A very small body with the mass m and the virtually zero initial velocity v0 = 0 slides down on a fixed smooth spherical surface with the radius R from topmost point.

Question

A very small body with the mass m and the virtually zero initial velocity v0 = 0 slides down on a fixed smooth spherical surface with the radius R from topmost point.

-Find the momentum vector p components px and py at the moment, when the body leaves contact with the sphere

Answer 1

We will use the law of conservation of mechanical energy and Newton's second law.

Let's chose the zero's level of energy the level, that is parallel to the ground and goes through the point 2, at which the body is leaving the contact with the surface. Point 1 is on top of the spherical surface. Then energy at point 1 is equal to energy at point 2; we can write

E₁ = E₂

At point 1 the body has only potential energy E₁ = mgh and at point 2 only kinetic energy mv²/2, or

mgh = mv²/2

From this equation

v² = 2gh (1)

From the geometry point of view

h = R(1 - cosΘ),

here R - radius of the sphere and Θ is the angle between the vertical line and radial line that goes through the center of the sphere and through point 2. Putting everything into equation (1), we get

v² = 2gR(1 - cosΘ) (2)

Let's use the Newton's Second Law to find cos Θ.

For body at point 2 we can write

mg + N = ma (3)

If we select the coordinate axis x in the direction to the center of the sphere along its radius and going through point 2, then in scalar form we get:

mgcosΘ - N = ma (4)

Normal reaction force N is greater or equal 0. The body will leave the sphere when N = 0, and hence we can write:

mgcosΘ = ma

or a = gcosΘ (5)

a is centripetal acceleration

a = v²/R (6)

Combining formulas (2), (5) and (6), we obtain:

gRcosΘ = 2gR - 2gRcosΘ

It is easy to see that

cos Θ = 2/3

Then using trigonometric unit (sin²Θ + cos²Θ = 1), we can find the sinΘ;

sin Θ = √5/3

The momentum vector p is tangent to the sphere and has the same direction as velocity v.

Component p_x = mvcosΘ and component p_y = - mvsinΘ. Using the values of corresponding trigonometric functions, we finally can write:

p_x = (2/3)mv p_y = - (√5/3)mv