In rotational motion, when the radius is decreased, does velocity increase or decrease?
Asked by: Lisa
Answer
There is no single answer to this question. The
answer is, 'it depends on the force'. Let us see
why.
First, lets take the case of a rock tied to a
string. In this case, the centripetal force is
supplied by the string, and is rather arbitrary,
as long as the string doesn't snap.
If you change the radius by shortening the string
slowly, the speed (velocity is not technically
appropriate here, that's a vector) will increase.
Since you're only pulling on the string, the
torque on the rock with respect to the rotation
center is zero, and angular momentum will not
change. Since L = mvr is then a constant of the
motion, the product of v and r is L/m, hence r
and v would be inversely proportional.
However, the nature of the force itself may
dictate what motions are possible, and hence
what the relation between v and r must be for
rotational motion involving that force. One
example would be gravitation, which applies to
the motion of satellites. In this case a simple
calculation yields that v^{2}r=GM, where G is the
gravitational constant, and M is the mass of the
earth (given the satellite is in earth orbit).
In this case, again, velocity increases as radius
is decreased, but to do that and remain in orbit,
one must fire engines to speed the satellite up!
Now it seems as if v always increases as r
decreases. But it does not. Now take the example
of a mass rotating tied to a spring, which
supplies a force F=-kr. A simple calculation
then shows that kr^{2} = mv^{2}, or that the angular
frequency, w is just sqrt(k/m). Hence, v = wr,
and since w is a constant independent of radius,
v and r are directly proportional. Increased r
will lead to increased v. In fact, one would have
to add energy to the system to increase r.
So, all in all, rotational motion itself does not
put a constraint on how r and v are related.
All it brings is one equation, which says the
centripetal force is mv^{2}/r. The second side of
the equation is supplied by the nature of the
force.
Answered by: Yasar Safkan, Ph.D. M.I.T., Software Engineer, Istanbul, Turkey
This is a good question and it is one that you will find has many answers depending on the
conditions you provide when you change the radius of your circle. You could go to any high school
physics book and find the equations that describe circular motion and from them you could get your
answer. But, this is PhysLink! Let's try another way to get your answer.
See if you can get a tube about the size of a straw. A straw itself will not hold up well to this
experiment, so see if you can get a tube of glass or plastic. The length of the tube should be at
least the width of your fist, so that the tube sticks out from the top and bottom of your closed
hand. Now, pass a string about a meter long through the tube. At the bottom of the string tie a
mass, about a half a kilogram (about a pound). At the top of the string tie a smaller mass. While
holding the tube in one hand start the top mass going in a circle and then by using the tube make
the top mass go faster and faster.
Is this not too cool?! Look what happens! As the top mass goes faster and faster the bottom mass
rises up! This means that the radius of the top mass's orbit is getting larger and larger. But,
is it going faster and faster? Now, let the bottom mass drop. As the radius of the circle
decreases what happens to is velocity? Does this surprise you?
But the fun does not stop here! What would happen if you got the top mass going at a constant
speed with a constant radius and someone else pulled down on the bottom mass? Consider what you
would have to do to keep the radius the same. Would you have to increase or decrease the speed of
the top mass?
How about this: Suppose you got the top mass to go around at a given speed with a measured radius.
You could do this by putting a paperclip on the string so that it just touches the bottom of the
tube when the top mass has the radius you want. The speed could be measured by counting the number
of times the mass goes around in, say, ten seconds. ( the Speed would be 2pi times r divided by
ten seconds). Then try this again with a different mass on the bottom and again with a different
mass on the top. You will quickly see that the speed of an object going in a circle in influenced
by a lot more than just the radius of its circle!
I hope you have fun with this and that you learn the answer to your question!
Answered by: Mr. Tom Young, B.S. Science Teacher, Whitehouse High School
'I believe there is no philosophical high-road in science, with epistemological signposts. No, we are in a jungle and find our way by trial and error, building our road behind us as we proceed.'