Question

A steel hemispherical shell forms a dome over a silo the radius R is 3.0 cm a hollow sphere with inner radius r1 is 20.0 cm outer radius r2 is 40.0 cm and mass m is 4.0 kg is placed at the precise top of the dome and allowed to roll

Find the angle from the initial vertical position that the sphere looses contact with the dome?

Answer 1

here,

If it can be assumed that there is no friction involved you can use the equivalence of kinetic and potential energy:

mv^2/2 = mgh

Where
m = mass of skier
v =speed he is travellíng
g = acceleration due to gravity = 9.81 m/s^2
h = VERTICAL distance travelled by skier measured from his point of rest at the very top of the sphere.

The skier's mass cancels out so we can write:
v^2/2 = gh

Let's call A the angle measured between the vertical and a line drawn from the centre of the sphere to the location of the skier at any moment. When the skier starts off at the top of the sphere A = 0.

When does the skier actually fall off the sphere? The skier keeps moving in a circle around the sphere for as long as the force acting towards the centre of the sphere, the centripetal force, is sufficient to make him move in this circle. The force is provided by the component of the gravitational force that acts on the skier towards the centre of the circle.

The centripetal force needed to cause circular movement is F = mv^2/r
The component of the gravitational force which acts on the skier towards the centre of the sphere is:
G = mgcos(A).

Until the skier leaves the sphere:

G >=F

When the skier is right on top of the sphere, A = 0 and cos(A) = 1, the full gravitational force acts on tthe skier. As he moves round the sphere the gravitational force acting towards the centre of the circle gets smaller and smaller until:

F = G
At this point the skier loses contact with the sphere, so putting F = G results in:

mv^2/r = mgcos(A)

The skier's mass cancels out so:
v^2/r = g.cos(A)

Now we'll go back to kinetic and potential energy:

The vertical height of the skier above the centre line (the equator) of the sphere is given by
r.cos(A)
while the distance of the skier from the top of the sphere is
h = r - r.cos(A) = r(1-cos(A))

The relationship resulting from the equivalence between kinetic and potential energy (see above) is:
v^2/2 = gh, or v^2 = 2gh
Substitute for h to get

v^2 = 2g.r(1 - cos(A))

Now substitue v^2 into the condition resulting from the centripetal force
condition above ( v^2/r = g.cos(A) ) :

so
2g.r(1-cos(A))/r = g.cos(A)

Simplifying we get:

2(1 - cos(A)) = cos(A)
2 - 2.cos(A) = cos(A)
3 cos(A) = 2
cos(A) = 2/3

A = arccos(2/3) = 48.19 dgree

the angle from the initial vertical position that the sphere looses contact with the dome is 48.19 degree