If you roll a particle of a large spherical shape, at what point would it fall off?

Question

Assuming there's no resistance to motion, and the particle starts on the top and is nudged just a tiny bit, at what angle would it fall off the sphere?

Answer 1

Assuming frictionless ideal conditions, with no rotational inertia of the particle, we let radius of the sphere = r, outward centripedal acceleration = a, velocity of the particle on the surface = v, the angle of the particle's position above horizon relative to the center of the sphere = θ, acceleration due to gravity = g, and finally the drop in elevation of the particle = d. Then the surface velocity is, because energy is conserved,

v = √(2gd)

but d = R(1 - Sin(θ)), so that we have:

v² = 2gR(1 - Sin(θ))

The centripetal acceleration is then:

a = v² / R = 2g(1 - Sin(θ))

However, the radial component of the gravitational acceleration of the particle is

g' = g Sin(θ)

so that they are in balance at the point the particle leaves the surface, which means that

g Sin(θ) = 2g(1 - Sin(θ))

3 Sin(θ) = 2

θ = ArcSin(2/3) = 41.8 degrees approximately

The interesting thing about this is that it's independent of radius or gravitational acceleration.

Addendum: Alexander below gets a thumbs up for the expanded answer that includes rotational inertia.

Answer 2

Probably more of a physics question.

Cohesion, the property of matter to "stick" to other matter, would come into play, as well as your definition of "just a tiny bit" of a nudge, and the size of the particle. High cohesion, small nudge, small particle, and the particle would roll down the surface of the sphere until it reached halfway down and there was no surface under the particle to support it, so the angle would be 90 degrees. Low cohesion, big nudge, big particle, and it would go flying off like a shot, so the angle would be 0 degrees. More information is needed.

Answer 3

This question belongs to physics, really.

The 'particle' will separate from the surface of
the sphere when the force of gravity is no longer
sufficient to provide enough centripetal force.

That is v²/R = g sinΘ.
The dependence of velocity v on latitude Θ is
somewhat tricky. If you have a block sliding, then
v² = 2gR(1-sinΘ) and eqation is

2(1-sinΘ) = sinΘ, sinΘ = 2/3, Θ = 41.8 deg


If you have a solid ball rolling, then
v² = 2 * (5/7) *gR(1-sinΘ) and eqation is

10/7(1-sinΘ) = sinΘ, sinΘ = 10/17, Θ = 36 deg

The latter case in practice never happens,
because few moments before separation the
ball will start sliding no matter how good is the friction.

Answer 4

At a right angle, or 90 degrees based on a tanget to the sphere.