# Work-Energy Problem

A wire, which passes through the hole in a small bead, is bent in the form of quarter of a circle. The wire is fixed vertically on ground as shown in the figure. The bead is released from near the top of the wire and it slides along the wire without friction. As the bead moves from A to B, the force it applies on the wire is

• (A) always radially outwards
• (B) always radially inwards
• (C) radially outwards initially and radially inwards later
• (D) radially inwards initially and radially outwards later

I thought the correct answer will be (B). But, it's not. The correct answer is (D). I've tried hard but I cannot understand how to arrive at this conclusion. Please help me see how it follows.

• It really depends on how fast the bead moves and how large the radius of the wire is Feb 3, 2016 at 6:43
• This is a variation of a marble rolling down a large sphere problem where one has to find the point where the marble loses contact with the sphere. As at B the net force on the bead is downwards and the bead is accelerating down the wire the bead will always have a position where the wire exerts no force on it? Feb 3, 2016 at 7:42
• Thank you. Can you tell me about when will that point arrive. Feb 3, 2016 at 13:49
• How is this a work-energy problem?
– nasu
Nov 12, 2020 at 0:38

Think about the problem from the point where you neglect the wire and just analyse the motion.

You have gravity and at any other point the force has to balance gravity that you end up in a circular motion.

In A its towards the origin whereas in B there is no force (except gravity).

Different from the classical space-orbit or car-in-a-curve problem is here the relative orientation of the gravitational force.

You should use 1. Conservation of energy and 2. Centripetal force to the problem. I think you can figure out that when the angle subtended by the bead along with x-axis is between 0 and $sin^{-1}(2/3)$, the normal force acting on the bead due to the wire is inwards. Hence, the normal force on the wire would be outwards. It is an easy problem, so I am not writing the full solution, I hope you got the hints to solve it yourself.

• Can't really understand sorry. Why radially outwards later ? Feb 3, 2016 at 13:44
• try solving the problem, with gravity, Normal force acting on the bead. You know the expression for the centripetal force. You know the velocity of the bead at an angle using conservation of energy. So you can compute when the Normal force becomes negative given the angle, velocity. Feb 4, 2016 at 5:22

A suggestion; since the wire is in a circle, use polar coordinates. Resolve the force of gravity into radial and tangential components; the force of constraint of the wire on the bead is only in the radial direction (assuming no friction). Use the standard equations for acceleration in the radial and tangential directions for polar coordinates (found in any good physics textbook) to evaluate the dynamics of the motion.