A small ball rolls around a horizontal circle at height y inside the cone shown in Figure P7.40. Find an expre?

A small ball rolls around a horizontal circle at height y inside the cone. Find an expression of the ball’s speed in terms of y and g only.

a- the radius of the cone

h- the height of the cone

y- the height where the ball is rolling around in the cone.

2 Answers

  • I assume the cone is inverted so the apex is at zero height.

    The two forces on the ball are its weight “mg” straight down and the normal force “N” exerted by the surface, in a direction perpendicular to the surface. The normal force will make angle theta with the horizontal circle. The component along the radius is the centripetal force causing the circular path. The force eqs are;

    Ncos(theta) = mv^2/R

    NSin(theta) = mg

    Divide eqs to elliminate “N”;

    Tan(theta) = gR/v^2

    But “theta” is also the half apex angle so;

    Tan(theta) = a/h

    v^2 = gRh/a

    The Radius is related to the distance “y” from the vertx by ratio & proportion;

    R/y = a/h

    y = Rh/a

    Sub this in the velocity eq;

    v = SqRt(gy)

  • I consistently experience dizzy as quickly as I run around in circles. Plus it form of feels that recently, I seem doing the circle dance all the time anymore. around and around i am going. with a bit of luck, i will get off this loopy Merry go around quickly. :o)

Leave a Comment