center is given by the formula: \[I_{sphere} = \frac{2}{5}MR^2\] Where M is the mass of the sphere and R is the radius. For the main body of the satellite, M = 10000 kg and R = 2.0 m. Now we can calculate the moment of inertia of the main body: \[I_{main} = \frac{2}{5}(10000 \mathrm{kg})(2.0 \mathrm{m})^2\]
02

Calculate the moment of inertia of the antennas

The moment of inertia of a rod rotating about an axis through one end, perpendicular to the length, is given by the formula: \[I_{rod} = \frac{1}{3}mL^2\] Where m is the mass of the rod and L is the length. For each antenna of the satellite, m = 10 kg and L = 3.0 m. Now we can calculate the moment of inertia of each antenna: \[I_{antenna} = \frac{1}{3}(10 \mathrm{kg})(3.0 \mathrm{m})^2\] Since there are two antennas, the total moment of inertia of the antennas is: \[I_{total_{antennas}} = 2 * I_{antenna}\]
03

Calculate the total moment of inertia of the satellite

To find the total moment of inertia (I_total) of the satellite, we sum the moment of inertia of the main body and the moment of inertia of the antennas: \[I_{total} = I_{main} + I_{total_{antennas}}\]
04

Calculate the angular momentum of the satellite

The angular momentum (L) of an object is given by the formula: \[L = I\omega\] Where I is the moment of inertia of the object and ω is the angular speed. The satellite is spinning at 6.0 rev/s. We need to convert this to radians/second. Since there are \(2\pi\) radians in a revolution, the conversion is: \[\omega = 6.0\,\mathrm{rev/s} * 2\pi\,\mathrm{rad/rev} = 12\pi\, \mathrm{rad/s}\] Now, we can find the angular momentum of the satellite: \[L = I_{total} * \omega\] This will give us the angular momentum of the satellite, which is the final answer.

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Key Concepts

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

Moment of Inertia
The moment of inertia is a fundamental concept in rotational dynamics. It describes how difficult it is to change the rotational state of an object. Simply put, it tells us where the mass of an object is distributed relative to the axis of rotation. Different shapes and mass distributions have different formulas for calculating their moment of inertia.

For example, for a sphere with mass concentrated evenly, the moment of inertia when rotating about its center is \( I_{sphere} = \frac{2}{5}MR^2 \). Meanwhile, for a rod rotating about an axis at one end, the moment of inertia is \( I_{rod} = \frac{1}{3}mL^2 \). In our satellite problem:
  • The main body is a sphere, so its formula applies.
  • The antennas are best approximated as rods rotating about one end.
To find the total moment of inertia for the satellite, you sum up all contributions—both from the main body and the antennas.
Rotational Dynamics
Rotational dynamics deals with the motion of objects when they rotate. The forces and torques acting on rotating bodies affect their angular speed and acceleration. A key aspect of this is understanding how torque influences angular momentum.

In this context, the angular momentum \( L \) for a rotating object is given by the product of its moment of inertia \( I \) and its angular velocity \( \omega \). The formula is expressed as:
  • \( L = I\omega \)
This concept is essential in many applications, including in satellite mechanics where the angular momentum determines how much effort is required to maintain or change the spin of a satellite. Similar principles apply to any rotating system, from bicycle wheels to planet Earth itself.
Satellite Mechanics
Satellite mechanics is about understanding the motion and control of satellites. Satellites can spin to maintain stability in orbit using the principles of angular momentum and rotational dynamics.

In our problem, the satellite is a composite system of a spherical main body and antennas. By analyzing each part's moment of inertia and the overall angular velocity, we can predict how it will behave in space.
This behavior includes maintaining orientation, controlling spin rates, and managing torque applied by onboard devices or external forces. These mechanics ensure that satellites fulfill their function, whether it's gathering weather data, enabling communications, or navigating Earth's environment.
Angular Speed Conversion
The angular speed of an object is how fast it rotates or spins, given in revolutions per second (rev/s), degrees per second, or radians per second (rad/s). To use these units effectively in formulas, we often need to convert between them.

In physics, radians are preferred because they simplify equations involving trigonometric functions. There are \( 2\pi \) radians in one revolution, which is a crucial conversion factor. For instance, converting 6.0 rev/s to radians per second involves:
  • \( \omega = 6.0 \text{ rev/s} \times 2\pi \text{ rad/rev} = 12\pi \text{ rad/s} \)
By converting to radians, we ensure compatibility with other physical calculations, particularly when determining quantities like angular momentum.

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Most popular questions from this chapter

A cylinder with rotational inertia \(I_{1}=2.0 \mathrm{kg} \cdot \mathrm{m}^{2}\) rotates clockwise about a vertical axis through its center with angular speed \(\omega_{1}=5.0 \mathrm{rad} / \mathrm{s} .\) A second cylinder with rotational inertia \(\quad I_{2}=1.0 \mathrm{kg} \cdot \mathrm{m}^{2} \quad\) rotates counterclockwise about the same axis with angular speed \(\omega_{2}=8.0 \mathrm{rad} / \mathrm{s} .\) If the cylinders couple so they have the same rotational axis what is the angular speed of the combination? What percentage of the original kinetic energy is lost to friction?

Two particles of equal mass travel with the same speed in opposite directions along parallel lines separated by a distance \(d .\) Show that the angular momentum of this two-particle system is the same no matter what point is used as the reference for calculating the angular momentum.

Three children are riding on the edge of a merry-go-round that is \(100 \mathrm{kg}\), has a 1.60 -m radius, and is spinning at \(20.0 \mathrm{rpm}\). The children have masses of \(22.0,28.0,\) and 33.0 kg. If the child who has a mass of 28.0 kg moves to the center of the merry-go-round, what is the new angular velocity in rpm?

A solid cylinder of radius \(10.0 \mathrm{cm}\) rolls down an incline with slipping. The angle of the incline is \(30^{\circ} .\) The coefficient of kinetic friction on the surface is \(0.400 .\) What is the angular acceleration of the solid cylinder? What is the linear acceleration?

A uniform rod of mass 200 g and length 100 cm is free to rotate in a horizontal plane around a fixed vertical axis through its center, perpendicular to its length. Two small beads, each of mass 20 g, are mounted in grooves along the rod. Initially, the two beads are held by catches on opposite sides of the rod's center, \(10 \mathrm{cm}\) from the axis of rotation. With the beads in this position, the rod is rotating with an angular velocity of \(10.0 \mathrm{rad} / \mathrm{s}\). When the catches are released, the beads slide outward along the rod. (a) What is the rod's angular velocity when the beads reach the ends of the rod? (b) What is the rod's angular velocity if the beads fly off the rod?

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