In an airless void, the answer is no—the mass terms of the force-due-to-gravity and the acceleration-due-to-force equations cancel out, and weight has nothing to do with the speed of the falling object.
In the presence of air resistance, however… the force from air resistance depends on how much air the object hits (which in turn depends on the shape of the object), and how fast relative to the object the air is moving. The force applied by air resistance is independent of the mass (but dependent on the shape and speed of the object) - but the acceleration caused by that force is dependant on the mass (f=ma). Therefore, the acceleration due to air resistance does depend partially on the mass of the object.
Okay, so not quite “nothing”, but mass is not the most important factor to consider in these equations...
I don’t know how you decide what’s more and what’s less important in physics equations :-/
If I tell you I dropped a sphere two inches in diameter from 200 feet up, can you calculate its speed at the moment it hits the ground? Without knowing its weight, I don’t think you can.
I don’t know how you decide what’s more and what’s less important in physics equations :-/
Predictive power. The more accurate a prediction I can make without knowing the value of a given variable, the less important that variable is.
If I tell you I dropped a sphere two inches in diameter from 200 feet up, can you calculate its speed at the moment it hits the ground? Without knowing its weight, I don’t think you can.
Ugh, imperial measures. Do you mind if I work with a five-centimetre sphere dropped from 60 metres?
A sphere is quite an aerodynamic shape; so I expect, for most masses, that air friction will have a small to negligible impact on the sphere’s final velocity. I know that the acceleration due to gravity is 9.8m/s^2, and so I turn to the equations of motion; v^2 = v_0^2+2*a*s (where v_0 is the starting velocity). Starting velocity v_0 is 0, a is 9.8, s is 60m; thus v^2 = (0*0)+(2*9.8*60) = 1176, therefore v = about 34.3m/s. Little slower than that because of air resistance, but probably not too much slower. (You’ll also notice that I’m not using the radius of the sphere anywhere in this calculation). It’s an approximation, yes, but it’s probably fairly accurate… good enough for many, though not all purposes.
Now, if I know the mass but not the shape, it’s a lot harder to justify the “ignore air resistance” step...
The speed as a function of time for an object with a constant drag coefficient dropping vertically is known and it is a direct function of mass. If I learned anything from making potato guns, it’s that in general, dragless calculations are pretty inaccurate. You’ll get the trend right in many cases with a dragless calculation, but in general it’s best to not assume drag is negligible unless you’ve done the math or experiment to show that it is in a particular case.
Huh. I thought the fact that it got continually and monotonically bigger until a given point and then monotonically smaller meant at least some aerodynamics in the shape. I did not even consider the wake...
The speed as a function of time for an object with a constant drag coefficient dropping vertically is known and it is a direct function of mass.
Well. I stand corrected, then. Evidently drag has a far bigger effect than I gave it credit for.
...proportional to the square root of the mass, given all oher factors are unchanged, I see.
It’s better than a flat plate perpendicular to the flow. Most people seem to not expect that the back of the object affects the drag, but there’s a large low pressure zone due to the wake. With high pressure in the front and low pressure in the back (along with a somewhat negligible skin friction contribution), the drag is considerable. So you need to target both the front and back to have a low drag shape. Most aerodynamic shapes trade pressure drag for skin friction drag, as the latter is small (if the Reynolds number is high).
For “an aerodynamic shape” my intuition first gives me a stylized drop: hemispheric in the front and a long tail thinning to a point in the back. But after a couple of seconds it decides that a spindle shape would probably be better :-)
The “teardrop” shape is pretty good, though the name is a fair bit misleading as droplets almost never look like that. Their shape varies in time depending on the flow conditions.
Not quite sure what you mean by spindle shape, but I’m sure a variety of shapes like that could be pretty good. For the front, it’s important to not have a flat tip. For the back, you’d want a gradual decay of the radius to prevent the fluid from separating off the back, creating a large wake. These are the heuristics.
Which shape objects have minimum drag is a fairly interesting subject. The shape with minimum wave drag (i.e., supersonic flow) is known, but I’m not sure there are any general proofs for other flow regimes. Perhaps it doesn’t matter much, as we already know a bunch of shapes with low drag. The real problem seems to be getting these shapes adopted, as (for example) cars don’t seem to be bought on rational bases like engineering. This should not be surprising.
cars don’t seem to be bought on rational bases like engineering.
Of course, but I don’t see it as a bad thing. Typically when people buy cars they have a collection of must-haves and then from the short list of cars matching the must-haves, they pick what they like. I think it’s a perfectly fine method of picking cars. Compare to picking clothes, for example...
You’re doing the middle-school physics “an object dropped in vacuum” calculation :-) If you want to get a number that takes air resistance into account you need college-level physics.
So, since you’ve mentioned accuracy, how accurate your 34.3 m/s value is? Can you give me some kind of confidence intervals?
You’re doing the middle-school physics “an object dropped in vacuum” calculation :-)
Yes, exactly. Because for many everyday situations, it’s close enough.
So, since you’ve mentioned accuracy, how accurate your 34.3 m/s value is? Can you give me some kind of confidence intervals?
No, I can’t. In order to do that, I would need, first of all, to know how to do the air resistance calculation—I can probably look that up, but it’s going to be complicated—and, importantly, some sort of probability distribution for the possible masses of the ball (knowing the radius might help in estimating this).
Of course, the greater the mass of the ball, the more accurate my value is, because the air risistance will have less effect; in the limit, if the ball is a hydrogen balloon, I expect it to float away and never actually hit the ground at all, while in the other limit, if the ball is a tiny black hole, I expect it to smash into the ground at exactly the calculated value (and then keep going).
And thus we get back to the question of what’s important in physics equations.
But let’s do a numerical example for fun.
Our ball is 5 cm in diameter, so its volume is about 65.5 cm3. Let’s make it out of wood, say, bamboo. Its density is about 0.35 g/cm3 so the ball will weigh about 23 g.
Let’s calculate its terminal velocity, that is, the speed at which drag exactly balances gravity. The formula is v = sqrt(2mg/(pAC)) where m is mass (0.023 kg) , g is the same old 9.8, p is air density which is about 1.2 kg/m3, A is projected area and since we have a sphere it’s 19.6 cm2 or 0.00196 m2, and C is the drag coefficient which for a sphere is 0.47.
So the terminal velocity of a 5 cm diameter bamboo ball is about 20 m/s. That is quite a way off your estimate of 34.3 and we got there without using things like hollow balls or aerogel :-)
To be fair, a light ball is exactly where my estimate is known to be least accurate. Let’s consider, rather, a ball with a density of 1 - one that neither floats nor sinks in water. (Since, in my experience, many things sink in water and many, but not quite as many, things float in it, I think it makes a reasonable guess for the average density of all possible balls). Then you have m=0.0655kg, and thus:
...okay, if it was falling in a vacuum it would have reached that speed, but it’s had air resistance all the way down, so it’s probably not even close to that. (And it it had been dropped from, say, 240m, then I would have calculated a value of close on 70 m/s, which would have been even more wildly out).
So, I will admit, it turns out that mass is a good deal more important than I had expected—also, air resistance has a larger effect than I had anticipated.
Nothing? Are you quite sure about that? :-)
Hmmm… let me consider it.
In an airless void, the answer is no—the mass terms of the force-due-to-gravity and the acceleration-due-to-force equations cancel out, and weight has nothing to do with the speed of the falling object.
In the presence of air resistance, however… the force from air resistance depends on how much air the object hits (which in turn depends on the shape of the object), and how fast relative to the object the air is moving. The force applied by air resistance is independent of the mass (but dependent on the shape and speed of the object) - but the acceleration caused by that force is dependant on the mass (f=ma). Therefore, the acceleration due to air resistance does depend partially on the mass of the object.
Okay, so not quite “nothing”, but mass is not the most important factor to consider in these equations...
I don’t know how you decide what’s more and what’s less important in physics equations :-/
If I tell you I dropped a sphere two inches in diameter from 200 feet up, can you calculate its speed at the moment it hits the ground? Without knowing its weight, I don’t think you can.
Predictive power. The more accurate a prediction I can make without knowing the value of a given variable, the less important that variable is.
Ugh, imperial measures. Do you mind if I work with a five-centimetre sphere dropped from 60 metres?
A sphere is quite an aerodynamic shape; so I expect, for most masses, that air friction will have a small to negligible impact on the sphere’s final velocity. I know that the acceleration due to gravity is 9.8m/s^2, and so I turn to the equations of motion; v^2 = v_0^2+2*a*s (where v_0 is the starting velocity). Starting velocity v_0 is 0, a is 9.8, s is 60m; thus v^2 = (0*0)+(2*9.8*60) = 1176, therefore v = about 34.3m/s. Little slower than that because of air resistance, but probably not too much slower. (You’ll also notice that I’m not using the radius of the sphere anywhere in this calculation). It’s an approximation, yes, but it’s probably fairly accurate… good enough for many, though not all purposes.
Now, if I know the mass but not the shape, it’s a lot harder to justify the “ignore air resistance” step...
(Engineer with a background in fluid dynamics here.)
A sphere is quite unaerodynamic. Its drag coefficient is about 10 times higher than that of a streamlined body (at a relevant Reynolds number). You have boundary layer separation off the back of the sphere, which results in a large wake and consequently high drag.
The speed as a function of time for an object with a constant drag coefficient dropping vertically is known and it is a direct function of mass. If I learned anything from making potato guns, it’s that in general, dragless calculations are pretty inaccurate. You’ll get the trend right in many cases with a dragless calculation, but in general it’s best to not assume drag is negligible unless you’ve done the math or experiment to show that it is in a particular case.
Huh. I thought the fact that it got continually and monotonically bigger until a given point and then monotonically smaller meant at least some aerodynamics in the shape. I did not even consider the wake...
Well. I stand corrected, then. Evidently drag has a far bigger effect than I gave it credit for.
...proportional to the square root of the mass, given all oher factors are unchanged, I see.
It’s better than a flat plate perpendicular to the flow. Most people seem to not expect that the back of the object affects the drag, but there’s a large low pressure zone due to the wake. With high pressure in the front and low pressure in the back (along with a somewhat negligible skin friction contribution), the drag is considerable. So you need to target both the front and back to have a low drag shape. Most aerodynamic shapes trade pressure drag for skin friction drag, as the latter is small (if the Reynolds number is high).
For “an aerodynamic shape” my intuition first gives me a stylized drop: hemispheric in the front and a long tail thinning to a point in the back. But after a couple of seconds it decides that a spindle shape would probably be better :-)
The “teardrop” shape is pretty good, though the name is a fair bit misleading as droplets almost never look like that. Their shape varies in time depending on the flow conditions.
Not quite sure what you mean by spindle shape, but I’m sure a variety of shapes like that could be pretty good. For the front, it’s important to not have a flat tip. For the back, you’d want a gradual decay of the radius to prevent the fluid from separating off the back, creating a large wake. These are the heuristics.
Which shape objects have minimum drag is a fairly interesting subject. The shape with minimum wave drag (i.e., supersonic flow) is known, but I’m not sure there are any general proofs for other flow regimes. Perhaps it doesn’t matter much, as we already know a bunch of shapes with low drag. The real problem seems to be getting these shapes adopted, as (for example) cars don’t seem to be bought on rational bases like engineering. This should not be surprising.
Of course, but I don’t see it as a bad thing. Typically when people buy cars they have a collection of must-haves and then from the short list of cars matching the must-haves, they pick what they like. I think it’s a perfectly fine method of picking cars. Compare to picking clothes, for example...
You’re doing the middle-school physics “an object dropped in vacuum” calculation :-) If you want to get a number that takes air resistance into account you need college-level physics.
So, since you’ve mentioned accuracy, how accurate your 34.3 m/s value is? Can you give me some kind of confidence intervals?
Yes, exactly. Because for many everyday situations, it’s close enough.
No, I can’t. In order to do that, I would need, first of all, to know how to do the air resistance calculation—I can probably look that up, but it’s going to be complicated—and, importantly, some sort of probability distribution for the possible masses of the ball (knowing the radius might help in estimating this).
Of course, the greater the mass of the ball, the more accurate my value is, because the air risistance will have less effect; in the limit, if the ball is a hydrogen balloon, I expect it to float away and never actually hit the ground at all, while in the other limit, if the ball is a tiny black hole, I expect it to smash into the ground at exactly the calculated value (and then keep going).
And thus we get back to the question of what’s important in physics equations.
But let’s do a numerical example for fun.
Our ball is 5 cm in diameter, so its volume is about 65.5 cm3. Let’s make it out of wood, say, bamboo. Its density is about 0.35 g/cm3 so the ball will weigh about 23 g.
Let’s calculate its terminal velocity, that is, the speed at which drag exactly balances gravity. The formula is v = sqrt(2mg/(pAC)) where m is mass (0.023 kg) , g is the same old 9.8, p is air density which is about 1.2 kg/m3, A is projected area and since we have a sphere it’s 19.6 cm2 or 0.00196 m2, and C is the drag coefficient which for a sphere is 0.47.
v = sqrt( 2 0.023 9.8 / (1.2 0.00196 0.47)) = 20.2 m/s
So the terminal velocity of a 5 cm diameter bamboo ball is about 20 m/s. That is quite a way off your estimate of 34.3 and we got there without using things like hollow balls or aerogel :-)
To be fair, a light ball is exactly where my estimate is known to be least accurate. Let’s consider, rather, a ball with a density of 1 - one that neither floats nor sinks in water. (Since, in my experience, many things sink in water and many, but not quite as many, things float in it, I think it makes a reasonable guess for the average density of all possible balls). Then you have m=0.0655kg, and thus:
v = sqrt( 2 0.0655 9.8 / (1.2 0.00196 0.47)) = 34.0785 m/s
...okay, if it was falling in a vacuum it would have reached that speed, but it’s had air resistance all the way down, so it’s probably not even close to that. (And it it had been dropped from, say, 240m, then I would have calculated a value of close on 70 m/s, which would have been even more wildly out).
So, I will admit, it turns out that mass is a good deal more important than I had expected—also, air resistance has a larger effect than I had anticipated.
Nope, because in that case, your value of g would be significantly higher than 9.8 m/s^2.