about circular motion

Q: if a ball is attached to a string doing circular motion, by F=mv^2/r
F increases
m constant
r constant = length of the string
Thus v increase

I think you have reversed the "cause and result". The equation gives the magnitude of the necessary centripetal force. It shows that if v increases, then a larger centripetal force is needed to maintain the same circular path with radius r. Note that v is the "cause", and F is the "result".

In that sense, the increase of v is brought by an external agent. This external force will act in the same direction of v (as contrast to the centripetal force which acts perpendicular to v), such that there is an increase of v, and subsequently an increase of kinetic energy. Therefore, the external agent (force) has done work on the object to increase its kinetic energy.

Do Not confuse work done by an external force with work done by centripetal force.

This is because the centripetal force is ALWAYS perpendicular to the velocity (tangential) of the object performing circular motion.

Then, since the object remains the same distance from the centre of the circular orbit, the NET displacement of the object in radial direction, i.e. the direction of action of centripetal force, is ALWAYS zero.

Finally by work done = force x displacement (vectorial product), the work done by the centripetal force is zero.

2010-07-05 20:33:42 補充：
Yes, by conservation of energy, when v increases, K.E. also increases and the increase in energy should be due to work done by external force.

However, it should be noted that v is ALWAYS TANGENTIAL to the circular path. So, the acceleration direction is ALWAYS PERPENDICULAR to the centripetal force.

2010-07-05 20:33:46 補充：
Therefore, we can conclude that there should be another external force other than the centripetal force to make tangential acceleration. Hence the work done is by the tangential force but NOT the centripetal force.