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Stabilization of the Hubble Space Telescope. The Hubble Space Telescope is stabilized to within an angle of about 2 -millionths of a degree by means of a series of gyroscopes that spin at 19,200 rpm. Although the structure of these gyroscopes is actually quite complex, we can model each of the gyroscopes as a thin-walled cylinder of mass \(2.0 \mathrm{~kg}\) and diameter \(5.0 \mathrm{~cm},\) spinning about its central axis. How large a torque would it take to cause these gyroscopes to precess through an angle of \(1.0 \times 10^{-6}\) degree during a 5.0 hour exposure of a galaxy?

Short Answer

Expert verified
The required torque to cause the gyroscopes to precess through an angle of \(1.0 \times 10^{-6}\) degree during a 5.0 hour exposure of a galaxy is approximately \(1.0 \times 10^{-14} \, \text{N} \cdot \text{m}\).

Step by step solution

01

Identify the given parameters

We are given the mass of the gyroscope (\(m = 2.0 \, \text{kg}\)), diameter (\(D = 5.0 \, \text{cm} = 0.05 \, \text{m}\)), spinning speed (\(n = 19,200 \, \text{rpm}\)), precession angle (\(\theta = 1.0 \times 10^{-6}\, \text{degree} = 1.74 \times 10^{-11}\, \text{rad}\)), and exposure time (\(t = 5.0 \, \text{hours} = 18,000 \, \text{s}\)).
02

Calculate the moment of inertia of the gyroscope

The gyroscope is modeled as a thin-walled cylinder, which has a moment of inertia (\(I\)) of \(I = 0.5 m r^2\), where \(r = D/2 = 0.025 \, \text{m}\) is the radius. Substituting the given values, we get \(I = 0.5 \times 2.0 \, \text{kg} \times (0.025 \, \text{m})^2 = 0.000625 \, \text{kg} \cdot \text{m}^2\).
03

Calculate the angular momentum of the gyroscope

Angular momentum (\(L\)) is given by \(L = I \omega\), where angular speed \(\omega = 2 \pi n / 60\). Substituting the given values, we obtain \(\omega = 2 \pi \times 19,200 \, \text{rpm}/60 = 2010.6 \, \text{rad/s}\). Thus, \(L = 0.000625 \, \text{kg} \cdot \text{m}^2 \times 2010.6 \, \text{rad/s} = 1.26 \, \text{kg} \cdot \text{m}^2/\text{s}\)
04

Calculate the torque needed for precession

The torque (\(\tau\)) required for precession is given by \(\tau = L \Delta \theta / \Delta t\). Substituting the calculated and given values, we get \(\tau = 1.26 \, \text{kg} \cdot \text{m}^2/\text{s} \times 1.74 \times 10^{-11} \, \text{rad} / 18,000 \, \text{s} = 1.0 \times 10^{-14} \, \text{N} \cdot \text{m}\)

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Key Concepts

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

Understanding Gyroscopes in Spacecraft Stabilization
Gyroscopes are fascinating devices that play a crucial role in the stabilization of spacecraft such as the Hubble Space Telescope. These devices leverage the principles of physics to maintain a stable orientation in space. A gyroscope typically consists of a spinning wheel or rotor, which can be thought of as a thin-walled cylinder in more complex systems like the Hubble's. The core functionality of a gyroscope lies in its ability to resist changes in its axis of rotation due to its angular momentum.

This resistance is what allows the Hubble Space Telescope to remain stable when observing distant galaxies. While the gyroscope spins at high speeds—in Hubble's case, at 19,200 revolutions per minute (rpm)—the conservation of angular momentum keeps the telescope pointed accurately. This is essential for obtaining clear images from space, free of distortions caused by unwanted motion.
Torque Calculation for Gyroscope Precession
Torque is a fundamental concept in physics, especially when analyzing the behavior of rotating objects like gyroscopes. In our exercise, we learned how to calculate the torque required for a gyroscope to precess or slowly change its axis of rotation over a specified time period.

The required torque for precession is determined using the formula \( \tau = L \Delta \theta / \Delta t \), where \( L \) is the angular momentum, \( \Delta \theta \) is the change in angle, and \( \Delta t \) is the time duration. This means that the angular momentum and the intended precession angle dictate how much force we need to apply.
  • For the Hubble's gyroscope, the required torque was found to be exceedingly small, around \( 1.0 \times 10^{-14} \) Newton meters.
  • This tiny torque showcases the precision needed in space operations and the efficiency of using gyroscopes in maintaining stability.
Angular Momentum and its Role in Gyroscopes
Angular momentum is the rotational equivalent of linear momentum and is a vector quantity that signifies the quantity of rotation of an object. For the Hubble's gyroscopes, angular momentum is critical. It is given by the product of the moment of inertia and angular velocity, expressed as \( L = I \omega \).

In the context of the Hubble Space Telescope, maintaining a high angular momentum through rapid spinning (19,200 rpm) is essential for effective stabilization. The greater the angular momentum, the stronger the resistance to changes in orientation, thereby reducing the chances of undesired rotations during long exposure imaging of galaxies. This resistance helps ensure that the telescope can capture precise and focused images.
  • This property is why gyroscopes are integral to many technologies that require orientation control, like spacecraft and drones.
Understanding the Moment of Inertia in Gyroscopes
The moment of inertia is a measure of an object's resistance to changes in its rotation. It depends on the distribution of mass relative to the axis of rotation. For a thin-walled cylinder, like the gyroscope in our problem, the moment of inertia \( I \) is calculated using the formula \( I = 0.5 m r^2 \), where \( m \) is the mass and \( r \) is the radius of the cylinder.

Understanding the moment of inertia is crucial for designing gyroscopes to stabilize spacecraft. In the Hubble's case, a smaller moment of inertia means the gyroscope can spin faster, increasing its angular momentum without a proportionate increase in size or weight.
  • The calculated moment of inertia for one of Hubble's gyroscopes was \( 0.000625 \, \text{kg} \cdot \text{m}^2 \), indicating how lightweight yet effective the device is.
  • The design must strike a balance between the size and rotational capabilities to ensure optimal performance in space.
This concept not only helps in understanding gyroscopes but is also applied in various fields, including engineering and robotics, where rotational dynamics are essential.

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Most popular questions from this chapter

Block \(A\) rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block \(B\) is suspended from the free end of the rope. The light rope that connects the two blocks does not slip over the surface of the pulley (radius \(0.080 \mathrm{~m}\) ) because the pulley rotates on a frictionless axle. The horizontal surface on which block \(A\) (mass \(2.50 \mathrm{~kg}\) ) moves is frictionless. The system is released from rest, and block \(B\) (mass \(6.00 \mathrm{~kg}\) ) moves downward \(1.80 \mathrm{~m}\) in \(2.00 \mathrm{~s}\). (a) What is the tension force that the rope exerts on block \(B ?\) (b) What is the tension force on block \(A ?\) (c) What is the moment of inertia of the pulley for rotation about the axle on which it is mounted?

A uniform marble rolls down a symmetrical bowl, starting from rest at the top of the left side. The top of each side is a distance \(h\) above the bottom of the bowl. The left half of the bowl is rough enough to cause the marble to roll without slipping, but the right half has no friction because it is coated with oil. (a) How far up the smooth side will the marble go, measured vertically from the bottom? (b) How high would the marble go if both sides were as rough as the left side? (c) How do you account for the fact that the marble goes higher with friction on the right side than without friction?

A \(12.0 \mathrm{~kg}\) box resting on a horizontal, frictionless surface is attached to a \(5.00 \mathrm{~kg}\) weight by a thin, light wire that passes over a frictionless pulley (Fig. E10.16). The pulley has the shape of a uniform solid disk of mass \(2.00 \mathrm{~kg}\) and diameter \(0.500 \mathrm{~m} .\) After the system is released, find (a) the tension in the wire on both sides of the pulley, (b) the acceleration of the box, and (c) the horizontal and vertical components of the force that the axle exerts on the pulley.

A \(50.0 \mathrm{~kg}\) grindstone is a solid disk \(0.520 \mathrm{~m}\) in diameter. You press an ax down on the rim with a normal force of \(160 \mathrm{~N}\) (Fig. \(\mathbf{P} 10.58\) ). The coefficient of kinetic friction between the blade and the stone is \(0.60,\) and there is a constant friction torque of \(6.50 \mathrm{~N} \cdot \mathrm{m}\) between the axle of the stone and its bearings. (a) How much force must be applied tangentially at the end of a crank handle \(0.500 \mathrm{~m}\) long to bring the stone from rest to 120 rev \(/ \min\) in \(9.00 \mathrm{~s} ?\) (b) After the grindstone attains an angular speed of 120 rev \(/ \mathrm{min}\), what tangential force at the end of the handle is needed to maintain a constant angular speed of 120 rev \(/ \min ?\) (c) How much time does it take the grindstone to come from \(120 \mathrm{rev} / \mathrm{min}\) to rest if it is acted on by the axle friction alone?

A large uniform horizontal turntable rotates freely about a vertical axle at its center. You measure the radius of the turntable to be \(3.00 \mathrm{~m} .\) To determine the moment of inertia \(I\) of the turntable about the axle, you start the turntable rotating with angular speed \(\omega\), which you measure. You then drop a small object of mass \(m\) onto the rim of the turntable. After the object has come to rest relative to the turntable, you measure the angular speed \(\omega_{\mathrm{f}}\) of the rotating turntable. You plot the quantity \(\left(\omega-\omega_{\mathrm{f}}\right) / \omega_{\mathrm{f}}\) (with both \(\omega\) and \(\omega_{\mathrm{f}}\) in rad \(\left./ \mathrm{s}\right)\) as a function of \(m\) (in kg). You find that your data lie close to a straight line that has slope \(0.250 \mathrm{~kg}^{-1}\). What is the moment of inertia \(I\) of the turntable?

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