What happens in non-uniform circular motion?

AI Thread Summary
In non-uniform circular motion, a particle can maintain circular motion if the radial acceleration is always perpendicular to the velocity. When a tangential component of acceleration is introduced, the particle's tangential velocity changes, but it can still remain in circular motion under specific conditions. If the acceleration is non-zero and always directed towards the center, circular motion is preserved. However, if the speed varies, the presence of a non-radial acceleration component can lead to a change in the motion path. Thus, the conditions for maintaining circular motion depend on the nature of the acceleration components involved.
BomboshMan
Messages
19
Reaction score
0
Hi,

Say there's a particle moving with just a radial component of acceleration, this will stay in circular motion because the acceleration is always perpendicular to the velocity. But if you introduce a tangential component of velocity, according to my book the particle stays in circular motion but it's tangential velocity changes. Why does this happen instead of the particle just moving in a path that isn't circular? Like an oval or something, seeing as the net acceleration no longer always points to the same place (centre of a circle).

Thanks
 
Physics news on Phys.org
Say there's a particle moving with just a radial component of acceleration, this will stay in circular motion because the acceleration is always perpendicular to the velocity.
This is true for a very special value of acceleration only.
You don't have to get a circular motion.
 
Last edited:
1. IF the acceleration is always perpendicular to the velocity, and non-zero, THEN you have circular motion.
Basically, as mfb says you, have muddled it.

2. However: If you make the PREMISE that you have circular motion, then it follows that if the speed is constant, your acceleration is strictly radially directed, but if the speed is non-constant, then you have a non-zero, non-radial acceleration component.
 
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top