# A pearl that moves in a smooth vertical hoop

I wanted to ask about the situation of a pearl that moves in a smooth vertical hoop in circular motion as described in the following sketch.

According to a simulation found in the internet , a moment after this situation , when the pearl reached the point C , the pearl keeps moving in circular motion. What causes the pearl to keep moving the in circular motion after reaching point C? (The reference frame is Inertial frame)

We already know that a(r) = V ^ 2 / Radius , at point C , v = 0 , therefore a(r) = 0. We also know that a(r) is the only acceleration of the pearl at point C.

So what makes the pearl keep moving in circular motion ? From the equations we can conclude that the pearl should stop moving at that point but according to the simulation , that's not true.

• Link please. And also your direction of N is opposite. – evil999man Apr 12 '14 at 11:35
• That's what special about pearl , even look in the internet , when normal equals zero , the normal force changes direction. and i prefer not to link the simulation , the simulation found in our school site and i don't want to publish it . this sketch contains all the information you need to answer my question. – Michael Apr 12 '14 at 11:41
• I don't see why Normal reaction is upwards. – evil999man Apr 12 '14 at 11:42
• Look , the pearl has internal and external surface. At some point between the most horizontal point and the point C , the normal equals 0 , and his direcion becomes upward because the normal force that was acting on the pearl from the internal surface now becomes 0 , and the normal force from the external surface gets value. – Michael Apr 12 '14 at 11:44
• The pearl is over the loop...ok – evil999man Apr 12 '14 at 11:45

Unfortunately I cannot comment due to insufficient reputation, so here a comment on the question. There are three cases:

1. $\frac{1}{2}mv_A^2>2mgR$ In this case the pearl has a velocity $v>0$ in the top point and will continue its movement.
2. $\frac{1}{2}mv_A^2<2mgR$ In this case the pearl won't reach the top and will oscillate around point $A$.
3. $\frac{1}{2}mv_A^2=2mgR$ In this case the pearl will indeed reach the top with zero velocity and remain there forever in an unstable equilibrium. However, if you write a simulation, many things can happen that prevent this case to occur.

a) The rounding errors cause the velocity at the top to be non-equal zero.

b) The positions of the pearl are calculated in small steps (that's how numerical calculations work) and the top is not amongst them. Then the pearl springs from a point right before the top to a point right after the top.

c) The initial conditions are given with insufficient accuracy (again rounding problem).

In short, the problem is not a physical but a computational one. So, as mentioned by Awesome, we need the code to give you the reason for the inaccuracy.

• Now you can. Feels good, doesn't it? – evil999man Apr 12 '14 at 12:22
• Indeed, let's use the opportunity! :) – Photon Apr 12 '14 at 12:33