If I am on a merry-go-round with a friend, and the friend is closer to center than I am, are they going faster then I am?
No, you are going faster. The merry-go-round is rotating with uniform circular motion - that means that it rotates at a constant angular speed.
Now lets say your friend is a distance r from the centre which is also equal to the radius of her circular path and you are a distance R.
Distance travelled in one period:
Friend = 2*pi*r
You = 2*pi*R
which is the circumference of your respective circular paths.
Lets say that the time period of one complete circle is T:
Now by the definition of angular speed
w=2*pi/T where w is the angular speed in radians/s
So this means that the speed is:
Since you are circling with uniform circular motion w is the same for both of you meaning that the closer you are to the centre the slower your linear speed is.
Martin Archer, Physics student, Imperial College London, Uk
'The strength and weakness of physicists is that we believe in what we can measure. And if we can't measure it, then we say it probably doesn't exist. And that closes us off to an enormous amount of phenomena that we may not be able to measure because they only happened once. For example, the Big Bang. ... That's one reason why they scoffed at higher dimensions for so many years. Now we realize that there's no alternative... '