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

Answer:

$mg +\frac{mv^{2}}{r}$

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

$N - W = \frac{mv^{2}}{r}$

$N - mg = \frac{mv^{2}}{r}$

$N = mg +\frac{mv^{2}}{r}$