Gyroscopes and the torque required for precession

In summary, a torque of 18000 N would be necessary to cause the Hubble Space Telescope to precess through an angle of 1.30×10−6 degrees during a 5.00 hour exposure of a galaxy.
  • #1
physicsguy128
2
0

Homework Statement



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 1.92×10^4 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 2.00kg and diameter 5.00cm , spinning about its central axis.

How large a torque would it take to cause these gyroscopes to precess through an angle of 1.30×10−6 degrees during a 5.00 hour exposure of a galaxy?

Homework Equations



Torque = [itex]\frac{dL}{dt}[/itex] (1) , Ω = [itex]\frac{Δ∅}{Δt}[/itex] (2)

The Attempt at a Solution



For a gyroscope torque is a a rate of change of momentum (1). So,

Torque = [itex]\frac{Angular momentum before - Angular momentum after}{5X60X60}[/itex]

Torque = [itex]\frac{I(ωbefore) - I(ωafter)}{18000}[/itex]

This reduces down to Torque = [itex]\frac{∏}{90000}[/itex] - [itex]\frac{ωafter}{800X18000}[/itex]

I am at a loss of how to find ω after, could anyone shed some light? Cheers
 
Last edited:
Physics news on Phys.org
  • #2
I think you need to look up precession. I think precession is when the object keeps rotating at the same speed, but the axis of rotation moves round in a cone-like shape as well. Much like a spinning top, when its axis is not perfectly vertical, its axis will wobble around the vertical direction, due to the torque applied by gravity.
 
  • #3
If you still need help on this problem, I figured it out. I used the formula Ω = torque/angular momentum.
As we know, angular momentum is simply the moment of Inertia * the angular velocity. In the problem you are given the angular velocity as 1.92*10^4 rpm. convert that to radians/second.

Now you need to find the moment of Inertia. Since it's a thin-walled cylinder, you know I=mr^2

As for omega you are given the degrees and the time. you have to convert degrees into radians and time into seconds. Divide your radians/secs and then multiply your answer by the moment of Inertia and angular velociy. That will give you your answer as:
Ω = [itex]\tau[/itex]/L
Ω = [itex]\tau[/itex]/I[itex]\omega[/itex]
[itex]\tau[/itex] = ΩI[itex]\omega[/itex]
 
  • #4
Thanks, what i was doing wrong was to convert from revs / s to radians / s was dividing by 2∏ when actually i should have multiplied by ∏! oops! Cheers
 
  • #5


I would first like to commend the student for their attempt at solving the problem and using the appropriate equations. However, there are a few areas that need clarification and improvement.

Firstly, the equation for torque (1) is missing a crucial factor, which is the distance from the axis of rotation to the point where the force is applied. This is typically represented by the variable "r" and should be included in the equation.

Secondly, the equation for angular velocity (2) is not relevant to this problem as it refers to the change in angular velocity over a certain time period, whereas the problem is asking for the torque required to cause precession, which is a different phenomenon.

To solve this problem, we need to first calculate the moment of inertia (I) of the gyroscope, which is a measure of its resistance to rotational motion. This can be calculated using the formula I = ½MR^2, where M is the mass and R is the radius of the gyroscope.

Once we have the moment of inertia, we can use the equation for precession torque, which is given by T = IΩ, where Ω is the angular velocity of the precession. In this case, we are given the angle of precession (1.30×10^-6 degrees) and the time period (5.00 hours), so we can calculate the angular velocity (Ω) using the equation Ω = Δ∅/Δt.

Finally, we can substitute the values for moment of inertia and angular velocity into the equation for torque and solve for the torque required to cause precession.

It is important to note that the actual torque required may vary depending on the specific design and construction of the gyroscopes, but this calculation provides a good estimate based on the given information. Additionally, this problem assumes ideal conditions and does not take into account any external forces or variables that may affect the precession of the gyroscopes.
 

What is a gyroscope?

A gyroscope is a device that is used to measure and maintain orientation and angular velocity. It typically consists of a spinning wheel or disk that is mounted on an axis, allowing it to rotate freely in any direction. Gyroscopes are commonly used in navigation, stabilization, and various scientific and engineering applications.

How does a gyroscope work?

Gyroscopes work based on the principle of angular momentum. When the axis of a spinning gyroscope is tilted, the angular momentum causes it to resist the change in orientation, resulting in a precession motion. This allows the gyroscope to maintain its orientation and detect any changes in orientation.

What is precession?

Precession is the circular motion of the axis of a spinning object when a torque is applied to it. In the case of a gyroscope, the precession motion is caused by the angular momentum of the spinning wheel or disk.

What is the torque required for precession?

The torque required for precession depends on the angular momentum and the rate of precession desired. It can be calculated using the equation: torque = angular momentum x precession rate. The greater the angular momentum, the larger the torque required for precession.

What are some real-life applications of gyroscopes?

Gyroscopes have numerous real-life applications, including navigation systems in airplanes, ships, and spacecraft, stabilization of cameras and drones, and measuring devices in various scientific and engineering instruments. They are also used in smartphones and gaming consoles for motion sensing and control.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
915
  • Mechanics
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
19K
  • Classical Physics
Replies
2
Views
815
Replies
11
Views
2K
Replies
6
Views
2K
  • Advanced Physics Homework Help
Replies
3
Views
1K
  • Classical Physics
Replies
7
Views
2K
Replies
14
Views
5K
Back
Top