BillyJoe said:
wdsmith
I don't think this quite accounts for the sensation of being pulled off the merry-go-round. If you are in orbit around the Earth, you won't get the sensation of being pulled off into space. Why not?
It seems to me the reason is that gravity acts on your body as a whole. Or, in other words, gravity acts on every part of your body. The merry-go-round, on the other hand, acts on only a part of your body.
On the merry-go-round, imagine that you are standing close to a bar fixed to the merry-go-round. The merry-go-round suddenly starts spinning very fast. You grab hold of the bar and your hands stay where they are fixed onto the bar. The rest of your body flings outwards. Your feet feel as if they are pulled out from under you and and end up about 7 ft away from the bar.
So I think it is the fact that the force of the merry-go-round is exerted directly on only one part of your body, whilst the rest of your body flings outwards, that you feel as if there is a force being applied to you trying to pull you off the merry-go-round. And you hold firmly onto the bar in a desperate attempt to stop this force from pulling you off.
Otherwise why wouldn't you feel the same sensation whilst in orbit around the Earth (for simplicity, imagine it is just you in orbit around the Earth).
BillyJoe
Not quite, Billyjoe:
The acceleration is acting on your whole body. Your arms are exerting the force that keeps you aboard. Do a free-body diagram...
The difference in forces is a direct result of the distance from center. Your legs are say, 2 meters further from the center of rotation than your hands. In a 150km orbit, that is a small effect. On a 3 meter merry-go-round, it is significant.
And wdsmith:
Yes, there are forces acting on the body--but they are in equilibrium.
Nobody said a thing about centrifugal
force--it is acceleration being discussed. (you can quibble over that all you want)
Centrepital acceleration is a result of the circular motion of a body. The mathematical value is r*w^2, where r is the radius of the circle, and w is the circular frequency (Radians/second)
Since centrepital acceleration forces the body to change direction (This is the gravitational acceleration acting on the body), the body's velocity is constantly changing (Velocity is a vector). The time rate of change of velocity is acceleration. This is the centrifugal acceleration, and it acts outward (away from the center of rotation
this circular velocity is always tangential to the circle. it is the straight line velocity should the centrifugal acceleration fail, and it acts opposite the centrepital acceleration. When r*w^2=g (where g is the local gravitational acceleration), then the sum of forces F=0.00
Should one of the two fail (ie, w-->0, or g-->0) then the object will travel in a straight line, either toward the center of mass (where w-->0), or tangentially off into space (where g-->0)
since the two accelerations are absolutely related to each other (assume g=constant,or reasonably so for some particular mass), if we increase the orbital velocity (w), then
r must get bigger unless another force acts on things to maintain equilibrium.
http://teacher.pas.rochester.edu/phy121/LectureNotes/Chapter14/Chapter14.html
also:
http://solarsails.jpl.nasa.gov/introduction/how-sails-work.html
has a pretty good diagram...
Now there are a number of folks who will take violent exception to my terminology, and my engineering poit-of-view on the world. Have at it. The princple remains...