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
Our server costs have gone up and our advertising revenue has gone down. You do the math! If you find our site useful, consider donating to keep us going. Thanks!
'On the mountains of truth you can never climb in vain: either you will reach a point higher up today, or you will be training your powers so that you will be able to climb higher tomorrow.'