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?

Asked by:
Nunya

Answer

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:

Speed:
Friend= 2*pi*r/T
You= 2*pi*R/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:
Friend= rw
You= Rw

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.
Answered by:
Martin Archer, Physics student, Imperial College London, Uk

'Physicists like to think that all you have to do is say, these are the conditions, now what happens next?'