No, it is *not.* Uniform circular motion refers to constant speed. The speed varies around the loop. a does not equal v^2/r anywhere except at the top or bottom of the loop.
From http://en.wikipedia.org/wiki/Mechanics_of_planar_particle_motion#Polar_coordinates_in_an_inertial_frame_of_reference , I see where we can come to this conclusion. v = rdot rhat + r theta-dot theta-hat and a = (r-double-dot - r theta-dot^2) rhat + ( other terms ) theta-hat. For...
Studying the text, it seems to me that in the derivation of the formula, one is using the fact that |delta v| / |v| = |delta r| / |r|, where all the letters represent vectors. This coems from the fact that the speed is constant. In the problem at hand, could it be be we are using the fact that...
Oops! That title should read constant *speed.*
1. An object with mass m starts at rest at height h. It slides down a frictionless incline. At the bottom, it enters a vertical circular loop-the-loop of radius r. It just touches the track at the top of the loop. What is h in terms of m...