Gyroscope- Angular Speed, Rotational Kinematics, Precession

Click For Summary
SUMMARY

The discussion focuses on calculating the time required to accelerate a gyroscope to a specific angular speed and determining the torque needed for precession. The gyroscope in question has a mass of 6.50×104 kg and a radius of 2.30 m, rotating at 500 revolutions per minute (rev/min). The power input is 7.45×104 W. The moment of inertia is calculated using the formula I=1/2mr2, resulting in 171925 kg·m2. The torque required for acceleration is found to be 1422.84 N·m, but the user struggles with the precession calculations, indicating a need for a deeper understanding of angular momentum and energy approaches.

PREREQUISITES
  • Understanding of rotational kinematics and angular speed
  • Familiarity with the moment of inertia formula for solid cylinders
  • Knowledge of power, torque, and their relationships in rotational dynamics
  • Basic principles of gyroscopic precession
NEXT STEPS
  • Study the energy approach to rotational dynamics, specifically K=1/2Iw2
  • Learn about gyroscopic precession and its mathematical derivation
  • Explore the relationship between torque and angular momentum in rotational systems
  • Practice problems involving the acceleration of gyroscopes and the effects of varying power inputs
USEFUL FOR

Students and professionals in physics, mechanical engineering, and aerospace engineering who are working with gyroscopic systems and require a solid understanding of rotational dynamics and kinematics.

jcwhitts
Messages
9
Reaction score
0
Gyroscope-- Angular Speed, Rotational Kinematics, Precession

Homework Statement



A Gyro Stabilizer. The stabilizing gyroscope of a ship is a solid disk with a mass of 6.50×10^4 kg; its radius is 2.30m, and it rotates about a vertical axis with an angular speed of 500 rev/min.

1) How much time is required to bring it up to speed, starting from rest, with a constant power input of 7.45×10^4 W?

2)Find the torque needed to cause the axis to precess in a vertical fore-and-aft plane at an angular rate of 1.00 degree/sec.



Homework Equations


I may be incorrect as I apparently keep coming up with the wrong answer. However, the moment of inertia for a solid cylinder I=1/2mr^2. P=power, I=moment of inertia, w=angular speed, a=angular acceleration, T=torque.

P=Tw. Ia=T. W=at.

For the second question, I'm a lot more lost. I'm guessing it's Torque/angular momentum but I continue to get incorrect answers.

The Attempt at a Solution



Ok, perhaps it's an issue with me actually understanding what is given because I would assume you would need a length from a pivot in order to work with a gyroscope. Since there is no such info in the problem, I assume that the disk is rotating directly on the pivot point.

I started with finding I. I=1/2(6.5*10^4)(2.3^2). I get 171925. Next, I use the Power equation to find the torque, using the given power and target angular speed. 7.45*10^4=T(500(rev/min). I convert 500 Rev/min to rads per second using 500*(2pi/60) to get 52.3599.

Then, solving for T, (7.45*10^4)/52.3599 to get 1422.84.

T=Ia. Using 1422.84=171925a, we find alpha to be .008276. Here is where I think there is something I may not be accounting for with torque due to the weight of the disc, but I assumed that since it is acting at the center of mass there is no torque. I tried the solution with torque being equal to mg + 1422.84, and the answer is still wrong.

Anyway, now that we have a, and given that the gyro starts from rest, w=at. So, 52.3599/.008276 = t, which =6327.

This answer is incorrect. Am I making incorrect assumptions somewhere as to using equations that wouldn't apply in this situation or something I am missing completely? Any help is much appreciated.

As for the precession part, that can wait until after I've figured out the first part.
 
Physics news on Phys.org


Any ideas?
 


Figured it out. My approach went off on a tangent because the angular velocity is not constant. Instead, you use the energy approach K=1/2Iw^2 and know that the work is equal to the change in kinetic energy, and Power is work over time. So, Power * change in time =W.

1/2Iw^2=P*t, divide by the given power.
 

Similar threads

  • · Replies 33 ·
2
Replies
33
Views
3K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 49 ·
2
Replies
49
Views
5K
Replies
10
Views
2K
  • · Replies 0 ·
Replies
0
Views
1K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 3 ·
Replies
3
Views
12K