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
'The mathematician's patterns, like the painter's or the poets, must be beautiful; the ideas, like the colours or the words, must fit together in a harmonious way. Beauty is the first test: there is no permanent place in the world for ugly mathematics.'