

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
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