Aball 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?

Explanation:

At the bottom of the pipe :

= Normal force acting in upward direction

 = mass of the ball

 = weight of the ball acting in downward direction = mg

 = speed of the ball at the bottom

 = radius of curvature

Force equation for the motion of ball is given as

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.

i really dont know but i will try okay.

so i think u would just add all them up and get your answer..