|
Main Forum Page
|
The Gyroscope Forum |
27 November 2024 10:58
|
|
Welcome to the gyroscope forum. If you have a question about gyroscopes in general,
want to know how they work, or what they can be used for then you can leave your question here for others to answer.
You may also be able to help others by answering some of the questions on the site.
|
Question |
| Asked by: |
Luis Gonzalez |
| Subject: |
Gyro Knowledge - Which Gyro has faster precession? |
| Question: |
Given 2 gyroscopes built exactly the same way with only one difference. The flywheels of the gyros have different mass (therefore have different weights on earth) even though they have the same dimensions. (The wheels are made of different materials).
Both gyros are spun with the same angular velocity.
Will their rate of precession be different or the same?
If there is a difference which one will display faster precession the lighter one or the heavier one?
Regards,
Luis
|
| Date: |
31 July 2005
|
| report abuse
|
|
Answers (Ordered by Date)
|
| Answer: |
Luis Gonzalez - 06/08/2005 16:23:32
| | | The word genius is an easy word to use but not as easy to live up to. Genius can be claimed by making up rules described by made up terms and by answering questions with riddles and changing the conversation.
A good indicator of genius-shortage is when relevant questions, whose answer should be known within a venue, remain unanswered; especially when an accurate answer is essential to the goals of the venue.
Luis
|
| Report Abuse |
| Answer: |
Elder Dr. & Mrs. Hsien-LuHuang - 08/08/2005 19:47:47
| | | Dear Mr. Gonzalez:
The basic equation governing gyroscopic motion of your gyros is: T= I ωs ωp where T is the applied torque, I is the Mass Moment of Inertia of a disk flywheel of radius R and mass M, ωs is the spin angular velocity of the disk flywheel, and ωp is the angular velocity of the resulting precession. The product (I ωs) is known as the Angular Momentum of the flywheel. Assume natural precession where the torque T is provided by the force of gravity at the surface of the earth and ωp is the resulting natural precession in the horizontal plane. [s and p are supposed to be subscripts to ω. Likewise in the following paragraphs, 1 and 2 are supposed to be subscripts to T, L, R, M, ω and I ]
Assume
a spin axis having a length of L1 from the pivot point with disk flywheel 1 having a Mass Moment of Inertia I1, mass M1, and a spin angular velocity ωs1, and
a spin axis having a length of L2 from the pivot point with disk flywheel 2 having a Mass Moment of Inertia I2, mass M2, and a spin angular velocity ωs2.
Assume that both spin flywheels are spinning in the same clockwise direction and are precessing in the horizontal plane and that the spin axis has negligible mass.
The resulting precessional angular velocity ωp in the horizontal plane due to the force Mg of gravity of the spinning disk flywheel on the axis, or due to the torque T which equals to Mg L around the pivot point where g is the gravitational acceleration.
For the disk flywheel 1, T1= I1 ωs1 ωp1, or ωp1=T1/ (I1 ωs1).
For the disk flywheel 2, T2= I2 ωs2 ωp2, or ωp2=T2/ (I2 ωs2).
For a disk flywheel of mass M and radius R, I=(1/2)MRR
Therefore, ωp1=T1/ (I1 ωs1)=M1gL1/(1/2)M1R1R1=gL1/(1/2)R1R1, and
ωp2=T2/ (I2 ωs2)=M2gL2/(1/2)M2R221=gL2/(1/2)R2R2.
If L1=L2=L, and R1 = R2=R, then,
ωp1=gL/(1/2)RR, and ωp2=gL/(1/2)RR=ωp1.
Therefore, the answer to your question is, there is no difference between the rate of precession of the lighter disk flywheel and that of the heavier disk flywheel if the two gyroscopes are built exactly the same way with only one difference in the weight of the disk flywheel as desribed in your question.
Proverbs 1:7, The fear of the LORD is the beginning of knowledge.
Proverbs 9:10, The fear of the LORD is the beginning of wisdom.
Psalms 111:10, The fear of the LORD is the beginning of wisdom.
May our LORD GOD bless you all.
Elder Dr. & Mrs. Hsien-Lu & Hui-Lien Peng Huang
|
| Report Abuse |
| Answer: |
Elder Dr. & Mrs. Hsien-LuHuang - 08/08/2005 20:11:12
| | | Dear Mr. Gonzalez:
Sorry for the typo error in our derivation of equations.
The original one was,
Therefore, ωp1=T1/ (I1 ωs1)=M1gL1/(1/2)M1R1R1=gL1/(1/2)R1R1, and
ωp2=T2/ (I2 ωs2)=M2gL2/(1/2)M2R221=gL2/(1/2)R2R2.
If L1=L2=L, and R1 = R2=R, then,
ωp1=gL/(1/2)RR, and ωp2=gL/(1/2)RR=ωp1.
The corrected one should be,
Therefore, ωp1=T1/ (I1 ωs1)=M1gL1/(1/2)M1R1R1ωs1=gL1/(1/2)R1R1ωs1, and
ωp2=T2/ (I2 ωs2)=M2gL2/(1/2)M2R2R2ωs2=gL2/(1/2)R2R2ωs2.
If L1= L2 = L, R1 = R2 = R, and ωs1 = ωs2 = ωs, then,
ωp1=gL/(1/2)RRωs, and ωp2=gL/(1/2)RRωs=ωp1.
Thanks.
God bless you all.
Elder Dr. & Mrs. Hsien-Lu & Hui-Lien Peng Huang
|
| Report Abuse |
| Answer: |
Eric James ----- - 09/08/2005 08:14:28
| | | Luis and Elder Dr. & Mrs. Hsien-LuHuang,
That is the correct answer, but the math is very difficult to follow for most people (including me). A simpler explanation follows:
A gyroscope in precession desribes a cone based on the width of the gyro's mass, the diameter of the gyro, and the length of the axle from the pivot. The total mass of the gyro and the angular velocity determine how long it will precess, but the width of the gyro applying a torque on the axle through the bearings and the axle length, determines the offset of the gyroscopic couple and hence the precession speed.
Basically, the inner bearing load path circumference is smaller than the outer bearing load path, so the gyro must precess around the spindle to make up the difference. This difference is the same regardless of how fast the gyro spins or what its material properties are.
This is why precession remains constant even as the gyro slows down. As long as enough mass is sustaining enough angular momentum to apply enough torque to the bearings, the gyro will precess at a constant speed.
Anyway, I hope this helps.
Eric
|
| Report Abuse |
| Answer: |
Luis Gonzalez - 10/08/2005 23:15:05
| | | Despite using the same symbol for two different angular velocities, Dr. & Mrs. Huang managed to convey the solution accurately and elegantly.
In simple terms, the rate of precession is equal to the torque from gravity, divided by the angular momentum of the gyro flywheel. Both the Torque / Angular Momentum of the flywheel contain a factor M for the mass; therefore M cancels out from numerator and denominator and, since all other measures are the same, both gyros have the same rate of precession.
Eric’s explanation was not as easy to follow and I believe was in error stating that the rate of precession is independent from the angular velocity of the gyro flywheel. The fact is that a gyro’s rate of precession does speed up as the gyro slows down. This is reflected in the denominator (flywheel’s angular momentum) of the equation for precession.
The answer presented is correct only for a theoretical version of the problem where M reflects only the mass of the flywheel. A real world gyro requires a frame (to support the flywheel) which contributes to the mass in calculating the torque applied by gravity (the numerator of the equation). Both gyros received an increased gravitational torque (from the deadweight frame) but the ratio of the mass in the numerator to the mass in the denominator is different for each gyro. The gyro with the heavier flywheel has a smaller ratio of the masses ((M2 + a) / M2) than does the ratio of the gyro with the lighter flywheel ((M1 + a) / M1). This can be verified by plugging in different values for M while maintaining “a” constant. The result is that the heavier gyro should have slower precession for real gyros (with dead weight frames) that are exact in every way except for the mass of their fly wheels.
Thank you for the contributions,
Luis
P.S. Eric, what do you mean by “the width of the gyro's mass”??
|
| Report Abuse |
| Answer: |
Eric James ----- - 11/08/2005 08:03:47
| | | Luis,
I was expessing a solution based on geometry, rather than calculus.
A precessing gyro experiences torque (applied by gravity) that is in turn transferred into the gyro and its axis.
A finer (thinner) gyro will express this torque on a smaller area of the axle and will therefore have different attributes than a wider (thicker) gyro.
This torque is experienced on the lower side of the gyro, on the outer bearing at the axis, and on the upper side of the gyro on the inner bearing.
This differential causes the upper gyro couple to orbit around the pivot in a smaller circle than the lower gyro couple.
"Couples" define angular momentum. Opposite and equal couples define pure angular momentum (without linear momentum). Unequal couples define angular momentum and a net momentum.
The precessing gyro's couple is not equal, hence the gyro must rotate around the pivot to express the asymmetry.
Equally support the axle on both sides and the couple returns to an evenly aligned (symmetrical) configuration and precession stops.
The increase in speed you observe from a slowing gyro is caused by nutation that delivers angular momentum from the gyro directly into the axle/pivot system. This is why this increase is not apparent until the gyro rpms fall of significantly.
You are right to consider the frame. "Real world" constructions are full of chaotic contributions that are tough to fully analyze. Experimental evidence is always best.
If you don't believe me, experiment. You'll find that a thicker gyro with a given axle length (center to pivot) will precess faster than a thinner gyro with the same axle length.
A heavier (more massive) gyro will precess slower due to greater stability and less nutation, but this is due to having less chaos, not pure system geometry.
Nutation can deliver angular momentum to the system, but it is not the cause of precession. Nutation changes the angular relationship between the axle and the pivot and allows the axle to spin up. Basically, this is a bit of chaotic influence due to the need for an axle, a suport and a frictional pivot point.
Eric
|
| Report Abuse |
| Answer: |
Luis Gonzalez - 14/08/2005 17:24:19
| | | Eric,
When a flywheel spinning in space receives acceleration (torque) that affects the alignment of its axis of rotation, the flywheel (and axis of rotation) will change at 90 degrees (from the direction that the applied torque would have affected change to a non-spinning object). This same response (not a reaction) causes what we observe as precession in gyroscopes under the acceleration of gravity. The 90 degree response does not require interaction with any parts of a supporting structure such as bearings or such.
Even though gyros are not fully explored, precession is sufficiently defined mathematically to deliver accuracy in guidance systems. The math tells us that (all else being equal) precession is slower for faster rotating gyros.
The Angular Velocity of the flywheel (along with radius and mass) is a factor of the angular momentum (the sum of the moments of inertia) that is the DENOMINATOR in the equation used to calculate the rate of precession. The concept is not intuitive but neither is the behavior of gyros. Balanced equipment able to deliver and measure different gyro speeds are needed to verify the math with accurate measurements.
Your statement that gyros with wider flywheels produce faster precession than their thinner counterparts is intriguing because the math does not appear to bear it out. However, it appears that as the flywheel becomes wider more factors of interaction should come into play because different slices of the flywheel will have different rates of precession (consider a flywheel whose width is greater than its diameter).
Fact: The rate of precession is equal to the torque from gravity, DIVIDED by the angular momentum of the gyro flywheel.
Thank you,
Luis
|
| Report Abuse |
| Answer: |
Eric James ----- - 15/08/2005 12:15:49
| | | Luis,
My apologies to you and Elder Dr. & Mrs. Hsien-LuHuang
You are right, I am wrong.
In my model I didn’t account for the changes in torque in relation to angular velocity (oops). I used to know that too. My only excuse is age… really… I used to be a drag racer. I had to know all about torque and angular velocity… wheels and flywheels and crankshafts and gears are all forms of gyros… (I continue, muttering indecipherably in my beer mug).
The more massive gyro will precess slower at a given angular velocity, but not just due to the frame/deadweight to gyro relationship. At the same angular velocity (rpm), the more massive gyro will have a greater moment of inertia than a lighter gyro turning at the same angular velocity (a gyro twice as massive should precess half as fast).
I was also wrong about the width thing, except that a wide gyro tends to have a smaller moment of inertia to a gyro of similar mass that is narrower and has a correspondingly greater diameter. Therefore it precesses faster, but not if it is spun up to the same moment of inertia (higher rpm)
Please see the following page for a better explanation:
http://www.infoline.ru/g23/5495/Physics/English/gyro_txt.htm
Anyway, to apply torque to the axis of a gyro does require structure (an axle at minimum).
I was unaware that precession was used in guidance systems. I thought the point was to avoid changes in gyro orientation.
I verified the math myself. I created a pretty decent gyro out of an old tape player flywheel. It did the job well enough for me to count time versus precession rate as the gyro slowed. There was indeed a marked increase in the precession rate as the math predicts.
Thanks for the education.
Eric
|
| Report Abuse |
| Add an Answer >> |
|
