# 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

'As long as men are free to ask what they must; free to say what they think; free to think what they will; freedom can never be lost and science can never regress. '

(

**J. Robert Oppenheimer**(

*1904-1966*)