Ball contain Partical Rotateed with very high RPM, How much force will be applayed on partical?

Answer 1

Is the particle on the ball's equator? If so, then the force acting on the particle is:

F = mω²r

where:
m = mass of particle;
ω = ball's rotation rate, measured in "radians" per unit time (1 radian = 1/(2π) revolutions)
r = radius of ball.

If the particle is not on the ball's equator, you have to multiply that by an additional factor, cosθ, where θ is the angular distance from the equator to the particle's location.

Often it's useful to express this force as a fraction of the particle's weight. You do this by dividing F by the particle's weight W:

Force relative to particle's weight
= F/W
= mω²r / W
= mω²r / (mg)
= ω²r / g
(where "g" is the acceleration due to gravity, 9.8m/s²)

Here is an example:

Say the ball's diameter is 4 inches (radius = 2 inches), and it is rotatating at 4000 RPM:

force relative to weight = ω²r / g
= (4000 RPM)²(2 inches) / (9.8 m/s²)
= (4000×2π radians/minute)²(2 inches) / (9.8 m/s²)
= (4000×2π radians/minute × 1 minute/(60sec))²(2 inches × 0.0254 meters/inch) / (9.8 m/s²)
= 909.5

This means the particle would feel a force equal to 909.5 times its own weight (that is: 909.5 "g-forces")