/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 The Hubble Space Telescope is st... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 kg and diameter 5.0 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 is \(1.212 \times 10^{-15} \, \text{N} \cdot \text{m}\).

Step by step solution

01

Determine the Moment of Inertia

The moment of inertia for a thin-walled cylinder spinning about its central axis is given by the formula \( I = m r^2 \), where \( m \) is the mass of the cylinder and \( r \) is the radius. Here, the mass \( m = 2.0 \, \text{kg} \) and the diameter is \( 5.0 \, \text{cm} = 0.05 \, \text{m} \), so the radius \( r = 0.025 \, \text{m} \). Thus, \( I = 2.0 \, \text{kg} \times (0.025 \, \text{m})^2 = 0.00125 \, \text{kg} \cdot \text{m}^2 \).
02

Calculate the Angular Velocity

The gyroscopes spin at \( 19,200 \) revolutions per minute (rpm). To convert this to radians per second (rad/s), use the formula \( \omega = \frac{2\pi \times \text{rpm}}{60} \). So, \( \omega = \frac{2\pi \times 19,200}{60} = 2,010.62 \, \text{rad/s} \).
03

Define the Precession Angle and Time

The gyroscope needs to precess through an angle of \( \Delta \theta = 1.0 \times 10^{-6} \) degrees. In radians, this is \( \Delta \theta = 1.0 \times 10^{-6} \times \frac{\pi}{180} = 1.745 \times 10^{-8} \, \text{rad} \). The precession has to occur over a time \( \Delta t = 5.0 \) hours, which is \( 18,000 \) seconds.
04

Use the Torque Precession Formula

The torque \( \tau \) needed to cause precession is given by \( \tau = I \alpha \), where \( \alpha \) is the angular acceleration. From precession, we have \( \alpha = \frac{\omega_{\text{precession}}}{\Delta t} \). The precession angular velocity can be described as \( \omega_{\text{precession}} = \frac{\Delta \theta}{\Delta t} \), where \( \Delta \theta \) and \( \Delta t \) are known from the previous steps.
05

Calculate the Angular Precession Velocity and Torque

\( \omega_{\text{precession}} = \frac{1.745 \times 10^{-8} \, \text{rad}}{18,000 \, \text{s}} = 9.694 \times 10^{-13} \, \text{rad/s} \). Now, substitute into \( \alpha = \omega_{\text{precession}} \) since it’s a constant precession rate: \( \alpha = 9.694 \times 10^{-13} \, \text{rad/s}^2 \). Finally, calculate torque \( \tau = I \alpha = 0.00125 \, \text{kg} \cdot \text{m}^2 \times 9.694 \times 10^{-13} \, \text{rad/s}^2 = 1.212 \times 10^{-15} \, \text{N} \cdot \text{m} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Hubble Space Telescope
The Hubble Space Telescope is a marvel of modern engineering, allowing astronomers to capture high-resolution images of distant galaxies, stars, and more. One critical component of its operation is the precision in pointing to specific celestial targets. Achieving this precision requires stabilization within extremely tight angular limits, incredibly around 2-millionths of a degree. This is where gyroscopes become essential. These devices are mounted on the telescope to ensure it holds an exact orientation while observing the cosmos through long exposures. By minimizing vibrations and external disturbances, the gyroscopes help maintain clear, sharp images necessary for astronomical studies.

The gyroscopes in the Hubble Space Telescope spin at an impressive rate of 19,200 revolutions per minute (rpm), functioning effectively as thin-walled cylinders that provide rotational stability. With a mass of 2.0 kg and a small diameter of just 5.0 cm, these gyroscopes ensure that the telescope achieves the required stability. They are fundamental for maintaining accuracy throughout various observational periods, sometimes extending over multiple hours.
Gyroscope
Understanding a gyroscope's role in stabilizing structures like the Hubble Space Telescope is crucial. A gyroscope is a device that uses the principles of angular momentum to help maintain balance and direction. Angular momentum is a concept from physics involving the rotation of objects. It is part of what makes gyroscopes so stable and reliable.

For the gyroscopes used in the Hubble, angular momentum arises from their fast rotations, which are maintained around their central axis. Simplified to a thin-walled cylinder, a gyroscope features a moment of inertia calculated by the formula:
  • Moment of Inertia, \( I = m r^2 \), where \( m \) is the mass and \( r \) the radius.
This moment of inertia, coupled with significant angular velocity (expressed in radians per second), helps gyroscopes resist any changes to their spinning motion unless acted upon by an external force like torque. This resistance is what gives the Hubble gyroscopes their stabilizing power.
Moreover, the gyroscopes’ ability to precess—that is, change their axis of rotation in response to torque—further enhances their stabilizing function, dampening deviations and ensuring precise targeting during telescopic observations.
Torque Precession
Torque precession is a fascinating phenomenon linked to gyroscopes. It explains how a spinning object like a gyroscope responds to external forces by adjusting its axis of rotation. In simple terms, torque is the rotational equivalent of linear force, crucial in altering the motion of rotating bodies.

In the case of the Hubble's gyroscopes, a tiny torque can induce precession—the gradual shift of the spinning axis over time. This is particularly evident when trying to maintain the telescope's orientation during observations, a task requiring incredible accuracy. To calculate the needed torque, the angular precession velocity is used:
  • \( \omega_{\text{precession}} = \frac{\Delta \theta}{\Delta t} \)
Where \( \Delta \theta \) is the angular change needed, and \( \Delta t \) is the time over which this change occurs. For constant precession, the angular acceleration \( \alpha \) is equated to precession velocity, simplifying torque calculation to \( \tau = I \alpha \), where \( I \) is the moment of inertia. This precession capability in gyroscopes is essential for their role in precision instruments like the Hubble Space Telescope, allowing for the necessary minor adjustments without manual interference.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The moment of inertia of the empty turntable is \(1.5 \mathrm{~kg} \mathrm{~m}^{2}\). With a constant torque of \(2.5 \mathrm{~N} \cdot \mathrm{m},\) the turntable-person system takes \(3.0 \mathrm{~s}\) to spin from rest to an angular speed of \(1.0 \mathrm{rad} / \mathrm{s}\). What is the person's moment of inertia about an axis through her center of mass? Ignore friction in the turntable axle. (a) \(2.5 \mathrm{~kg} \cdot \mathrm{m}^{2}\) (b) \(6.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\) (c) \(7.5 \mathrm{~kg} \cdot \mathrm{m}^{2}\) (d) \(9.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\).

A lawn roller in the form of a thin-walled, hollow cylinder with mass \(M\) is pulled horizontally with a constant horizontal force \(F\) applied by a handle attached to the axle. If it rolls without slipping, find the acceleration and the friction force.

You are designing a simple elevator system for an old warehouse that is being converted to loft apartments. A 22,500-N elevator is to be accelerated upward by connecting it to a counterweight by means of a light (but strong!) cable passing over a solid uniform disk- shaped pulley. The cable does not slip where it is in contact with the surface of the pulley. There is no appreciable friction at the axle of the pulley, but its mass is 875 kg and it is 1.50 m in diameter. (a) What mass should the counterweight have so that it will accelerate the elevator upward through 6.75 m in the first 3.00 s, starting from rest? (b) What is the tension in the cable on each side of the pulley?

An engine delivers 175 hp to an aircraft propeller at 2400 rev/min. (a) How much torque does the aircraft engine provide? (b) How much work does the engine do in one revolution of the propeller?

A thin, uniform metal bar, 2.00 m long and weighing 90.0 N, is hanging vertically from the ceiling by a frictionless pivot. Suddenly it is struck 1.50 m below the ceiling by a small 3.00-kg ball, initially traveling horizontally at 10.0 m/s. The ball rebounds in the opposite direction with a speed of 6.00 m/s. (a) Find the angular speed of the bar just after the collision. (b) During the collision, why is the angular momentum conserved but not the linear momentum?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.