What happens to a rolling ball when it falls?

Question

Let's say a ball is rolling down a roof and then falls off it to the ground.

From the top of the roof to the border of the roof potential energy transforms to kinetic energy (both translational and rotational energy).

But I can't see what happens when the ball falls off the border of the roof? At first the ball has potentional, rotational and translational energy. Which of those energies transform into which energies as the ball falls? Does rotational energy stay the same or does it transform? Does the ball still "roll / rotates" in the air or does it stop, does it still roll after it falls to the ground?

Answer 1

It's normal when thinking about these things to make some simplifying assumptions, such as

• There's no air resistance.
• There's total contact friction (surfaces aren't sliding against each other).
• Collisions are either perfectly elastic or perfectly inelastic.
• Let's also assume that the ramp ("roof") has a little bit of a curve to it such that at the moment the ball leaves the ramp its velocity is purely horizontal.

So what happens?

• At $t_0$, when the ball is motionless at the top of the ramp, it has potential energy $e_{p0} = e_{total}$.
• At $t_1$, when the ball is at the end of the ramp,
  • it has potential, rotational, and kinetic energy such that $e_{p1}+e_{r1}+e_{k1}=e_{total}$
  • per our assumption about the shape of the ramp, all of $e_{k1}$ is represented by the horizontal velocity $v_{h1}$
  • per our assumption about the friction between surfaces, $e_{r1}$ is such that the speed of rotation $\omega_{1}$ "matches" $v_{h1}$. In other words, at $t_1$, the point of the ball that's touching that last point on the ramp has no velocity, and the point on the ball opposite it is moving at $2v_{h1}$.
• From $t_1$ to $t_2$ (when the ball touches the ground) the only force acting on the ball is gravity, which is purely vertical. This means the rotation and the horizontal velocity don't change.
• At $t_2$, when the ball touches the ground, it will have lost all of its relevant potential energy and gained additional kinetic energy. $e_{k2}+e_{r2}=e_{total}$. Since its speed of rotation hasn't changed, neither has its rotational energy; $e_{r2}=e_{r1}$.
• There's been no horizontal force, so $v_{h2}=v_{h1}$, and there has been no torque, so $\omega_1=\omega_2$. These are still matched. The part of the ball that touches the ground has no horizontal velocity, so no sliding "wants" to happen, so friction isn't exerting any force. The only forces during impact are the normal force from the ground and the force of gravity, both of which (since this object is a sphere) will be through the center of the ball's mass. Therefore there's no torque.
• The collision with the ground may be elastic (the ball bounces) or not. Either way, the ball keeps moving/rolling horizontally exactly as it was when it left the ramp.

Answer 2

If you neglect the friction of the ball with the air, it will continue rotating whilst falling as the angular momentum of the ball is conserved if no torque is acting on it. It should still roll after it falls to the ground, but may have lost some of its speed due to friction with the ground.

Answer 3

Does rotational energy stay the same or does it transform? Does the ball still "roll / rotates" in the air or does it stop, does it still roll after it falls to the ground?

As the ball travels (falls) through the air, only if some torque acts on it, will its angular velocity $\omega$) change and thus its rotational kinetic energy $K_R=\frac12 I\omega^2$ will also change.

What happens when the ball hits the ground depends a lot on what the "ground" really is.

E.g. let's say the ball hits wet ice (the coefficient of friction is very small or zero). In that case, during collision no torque is exerted on the ball and its $K_R$ doesn't change.

But if there is friction between the ground and the ball then part of it's $K_R$ will be lost as heat due to friction work and another part will be converted into translative kinetic energy. Which part exactly cannot be calculated by a simple formula because there's also rebound of the ball.

Answer 4

The ball has picked up X-translational velocity (away from the center of the roof) while rolling down the roof. Neglecting air friction (low velocity) the X-velocity remains constant after departing the roof (like a bullet fired horizontally from a gun (while losing X-velocity due to air friction (high velocity))).

It has picked up negative Y-velocity while rolling down the roof. When it leaves the roof, it continues with that initial negative Y-velocity and begins to increase negative Y-velocity due to falling (gravity).

When you throw a football, you put spin on it to give it stability. Do you think the football continues to spin when it leaves your hand (is it continuing to fly with stability throughout its travel - or does it tumble)? Do you think the rolling ball on the roof continues to spin when it leaves the roof (without air friction, what would stop it from spinning)?

Until the ball falls to the ground the potential energy remains but decreases due to the diminishing elevation relative to the ground. Until it hits the ground the kinetic energy is increasing due to the increased velcity due to gravity. (Does this sound like a transformation to you?)

When the bouncing stops, it still has (some of) the rotational velocity (may have lost some small fraction with each bounce) and the X-velocity, until finally the rolling stops (obstacles (sand/dirt/grass) and pits (small variations in the ground surface) in the path). At that point, all the potential energy and kinetic energy throughout its travel will have been spent as friction (heat) (energy loss due to deformations of the ground surface (plowing material sideways and compacting material under it (like rolling up a micro-hill)) and the ball surface at impacts and rolling friction (imagine the micro-deformation at the bottom of the ball and having to move that micro-deformation on to the next continued movement forward (as heating of the ball surface)) bringing it to a standstill.