Deformation of a ball when it falls on a rigid surface

collisionelasticityforcesnewtonian-mechanicsstress-strain

Suppose, we drop a rubber ball on the floor. The ball collides with the floor, comes to rest and often bounces back up to lower height. I'm trying to look at this from the force perspective, not energy or momentum.

The ball, on hitting the ground, gets deformed, and being elastic, exerts some force on the ground to get back to its original state. Hence, the ground provides a normal reaction greater than the weight, which gives the ball an upward acceleration causing it to bounce back up.

However, what compresses the ball in the first place ? There is some stress acting on the ball, but where does it originate ?

I like to think of it as, the lower surface of the ball comes to rest first, while the center of mass keeps moving due to inertia. This creates the stress, that deaccelerates the center of mass, till it comes to a halt. Is this the source of the compression stress ? I'm not sure.

There seems to be another explanation : when the ball drops on the ground, even when the reaction matches the weight, the ball has a downward velocity. Hence the ground must provide a force greater than the weight, to stop this velocity. This weight compresses the ball, which tries to retain its shape, and thus apply a greater force on the ground, which in turn leads to greater reaction and more compression until the ball stops. At this point, this total reaction accelerates it backwards i.e. upwards.

Are these two explanations equivalent ? If not, what is the correct explanation ? From a purely force perspective, where does the compressive stress originate. If I consider the ball to be something like a mass on a spring, that suggests the first explanation is correct.

I think the the second explanation is needlessly convoluted, and that the first explanation is the correct one. It is the inertia and the acceleration of the top of the ball relative to the lower half, that creates the compression, that ultimately increases the normal reaction, causing the ball to bounce back up.

Is this correct ?

Best Answer

Both of the effects you describe take place. The centre of mass of the ball does continue to move downwards after the lowest point of the ball has hit the ground, which means that the ball is deformed. The reaction force subjected by the ground is greater than the weight of the ball- it has to be in order to accelerate the ball. You should be able to see that intuitively, as it accounts for the fact that a ball simply at rest on the ground is not kicked in to the air by a normal force equal to its weight. The reaction force increases with the speed of incidence of the ball, which accounts for the fact that the height of a ball's bounce will depend upon the height from which it drops.

Related Solutions

[Physics] Model/formula for bouncing ball

You say you are ignoring friction of the air, but assume a partially inelastic collision. That means we can break the problem into two parts:

While the ball is not in contact with the ground, the height at time $t$ after the last bounce at $t_0$ is given by $$h(t+t_0) = v_0 t - \frac12 g t^2$$ where $v_0$ is the velocity just after the bounce. This velocity will change from one bounce to the next.
During the impact, the ball will deform and there will be friction. The exact shape of the ball during the impact is hard to capture in a simple equation because it depends on the composition of the ball and the floor. As a simplifying assumption, consider the ball to be a "mass on a slightly lossy spring with a spring constant $k$"; then the ball will exhibit (approximately) damped simple harmonic motion during the impact: this means that the impact time will be independent of the impact velocity(!)

The coefficient of restitution is a parameter of a ball/surface, and reflects the fraction of velocity just after the bounce divided by the velocity just before. For a golf ball, this can be as high as 0.8; for a tennis ball it is around 0.75.

Using $\rho=0.75$, if you drop a ball from a height of 5 m (and putting $g=10~m/s^2$ for convenience) it would land after 1 second with a velocity of 10 m/s, and after the bounce (some short time $\Delta t$) it would come up again with a lower velocity $v_1 = 7.5 m/s$. The time spent in the air until the next bounce is given by $$t_1 = 2\frac{v_1}{g} = 2\rho \frac{v_0}{g}$$

While the ball is in contact with the ground, we can consider its center to move in an approximately sinusoidal fashion - as I said, the important thing is that the contact time is independent of the drop height (on the assumption of linear spring constant). In reality, as the ball deforms more, the spring constant will increase - which will actually make the contact time shorter as the impact velocity is higher. But for your simulation that is unlikely to matter.

To get from the above to a "mathematical model" you might want to modify this python code to your needs:

from math import sqrt
import matplotlib.pyplot as plt

h0 = 5         # m/s
v = 0          # m/s, current velocity
g = 10         # m/s/s
t = 0          # starting time
dt = 0.001     # time step
rho = 0.75     # coefficient of restitution
tau = 0.10     # contact time for bounce
hmax = h0      # keep track of the maximum height
h = h0
hstop = 0.01   # stop when bounce is less than 1 cm
freefall = True # state: freefall or in contact
t_last = -sqrt(2*h0/g) # time we would have launched to get to h0 at t=0
vmax = sqrt(2 * hmax * g)
H = []
T = []
while(hmax > hstop):
  if(freefall):
    hnew = h + v*dt - 0.5*g*dt*dt
    if(hnew<0):
      t = t_last + 2*sqrt(2*hmax/g)
      freefall = False
      t_last = t + tau
      h = 0
    else:
      t = t + dt
      v = v - g*dt
      h = hnew
  else:
    t = t + tau
    vmax = vmax * rho
    v = vmax
    freefall = True
    h = 0
  hmax = 0.5*vmax*vmax/g
  H.append(h)
  T.append(t)

print("stopped bouncing at t=%.3f\n"%(t))

plt.figure()
plt.plot(T, H)
plt.xlabel('time')
plt.ylabel('height')
plt.title('bouncing ball')

This code produces the following plot:

What makes a ball bounce back up

First is the normal force exerted on the floor by the ball (indirectly due to gravity).

Gravity is not the only effect here. The ball is moving downward when it reaches the floor. In order to stop moving downward, the ball has to undergo an acceleration upward. This acceleration requires a force: $a = F/m$. The strength of this force and acceleration depends on the speed of the ball (how much acceleration is required) and the materials (how far the ball can deform).

It may be helpful to think of the ball as a spring. The force the ball exerts depends on the deformation, just like the force exerted by a spring depends on the deformation.

When the ball first touches the floor, the normal force is small, so the ball is still accelerating downward. After a certain amount of deformation, the ball is exerting a force equal to its weight. But this doesn't stop the ball, it just stops the acceleration. The ball continues to move downward.

As it does it compresses further. The normal force exceeds the weight and the ball begins to slow down. This continues until the acceleration removes all of the ball's downward velocity. The spring forces are still greater than gravity though, so the upward acceleration continues. This is sufficient to give it an upward velocity that allows it to bounce off the floor.

So, the normal reaction should always be more than the weight of a falling body, to account for that extra velocity that remains, even after the net force acting on the body is 0.

Well, at the instant the net force is zero, the normal force would be equal to the weight. But the downward velocity that remains will cause the ball to move closer to the floor and for the forces to increase with time.

For a stationary object kept on a floor, the normal force is exactly equal to weight, as there is no 'extra downward velocity'. Are these assertions correct ?

Right. The system is in equilibrium. The weight is exactly opposed by the normal force, so the object does not accelerate.

Best Answer

Related Solutions

[Physics] Model/formula for bouncing ball

What makes a ball bounce back up

Related Question