Questions: From the top of a sphere of radius r a point mass m glides smoothly down (no friction). At what height does it detach itself from the surface of the sphere? Hint: the component of the gravitational force pointing towards the center of the sphere acts as centripetal force which is mv^2 / r for a circular path with radius r and velocity v. Furthermore, potential energy is transformed into kinetic energy. Combining these two conditions leads to the solution.

Solution Steps

At the point of detachment, the normal force (N) between the mass and the sphere becomes zero. The only force acting on the mass is gravity (mg).

The gravitational force (mg) can be resolved into two components: one tangential to the sphere (mg_sin θ) and one radial towards the center of the sphere (mg_cos θ), where θ is the angle between the vertical and the radius to the mass.

The radial component of gravity provides the centripetal force required for the circular motion of the mass until the detachment point. Therefore, mg*cos θ = mv²/r.

As the mass slides down, its potential energy is converted into kinetic energy. The loss in potential energy is mgh, where h is the vertical distance fallen. This equals the gain in kinetic energy, ½mv². So, mgh = ½mv².

The height fallen, h, can be related to the radius, r, and angle, θ: h = r - r*cos θ = r(1 - cos θ).

Substituting v² from the energy equation (v² = 2gh) into the centripetal force equation (mg_cos θ = mv²/r) gives: mg_cos θ = m(2gh)/r. Simplifying this and substituting h = r(1-cos θ), we get: cos θ = 2(1- cos θ) which simplifies to 3cos θ = 2.

Final Answer The height of detachment is (2/3)r