Rolling ball down ramp to calculate acceleration of gravity

@roger,

Then you're as wrong as he is. Just consider: a .45 caliber slug will travel different distances if the cartridge is loaded with different amounts of propellant. So it will take different amounts of time to hit the ground. But the same slug if dropped from shoulder height will always take a constant amount of time to freefall to the ground.

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These Physics threads get so ridiculous. There is a correct answer here to physics problems. The correct answer involves math. The correct answer gets lost and physics gets turned into a creative writing experiment.

Kolyo already suggested a correct mathematical answer (if this is a thread about physics). It involves calculating the potential energy at the top of the ramp and the kinetic energy (both translational and rotational) at the bottom.

The answer will be

PE = KEtranslational + KErotational + LE (where lost energy is friction)

Which will turn into

mgh = 1/2mv^2 + 1/5mv^2 (assuming the golf ball is spherical and uniform density)

Which simplfies to

g = 7/10(v^2/h)

This is the correct answer (if we are actually doing Physics). The rest is silliness.

Of course the energy lost to friction, and the possibility that the golf ball is not a uniform sphere will effect the answer.

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My point is that you can't do Physics without math... and if anyone tries, they are bullshitting you.

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@layman,

Let's look at the bullet problem a different way.

If you fire a projectile horizontally, and the muzzle velocity is fast enough, and if you ignore wind resistance and interposing objects, the bullet would go into orbit. It will never hit the ground.

If you fire it just a little slower than that, it won't go into orbit but it will travel nearly around the globe, with the ground falling away from it the whole while, because of the curvature of the Earth.

Even an ordinary high velocity rifle bullet, with a muzzle velocity of 2,800 miles per hour and fired at a 45 degree angle, would be able to travel 90 miles under such circumstances; quite far enough for the curvature of the Earth to make a measurable difference in drop time.

http://www.explainthatstuff.com/bullets.html

Of course, air resistance is a factor, so the actual range of such a bullet is much less. But your hypothetical example implicitly assumed zero air resistance and no intervening objects.

p

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@maxdancona,

maxdancona wrote: "Kolyo already suggested a correct mathematical answer (if this is a thread about physics). It involves calculating the potential energy at the top of the ramp and the kinetic energy (both translational and rotational) at the bottom."

Nope. Two ramps of different incline angles will involve a ball path of different distances if you measure from top to bottom of the ramp. A six foot board, if stood horizontally, involves a six foot path. A ramp with near zero incline will involve an arbitrarily long ball path from top to bottom, depending on how close to zero you set the incline. So you're comparing apples to oranges.

If you take a ramp with a 45 degree angle, and another ramp with a near zero incline, and measure how long it takes a ball to move 20 linear feet along the ramp, the second ramp will produce a much longer travel time for the same distance. Therefore acceleration is much smaller.

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@puzzledperson,

Quote:

Even an ordinary high velocity rifle bullet, with a muzzle velocity of 2,800 miles per hour and fired at a 45 degree angle, would be able to travel 90 miles under such circumstances; quite far enough for the curvature of the Earth to make a measurable difference in drop time.

Let's do some math.

The radius of the Earth is 3959 miles.

The circumference of the Earth is 24,901 miles

A distance of 90 miles represents an angle of (measured as chords from the starting point and the ending points to the Earth's center) 90/24,901 * 360 degrees = 0.0036 degrees

The angle between the horizontal to the Earth's surface at the starting point to the straight line between the starting and ending points is A/2. So that would be 0.0018 degrees.

So the fall off of the earth over 90 miles would be 90 * sin (0.0018 deg) = 0.00283 miles = 14.9 feet.

That doesn't seem like very much to me (given the distance of 90 miles).

Physics is math.

1 Reply

@puzzledperson,

I wrote: " If you fire a projectile horizontally, and the muzzle velocity is fast enough, and if you ignore wind resistance and interposing objects, the bullet would go into orbit. It will never hit the ground. If you fire it just a little slower than that, it won't go into orbit but it will travel nearly around the globe, with the ground falling away from it the whole while, because of the curvature of the Earth."

Actually, if you fire the projectile close enough to orbital velocity, you can get it to make an arbitrarily large number of orbits around the Earth; but it won't stay in orbit.

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@puzzledperson,

Can you express that in math? You aren't making much sense... but it doesn't matter since you can't do physics without math anyway.

The original problem didn't use "linear feet". It talked about the actual length of the ramp, but again we need to put this in terms of math to do the physics anyway... so let's do it.

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@puzzledperson,

Physics is math.

If you want to make this argument, we can calculate the orbital velocity. Then we can take some amount off of this and calculate the path the object will take and see where it will intercept the Earth.

(Actually the "orbital velocity" is an strange term. again, if we are going to discuss this we should define this mathematically so that we can be exact and get the correct answer).

If you are not willing to do the math, then you are just making stuff up.

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@maxdancona,

By the way, orbital math is not that difficult. If you are willing to do a little algebra we can work a lot of this out given a few physical laws.

If you are interested, I would be happen to go over this.

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@maxdancona,

maxdancona wrote: "14.9 feet doesn't seem like very much to me"

That's not math: that's your subjective impression. layman claimed that a rifle bullet takes the same amount of time to hit the ground when fired as it does when dropped from shoulder height. That's mathematically false, whether air resistance is considered or not. And even though *I* could tell that it implicitly assumed no air resistance, in any empirical trial air resistance is a dominating factor. So from an engineering, or simply common-sense standpoint, it's also misleading. If you're going to illustrate ideal principles using "real-world" circumstances (shooting rifles) you're inevitably going to err.

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@puzzledperson,

Quote:

Just consider: a .45 caliber slug will travel different distances if the cartridge is loaded with different amounts of propellant. So it will take different amounts of time to hit the ground.

Say what? We're not talking about how FAR it goes, that was the whole point. We're talking about how long it takes to fall 6', if that's the figure you want to use.

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@puzzledperson,

Sure, The 14.9 feet is math. The question about whether this is a "measurable" difference is not. In engineering we often look at how significant an effect is compared to the overall measurements. If you are measuring in terms of hours, a couple of hundred milliseconds difference is generally not considered important (there are exceptions of course), if you are measuring in seconds... a couple hundred milliseconds is very important.

We could calculate the time it will take for the bullet to drop that 14.9 feet compared to the time it takes to complete the entire 90 mile trip. It will be very small compared to the overall trip time.

Should we do the math?

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@puzzledperson,

Quote:

If you fire a projectile horizontally, and the muzzle velocity is fast enough, and if you ignore wind resistance and interposing objects, the bullet would go into orbit. It will never hit the ground.

OK, sure, but I was talking about a rifle.

Quote:

Even an ordinary high velocity rifle bullet, with a muzzle velocity of 2,800 miles per hour and fired at a 45 degree angle, would be able to travel 90 miles under such circumstances;

1. I wasn't talking about a 45 degree angle
2. 90 miles? I doubt that, without worryin about the math.
3. At 2800 mph, an object will travel less than .3 miles in 3/8 second.

Quote:

your hypothetical example implicitly assumed zero air resistance and no intervening objects.

Sure it did. It also assumed it wasn't passing over top of a volcano which would suddenly erupt and create a powerful "updraft" force. So?

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@maxdancona,

maxdancona wrote: " The original problem didn't use "linear feet". It talked about the actual length of the ramp."

Nope. As usual you can't comprehend what you read. The original question said that he rolled a golf ball down a ramp and used a yardstick and a digital timer. It didn't say that he rolled the ball the entire length of the ramp. If he knew the length of the ramp he wouldn't need to measure the ball path using a yardstick; only if the ball path was less than the full length of the ramp. The distance would be linear anyway.

maxdancona: "Can you express that in math?"

Can I express what in math?

"You aren't making much sense..."

Sorry, I can't follow that unless you express yourself in math. Or maybe you could specify in words specifically what you are claiming doesn't make sense, and what specifically doesn't make sense about it.

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@maxdancona,

Quote:

Physics is math.

I thought Feynman was your idol.

I thought that if you posted a video by Feynman, you might actually listen to what it said.

Are ya deaf, or just lacking in memory?

Don't you recall Feynman saying "Physics is NOT math?"

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@maxdancona,

I'm sorry, but "significant difference" isn't math. It's a subjective, contextual, and somewhat arbitrary characterization. Are you saying that X + 14.9 is mathematically the same as X, for any value of X?

The Earth is round. It doesn't matter whether the difference is 14.9 feet or .149 feet. They aren't mathematically the same. Case closed.

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@layman,

Oh, so NOW you want us to ignore the curvature of the Earth as well as the existence of its atmosphere?

There is no distance, however short, for which a mathematical identity exists between the time it takes the fired bullet to hit the ground, and the time it takes to reach the ground in freefall when dropped from the same initial height. If maxdancona concludes otherwise, he needs his circuitry checked.

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@puzzledperson,

Physics is math.

With any of these factors, you can calculate the effect of each factor as a percentage of the total answer.

If I am buying a sandwich, I will worry if the bill has an extra dollar added. If I am buying a new car, a single extra dollar isn't something I am going to worry about. Sure, this is subjective.... but it can also be quantified. A dollar is 10% of the price of a $10 sandwich. It is 0.01% of the price of a $10,000 car.

I can use math to determine that the curvature of the earth changes the time traveled by less than 0.01%. You can then subjectively decide if this is important to you or not (of course if your measurements aren't very accurate than you would be foolish to even consider it).

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@puzzledperson,

Quote:

Oh, so NOW you want us to ignore the curvature of the Earth

Of course. The difference is miniscule, and in no way changes the general point I was making. You want to say something like:

BUT IT TAKE LONGER TO FALL 6.00000000001 FEET THAN IT DOES 6.0000000000 FEET!!!!!

Could ya get just a little more picky there, ya think, PP? I mean, like, would that be possible, ya figure?

Let me rephrase the point, eh? It will fall the same distance in the same time. There, I didn't use the word "ground" for a common, everyday illustration of a simple point. Satisfied?

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Rolling ball down ramp to calculate acceleration of gravity

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