Question

A wire, which passes through a hole in a small bead, is bent in the form of a quarter of a circle. The wire is led 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

Answer 1

The correct option is D radially inwards initially and radially outwards later

When the ring will start sliding downward then at the initial stage its velocity will be less

So, it will require less centripetal force while it will get more component of its weight for the centripetal acceleration.So, the wire will exert force on the ring in the radially outward direction.Therefore the ring will exert force on the wire in the radially inward direction.

Similarly in the later stage the velocity of the ring will increase and the weight component in the radial direction will decrease.So, wire will exert force radially inward and ring will exert force on the wire in radially outward direction