From: Ronald Wong (ronwong@inreach.com)
Date: Thu Jan 24 2002 - 16:31:27 PST
Message-Id: <l03102801b8750b6d5d6b@[209.209.18.248]> Date: Thu, 24 Jan 2002 16:31:27 -0800 From: Ronald Wong <ronwong@inreach.com> Subject: RE: angular momentum
Regarding angular momentum, "MC elover" <mcelover@yahoo.com> said:
>i had an important revelation yesterday about angular
>momentum... as i try to explain why the spinny
>stool does its magic... angular momentum is an incredibly
>important topic!!! ... conservation of angular momentum
>could be the most important law in all of physics...
So why don't we see a lot of coverage in rotational motion in our high
school physics classes?
One reason is that there isn't enough time to fit it in with everything
else during the first semester, when mechanics is usually covered, and
still do it justice.
Another reason is the fact that once the students see how great the
similarity is between linear and rotational KINEMATICS (distance, velocity,
and acceleration are simply replaced by angular distance, angular velocity,
and angular acceleration - with their attendant symbols and units) they are
bored silly and find the "problems" redundant - to say the least. From
their standpoint, the equations of motion are basically the same -- so the
problems appear to be "more of the same".
A third reason is that the DYNAMICS of rotational motion really throws a
monkey wrench into the way they have to think about things.
True, the similarity they noticed in the kinematics seems to appear in the
dynamics as well: Balanced forces are present when you have uniform
velocity, and balanced torques are present when you have uniform ANGULAR
velocity; unbalanced forces produce an acceleration that is proportional to
the resultant force and unbalanced torques produce an ANGULAR acceleration
that is proportional to the resultant torque; and forces appear in pairs
that are equal in size but opposite in direction and etc....
Of course, in place of mass, we have, for rotational dynamics, the moment
of inertia but this is just a reflection of the fact that the angular
acceleration is a function of the amount of mass an object has AND the way
that mass is distributed relative to the axis of rotation.
Still, there is the similarity in that F(net) = mass X acceleration and
T(net) = moment of inertia X angular acceleration.
And this extends to momentum: F(net) X time = change in (mass X velocity) and
T(net) X time = change in (moment of inertia X angular velocity).
So what's the problem?
The problem arises when we compare the affect that F(net) has on the motion
of an object as opposed to T(net).
For linear motion, the acceleration (that is, the rate of CHANGE in
velocity) is in the same direction as the F(net). Whereas in rotational
motion, the angular acceleration (that is, the rate of CHANGE in angular
velocity) - while in the same direction as T(net) - is in a direction that
is PERPENDICULAR to the forces that are responsible for T(net). This means
that the angular velocity will change in a direction perpendicular to the
plane determined by the force and it's moment arm. (for the technically
inclined, this is due to the fact that F(net) is the result of a vector sum
whereas T(net) is the result of a vector cross-product where T = r X F. The
order of "r" and "F" is important. Rotate the fingers of your right hand
from "r" to "F" as you form a fist with your thumb perpendicular to your
wrist and your thumb will point in the direction of the torque and thus the
direction of the CHANGE in angular velocity).
In high school physics, we keep things simple. The F(net) is constant and
either collinear or perpendicular to the velocity. In the first case, the
velocity will increase or decrease along the line. In the second case, the
speed will remain unchanged while the direction is constantly changing.
Replace F(net) with T(net) and add the modifier, angular, to the velocity
and speed and you have the case for rotational dynamics.
Eric's question regarding how "the spinny stool does its magic" as well as
a later question by Neil regarding the gyroscope involve situations where
the T(net) is perpendicular to the angular velocity.
First, an attempt to help Neil in his quest for a "clear explanation..."
There are a number of ways a gyroscope can be examined. Consider the case
where the spinning gyroscope is simply placed on a horizontal surface (or
on top of a small pedestal, or on string stretched between one's hand, or
suspended from a string held in one's hand - the other tied to the bottom
of the gyroscope). For some reason, the gyroscope will start precessing
about a vertical line that passes through the point of contact as soon as
it's axis of rotation tilts away from the vertical.
Why the precession? Why doesn't it just fall over?
Let's say that the gyroscope is spinning in a counterclockwise direction
when viewed down the axis of rotation. Using the right-hand rule, the
angular velocity vector, collinear with the axis of rotation, is pointing
away from the point of contact. Think of an arrow representing the
angular-velocity vector with it's tail at the point of contact and it's
head extending out of the top of the gyroscope.
There are two forces acting on the gyroscope: The weight of the gyroscope
acting downward from it's center of mass and an equal force acting upwards
at the point of contact. If you apply the right-hand rule to these forces,
you'll see that the torque they produce is perpendicular at all times with
the angular velocity vector. This means that the CHANGE in angular velocity
is always going to be perpendicular to the angular velocity vector (this is
analogous to an object spinning about your head on the end of a string. The
centripetal force and it's attendant acceleration (the rate of change of
velocity) are always perpendicular to the tangential velocity so that the
speed remains unchanged while the direction of the tangential velocity is
constantly changing about the center of rotation).
The CHANGE in angular velocity is a vector quantity and to see it's effect
on the angular velocity, draw a small arrow in the direction of the torque
and place it's tail at the head of the angular velocity vector. Since it's
perpendicular to the angular velocity vector, the angular speed remains
unchanged while the direction of the angular velocity vector is constantly
changing about the center of rotation (which is a point on the vertical
line passing through the point of contact) - sound familiar?
If you've done the above correctly, you'll see that the top of the
gyroscope will end up moving counterclockwise about a vertical line through
the point of contact when viewed from above. This is called precession. The
force that would ordinarily cause the gyroscope to fall down (it's weight)
is working in conjunction with the reaction force at the point of contact
to produce a torque that causes the gyroscope to precess instead of fall
down.
This explanation applies to all of the other examples given above as well.
Regarding Eric's "spinny stool":
You can use the same wordy explanation given above for what happens here,
but a simpler one is available because of the fact that we basically have a
closed system and thus can use the "conservation of angular momentum law"
(it didn't apply in Neil's case because the angular momentum was constantly
changing due to the constant change in direction brought about by the
unbalanced torque - i.e. it wasn't a closed system).
The classic demo involves someone sitting on a chair in the middle of a
platform that is free to rotate about the vertical axis through it's
center. The platform is usually at rest (this is not a required condition).
The person holds a spinning bicycle wheel and notices that when he/she
flips it over, the platform is set in motion.
Let's start with the axis of the bicycle wheel vertical and the direction
of rotation clockwise as viewed from above. The angular momentum has a
magnitude equal to the wheel's moment of inertia times it's angular
velocity and it's coincident with the wheel's axis and pointing straight
down. This is the total momentum of the system and must remain the same
throughout the change.
When the wheel is completely inverted from it's original position, the
total momentum OF THE SYSTEM must still be clockwise and of the same
magnitude as before. But now the wheel is turning in a counterclockwise
direction when viewed from above. Thus, during the time the wheel was being
turned over the rest of the system must have started moving clockwise (to
make up for the decreasing clockwise rotation of the wheel) until, in the
end it has twice the angular momentum in the clockwise direction that the
wheel originally had (remember: assuming that the wheel still is rotating
with the same speed, it will have the same size angular momentum as before
but in the counterclockwise direction. Combining this with twice the amount
in the clockwise direction yields the original angular momentum in both
magnitude and direction).
If he/she now turns it back up to the original orientation, the rest of the
system will come to a halt since the wheel, by itself, now has all of the
original momentum as before (a twist to this demo is to hand the spinning
wheel off to someone once it's inverted. It is then turned "right side up"
and handed back to the spinning individual who then inverts the wheel
again and so on...).
Go to:
http://solomon.physics.sc.edu/~tedeschi/demo/demo15.html
Look at the diagrams and, if you have a fairly fast connection to the
internet, watch the movie (Question regarding the movie: What was the
original direction of rotation of the wheel?)
Let's test your understanding: If you have a tippe top (the kind that flip
upside down when you spin them fast enough) and give it an initial rotation
in the clockwise direction, what is the direction of rotation after it
flips upside down and how is this achieved?
Enough said.
Cheers.
ron
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