Question

A ball of mass m is released at the top edge of a frictionless half-pipe with a radius of curvature r. At the bottom of the pipe, the ball is moving with a speed v. What is the normal force acting on the ball at this point?

Answer 1

`mg +(mv^(2))/(r)`

Step-by-step explanation:

At the bottom of the pipe :

`N` = Normal force acting in upward direction

`m` = mass of the ball

`W` = weight of the ball acting in downward direction = mg

`v` = speed of the ball at the bottom

`r` = radius of curvature

Force equation for the motion of ball is given as

`N - W = (mv^(2))/(r)`

`N - mg = (mv^(2))/(r)`

`N = mg +(mv^(2))/(r)`