Department of Energy Argonne National Laboratory Office of Science NEWTON's Homepage NEWTON's Homepage
NEWTON, Ask A Scientist!
NEWTON Home Page NEWTON Teachers Visit Our Archives Ask A Question How To Ask A Question Question of the Week Our Expert Scientists Volunteer at NEWTON! Frequently Asked Questions Referencing NEWTON About NEWTON About Ask A Scientist Education At Argonne Rotational Motion
Name: James B.
Status: educator
Age: 30s
Location: N/A
Country: N/A
Date: Thursday, November 28, 2002


Question:
Re: gyroscope math and physics

I have salvaged the motors and disks from several 3 1/2 inch hard drives to act as demonstrations of gyroscopes. I am able to use from 1 to 5 disks as needed.

We know that adding more disks (ie rotational mass) increases the effectiveness" or "effect" of the gyroscope but to what extent athematically?......does doubling the rotational mass double the gyroscopic effect or is the relationship different than that? Also: I am able to vary the rotational speed from about 2500 rpm to about 7500 rpm. What is the mathematical relationship between RMP and gyroscopic effect?

In summary what can I predict mathematically between RPM and mass (what happens if I halve the mass and double the RPM's)?? Is there a simple formula to explain this question as there is, for example for a projectile F=MV^2?


Replies:
The angular momentum of a gyroscope (L), which you refer to as the "effect" of the gyroscope is given by L = Iw, where I is the "moment of inertia" of the gyroscope, and w is its angular velocity in radians/sec.

From this you can see that the gyroscopic effect is directly proportional to the rotational speed. You must multiply the rotational speed in revolutions per second by 2 pi to get the angular velocity. Thus, 7500 rpm is about 785.4 radians/sec.

The moment of inertia is more subtle. The moment of inertia of a mass point with mass m placed a distance r from the axis of revolution is I = mr^2. For a disk, you must break it up into tiny pieces, calculate the I for each, and sum them all up. For a uniform disk of mass M and radius R the result is I = MR^2/2.

Notice that if you double the radius of either a mass point or a disk without changing the total mass or the distribution of the mass, you quadruple the moment of inertia and so, for the same angular velocity. On the other hand doubling the mass while leaving the radius and mass distribution the same just doubles the angular momentum.

So if you halve the mass and double the RPM's, the angular momentum remains the same.

I don't know where you got F = Mv^2. The equation is incorrect as you can check by showing that the dimensions on the two sides are not the same. We know F = ma which has units kg m/s^2 whereas Mv^2 has units kg m^2/s^2.

Please let me know if this is not sufficiently clear or you would like more information.

Best, Dick Plano



Click here to return to the Physics Archives

NEWTON is an electronic community for Science, Math, and Computer Science K-12 Educators, sponsored and operated by Argonne National Laboratory's Educational Programs, Andrew Skipor, Ph.D., Head of Educational Programs.

For assistance with NEWTON contact a System Operator (help@newton.dep.anl.gov), or at Argonne's Educational Programs

NEWTON AND ASK A SCIENTIST
Educational Programs
Building 360
9700 S. Cass Ave.
Argonne, Illinois
60439-4845, USA
Update: June 2012
Weclome To Newton

Argonne National Laboratory