Physics: Is it cricket or just a law?
August 18, 2004 7:23 PM
Newtonian physics/Cricket question. [6kb GIF and full question inside]
I'm no pro at either cricket or physics, and this has me vexed.
Assuming that the blue and red balls have identical shape, volume, and 'elasticity' (meaning they return the same bounce), they should both be affected by air resistance equally (right?). So... I've ruled that out.
I also think that on a frictionless surface, the "rebound angle" (just what I'm calling it... not sure what the real name is) would be identical... (let me know if this is wrong).
Now, here's the question.
In a "REAL WORLD" setting, where there is a consistant amount of friction in common between the two (say... astroturf acting on a leather ball), which angle, "a" or "b" would be greater?
In other words... Does the heavier ball not bounce as high?
All help appreciated greatly, this has been working on me for a few days.
I'm no pro at either cricket or physics, and this has me vexed.
Assuming that the blue and red balls have identical shape, volume, and 'elasticity' (meaning they return the same bounce), they should both be affected by air resistance equally (right?). So... I've ruled that out.
I also think that on a frictionless surface, the "rebound angle" (just what I'm calling it... not sure what the real name is) would be identical... (let me know if this is wrong).
Now, here's the question.
In a "REAL WORLD" setting, where there is a consistant amount of friction in common between the two (say... astroturf acting on a leather ball), which angle, "a" or "b" would be greater?
In other words... Does the heavier ball not bounce as high?
All help appreciated greatly, this has been working on me for a few days.
The heavier ball is deformed more and deforms the surface more; not all that energy is available to re-accelerate the ball (some is lost as heat and sound, both during and after the ball-surface contact). So it doesn't bounce as high. But if you want to know the angle at which it bounces, you'll have to figure out the components of the loss of energy normal and parallel to the surface, which depend on the materials in a complex fashion.
For example; some of the interaction energy will go into spinning the ball, and the softer the materials and the longer they are in contact, the more it'll be - and the greater the difference in the effect will be on the two balls. (And then the rotating ball has a complex friction interaction with the air. Oy.)
(I'm familiar with this problem from pool, where the speed of the ball and the firmness of the cushion factor into the angle of return in a way that's difficult to predict.)
posted by nicwolff at 8:46 PM on August 18, 2004
For example; some of the interaction energy will go into spinning the ball, and the softer the materials and the longer they are in contact, the more it'll be - and the greater the difference in the effect will be on the two balls. (And then the rotating ball has a complex friction interaction with the air. Oy.)
(I'm familiar with this problem from pool, where the speed of the ball and the firmness of the cushion factor into the angle of return in a way that's difficult to predict.)
posted by nicwolff at 8:46 PM on August 18, 2004
I also think that on a frictionless surface, the "rebound angle" (just what I'm calling it... not sure what the real name is) would be identical... (let me know if this is wrong).
This is wrong, unless the surface and the ball are made of a supermaterial that doesn't compress at all, and even then it's only close to right if the surface object is much more massive than the balls are (e.g. it's a planet). At any rate, in this case, you pretty much have an "inelastic collision" (a googleable term). Key to such a collision is that all the kinetic energy in the ball and the surface object is portioned back out into kinetic energy in the ball and the surface object after the collision; the amounts differ, however.
In the real world, things do squish a bit (rendering elastic collisions), and those dynamics are really complicated. Some of the energy squishes the ball and ground, leaving a divot or making heat. Some of the energy makes noise. Some of the energy moves the ball and the surface object after impact. The equations balance in the end (all the energy is accounted for), but to guess what happens to any accuracy, it probably helps to have a supercomputer.
posted by tss at 8:55 PM on August 18, 2004
This is wrong, unless the surface and the ball are made of a supermaterial that doesn't compress at all, and even then it's only close to right if the surface object is much more massive than the balls are (e.g. it's a planet). At any rate, in this case, you pretty much have an "inelastic collision" (a googleable term). Key to such a collision is that all the kinetic energy in the ball and the surface object is portioned back out into kinetic energy in the ball and the surface object after the collision; the amounts differ, however.
In the real world, things do squish a bit (rendering elastic collisions), and those dynamics are really complicated. Some of the energy squishes the ball and ground, leaving a divot or making heat. Some of the energy makes noise. Some of the energy moves the ball and the surface object after impact. The equations balance in the end (all the energy is accounted for), but to guess what happens to any accuracy, it probably helps to have a supercomputer.
posted by tss at 8:55 PM on August 18, 2004
Assuming that the blue and red balls have identical shape, volume, and 'elasticity' (meaning they return the same bounce), they should both be affected by air resistance equally (right?).
I doubt that air resistance is a factor here, but that statement's not quite true. If they have identical surfaces and identical cross sections (and identical spin), then the forces due to the air resistance will be the same; but the respective accelerations will be different, since F = ma. Specifically, the acceleration of the heavier ball due to the air resistance will be half that of the lighter one.
At any rate, in this case, you pretty much have an "inelastic collision" (a googleable term). Key to such a collision is that all the kinetic energy in the ball and the surface object is portioned back out into kinetic energy in the ball and the surface object after the collision; the amounts differ, however.
What Tom wrote above is all correct, except I think he switched the nomenclature just to confuse us. An elastic collision is one that preserves energy, an inelastic one doesn't.
posted by Johnny Assay at 9:30 PM on August 18, 2004
I doubt that air resistance is a factor here, but that statement's not quite true. If they have identical surfaces and identical cross sections (and identical spin), then the forces due to the air resistance will be the same; but the respective accelerations will be different, since F = ma. Specifically, the acceleration of the heavier ball due to the air resistance will be half that of the lighter one.
At any rate, in this case, you pretty much have an "inelastic collision" (a googleable term). Key to such a collision is that all the kinetic energy in the ball and the surface object is portioned back out into kinetic energy in the ball and the surface object after the collision; the amounts differ, however.
What Tom wrote above is all correct, except I think he switched the nomenclature just to confuse us. An elastic collision is one that preserves energy, an inelastic one doesn't.
posted by Johnny Assay at 9:30 PM on August 18, 2004
All of these answers have been very, very helpful (and reinforced my presumption).
Let me elaborate a bit:
I have two balls, one is a (regular) tennis ball, weighing around 2.5 oz. The other is a special "cricket" tennis ball, weighing around 4.5 oz. When dropped from an identical height, they both bounce around 60% of the original distance.
I'm throwing both onto astroturf, at around 35 m/s, at an angle parallel with the earth's surface, and from a release point around 7.5 feet.
Now... I'll understand if this is completely unfathomablly incalculable (tss, and nicwolff's comments of the complexity of the exchange make perfect sense to me, and i'm not really getting my hopes up)... but, what would the difference in rebound (first bounce) height be?
I'm not looking for "exactly 1.125 inches" ... I'm looking for something like:
a: "imperceptable"
b: a little (1"- 4")
c: quite a bit (4" - 12")
I'll understand if this can't be simplified any further, and I have a very profound satisfaction with the new insight I have because of the answers already posted.
Y'all are great, by the way.
posted by cadastral at 9:31 PM on August 18, 2004
Let me elaborate a bit:
I have two balls, one is a (regular) tennis ball, weighing around 2.5 oz. The other is a special "cricket" tennis ball, weighing around 4.5 oz. When dropped from an identical height, they both bounce around 60% of the original distance.
I'm throwing both onto astroturf, at around 35 m/s, at an angle parallel with the earth's surface, and from a release point around 7.5 feet.
Now... I'll understand if this is completely unfathomablly incalculable (tss, and nicwolff's comments of the complexity of the exchange make perfect sense to me, and i'm not really getting my hopes up)... but, what would the difference in rebound (first bounce) height be?
I'm not looking for "exactly 1.125 inches" ... I'm looking for something like:
a: "imperceptable"
b: a little (1"- 4")
c: quite a bit (4" - 12")
I'll understand if this can't be simplified any further, and I have a very profound satisfaction with the new insight I have because of the answers already posted.
Y'all are great, by the way.
posted by cadastral at 9:31 PM on August 18, 2004
D'oh! I always get those two terms mixed up. Many apologies!
posted by tss at 10:17 PM on August 18, 2004
posted by tss at 10:17 PM on August 18, 2004
If the surface is frictionless, then the rebound angles will be the same for A and B, since friction is the only way to transfer horizontal momentum to the ball. It's basically the same as bouncing them straight up and down--the horizontal velocity plays no role, and remains unchanged.
Once you add friction, the horizontal velocity will change on impact. The ball will start spinning, and some of the ball's horizontal momentum will be lost to the ground. How much will depend on the force between the ball and the ground, the coefficient of friction, and the duration of the impact.
The vertical impulse to the heavier ball (integral of F*dt, equal to the change in momentum) over the duration of the impact should be twice that of the lighter ball. Assuming a simple constant coefficient of friction, the horizontal impulse on the heavy ball will be exactly twice that on the smaller ball. This means that both components of the momentum after impact are proportional to the mass of the ball, so the angle of reflection should be the same, independent of the ball's mass.
Of course, to solve the problem without these assumptions, yeah, you're going to need a supercomputer and some really good modeling skills.
posted by yarmond at 11:18 PM on August 18, 2004
Once you add friction, the horizontal velocity will change on impact. The ball will start spinning, and some of the ball's horizontal momentum will be lost to the ground. How much will depend on the force between the ball and the ground, the coefficient of friction, and the duration of the impact.
The vertical impulse to the heavier ball (integral of F*dt, equal to the change in momentum) over the duration of the impact should be twice that of the lighter ball. Assuming a simple constant coefficient of friction, the horizontal impulse on the heavy ball will be exactly twice that on the smaller ball. This means that both components of the momentum after impact are proportional to the mass of the ball, so the angle of reflection should be the same, independent of the ball's mass.
Of course, to solve the problem without these assumptions, yeah, you're going to need a supercomputer and some really good modeling skills.
posted by yarmond at 11:18 PM on August 18, 2004
a: "imperceptable"
Assuming you actually hit them both the same speed and angle.
posted by Nothing at 2:18 AM on August 19, 2004
Assuming you actually hit them both the same speed and angle.
posted by Nothing at 2:18 AM on August 19, 2004
Or b: a little (1"- 4"), with the heavier ball boucing higher, if the astroturf is thick or padded.
posted by Nothing at 2:22 AM on August 19, 2004
posted by Nothing at 2:22 AM on August 19, 2004
You can't tell how "bouncy" something is just by knowing how much it weighs.
Other things being equal, a heavier ball of the same volume will bounce higher, since air friction slows it less. But it's mainly the elasticity that matters. A thin cover of leather wouldn't change that much.
In your tennis ball question... if you fill a tennis ball with lead, it will probably bounce less. If you suspend a smallish piece of lead right in the centre of the ball using string to attach it to various points on the edge, you might be able to get it to bounce higher.
posted by sfenders at 5:18 AM on August 19, 2004
Other things being equal, a heavier ball of the same volume will bounce higher, since air friction slows it less. But it's mainly the elasticity that matters. A thin cover of leather wouldn't change that much.
In your tennis ball question... if you fill a tennis ball with lead, it will probably bounce less. If you suspend a smallish piece of lead right in the centre of the ball using string to attach it to various points on the edge, you might be able to get it to bounce higher.
posted by sfenders at 5:18 AM on August 19, 2004
You can't tell how "bouncy" something is just by knowing how much it weighs.
(perhaps I was unclear... I came to the conclusion that they were the same 'bouncyness' by dropping them and measuring the rebound, not through assumption)
Many thanks for all the answers. There's some conflicting advice there, but I'm sifting through it.
posted by cadastral at 6:12 AM on August 19, 2004
(perhaps I was unclear... I came to the conclusion that they were the same 'bouncyness' by dropping them and measuring the rebound, not through assumption)
Many thanks for all the answers. There's some conflicting advice there, but I'm sifting through it.
posted by cadastral at 6:12 AM on August 19, 2004
This thread is closed to new comments.
posted by riffola at 8:31 PM on August 18, 2004