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In this video

Khan academy explains conservation of energy for a falling object. He looks at an object falling perpendicularly from height h and computes its velocity at height zero. So far so good. But then he draws a curve that has several variations in height such that the object needs to climb back to certain height before falling back. At the end he computes again the velocity of the object and finds the same as object falling on the perpendicular.

Intuitively this seems wrong to me. To simplify his drawing I drew this picture:

enter image description here

I reason like this: The object falls to C but then it needs to climb back up to D and at D its vertical velocity is zero (because it changes direction, but not sure if this correct). So at B the velocity will be as if the object fell from D not from E. Is this correct? What is the best way of thinking about this problem?

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Conservation of energy doesn't care... the velocity you end up with is dependent only on the initial velocity and the height difference (assuming no friction). You can apply it for E-B or C-B or D-B, but each case will have a different initial velocity and height. – Chris Gerig Dec 30 '11 at 0:32
Is there a formula I can use for this case to find velocities at C and D? – Zeynel Dec 30 '11 at 13:24
Yes, the definition of conversation of energy, $KE+PE=\frac{1}{2}mv^2+mgh=constant$. – Chris Gerig Dec 30 '11 at 13:39

Assuming your diagram here is something like a ball rolling down a hill or a bead sliding on a wire (ignoring friction of course), you are right to say that the vertical velocity at D is 0, but this is irrelevant for the energy. What is relevant is the total velocity, and at D the total velocity there will be entirely horizontal, and given by the height difference $h_E - h_D$. For a simplifying example, consider a ball rolling down a quarter-pipe onto a flat surface. When the ball reaches the bottom, it has no vertical velocity, but only horizontal. The horizontal velocity will have the same magnitude as the veritcal velocity of a ball dropped straight down (with no intervening surface) from the same height.

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I am not disputing what happens in quarter-pipe, or an inclined plane situation where the ball is moving smoothly along an inclined plane and it is losing height at every moment. In this question, the ball falls to C and then climbs back up to D. Your example is not a "simplifying example" it is a different problem, the question is about how the body climbs up from C to D. On an inclined plane the ball does not climb up. – Zeynel Dec 30 '11 at 13:26
Again, intuitively, it appears to me that in this case we cannot assume no friction. How does the object change direction at C without friction? Assuming that it did and it started to move upward to D. What would make the object change direction at D instead of continuing to move upward in the direction of CD above D? In this case, assumption of no friction is equivalent to saying "assume that ECDB is an inclined plane" but ECDB is not an inclined plane. – Zeynel Dec 30 '11 at 13:26
Friction has nothing to do with it. The surface can exert a normal force while being frictionless. The shape of the path doesn't really matter, as long as the starting point is the highest point in the path, and the curvature constraint that Ron mentions is not violated. – Colin K Dec 30 '11 at 16:47

If the radius of curvature R at point D satisfies

$$v^2/R > g$$

(gravity is insufficient centripetal force) where v is the velocity computed from the conservation of energy, the sliding object will leave the ramp. This is why your intuition is upset-- if the object is moving fast enough, and the ramp is not sufficiently slowly curving, gravity will not keep the sliding object on the ramp. In the picture, the point D has a pretty small looking R.

Answering the title question

The title question is much more interesting than the example--- can energy be conserved when the constraints are non-differentiable? The answer is no, and a simple example is a cylinder that hits a step bump, and rises up.

Conservation of angular momentum at the contact point requires that if the cylinder doesn't bounce off, it loses energy at the bump. This is also true in other constrained systems with non-differentiable constraints, and the amount of energy loss is readily calculable from conservation principles alone, just from the form of the non-differentiability. This is a common olympiad style exercize in physics.

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I really doubt that the OP meant "smooth" in the sense that you (correctly) use it. This would be a good answer if it weren't almost certain to confuse the OP. – Colin K Dec 30 '11 at 16:44

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