# I Terminal velocity of two balls with different masses

1. Feb 7, 2019

### liroj

Two objects of the same shape (say balls) fall through the earth's atmosphere. For simplification, let's say that the air density is the same (some average sea level value) despite altitude change.
One ball has considerably bigger mass than the other, but besides that, they're the same.

Will the heavier ball accelerate faster even before the lighter ball reaches it's terminal velocity or will they accelerate at the same rate until that point, and only after the lighter ball one tops at it's terminal velocity and stops accelerating will the heavier ball win the "race" and continue accelerating further.

Some math/formulas behind the explenation would be helpful.

2. Feb 7, 2019

### kuruman

Suppose you release two identical balloons, one filled with air and one filled with lead shot. Which one do you think will hit the floor first?
As for the equations, just do a search on "free fall with air resistance". There is a whole lot out there including videos so take your pick. Terminal velocity is reached asymptotically and the mass has to do with how fast terminal velocity is reached.

3. Feb 7, 2019

### liroj

I understand that part but what puzzles me is the acceleration of both balls until the lighter one reaches it's terminal velocity. Will their speed be the same at that point or will the heavier ball accelerate faster the whole time?

4. Feb 7, 2019

### Staff: Mentor

Do dropped objects ever accelerate at a constant rate in air? What forces are acting? Are the forces constant?

5. Feb 7, 2019

### liroj

Acceleration drops as objects keep falling in the air. It reaches zero at terminal velocity.
Force of gravity is the same for the duration of the fall (if we simplify it, although it increases slightly as the object is closer to the Earth) and depends on the mass of the object.
Drag depends on shape (which is the same for both bodies) and on speed - it increases as speed increases.

6. Feb 7, 2019

### kuruman

Right. The acceleration of an object released from rest is initially g but once the object starts moving, its acceleration eventually drops to zero. How what the shape of v(t) vs t is depend on the shape and mass of the object, the air density etc.
But increasing drag decreases the speed which decreases drag. A balance is reached at terminal velocity at which point the drag force balances gravity and the net force is zero. A balloon filled with air reaches this point more quickly than a balloon filled with lead shot.

7. Feb 7, 2019

### liroj

Increasing drag because of increasing speed doesn't decrease speed of falling in the air. It decreases acceleration and that means it decreases THE INCREASE of speed until terminal velocity is reached. At no point does the speed (and therefore the drag) decrease.

Last edited: Feb 7, 2019
8. Feb 7, 2019

### kuruman

Yes of course. That's what I meant.

9. Feb 7, 2019

### liroj

But I'm not sure that air vs lead is a fair example because buoyancy takes over with small densities. I believe we should ignore buoyancy here...

10. Feb 7, 2019

### kuruman

I used air vs lead shot as two extremes to illustrate your condition that one mass is considerably bigger than the other, yet the volume is the same. So instead of air fill the balloon with water. Is the density of lead "considerably bigger" than the density of water?

11. Feb 7, 2019

### jbriggs444

Simplify even more. Make the one ball precisely half as dense as the other. Wait until the less dense ball is encountering 1/2 of its weight as air resistance.

For convenience, designate the masses of the two balls as m and 2m and the acceleration of gravity as g. We assume that the balls are indistinguishable except for their weight and density. Same shape, same size, same texture, same spin, etc.

Question 1: What is the net downward force on the less dense ball at this time.
Question 2: What is its resulting downward acceleration?

Now suppose that the more dense ball were travelling just as fast at this same time.

Question 3: What is the resulting drag force on the more dense ball?
Question 4: What is the net downward force on the more dense ball?
Question 5: What is its resulting downward acceleration?

Question 6: Are those two accelerations equal?

What does that suggest about the supposition that they were each travelling equally fast at this time.

Extra credit: Is there anything special about the "drag equal to half weight" point? What about "drag equal to 10% of weight"? What about drag equal to 1% of weight? Is there any non-zero downward velocity where the resulting accelerations of the two falling balls will be identical?

12. Feb 9, 2019

### liroj

I'm not skilled enough for the math needed here, sorry.
One more question - how does this fact that heavier objects do fall faster in air relate to the Galileo tower of Pisa (most likely thought) experiment... how did he come to his conclusion based on that - the heavier object should have fallen first...

13. Feb 9, 2019

### jbriggs444

When you do a real experiment, you look to see how it comes out. You do not dictate how it should have come out.

14. Feb 9, 2019

### liroj

So how do you explain his experiment?

15. Feb 9, 2019

### Staff: Mentor

It looks like they used two heavy balls with the same density (hence the heavier one has a bit more air resistance). Besides, they were mainly trying to refute that the ball that was 10 times lighter did not take 10 times as long to fall. The slight difference in landing time was probably ignored.

https://en.wikipedia.org/wiki/Galileo's_Leaning_Tower_of_Pisa_experiment

16. Feb 10, 2019

### liroj

Thanx!

17. Feb 10, 2019

### kuruman

I believe it was noticed but not understood hence ignored. As you mentioned, the thrust of all this was to refute the Aristotelian view. In his Dialogues Concerning Two New Sciences Galileo has the following passage, Aristotle says that “an iron ball of one hundred pounds falling from a height of one hundred cubits reaches the ground before a one-pound ball has fallen a single cubit.” I say that they arrive at the same time. You find, on [65] making the experiment, that the larger outstrips the smaller by two finger-breadths, that is, when the larger has reached the ground, the other is short of it by two finger-breadths; now you would not hide behind these two fingers the ninety-nine cubits of Aristotle, nor would you mention my small error and at the same time pass over in silence his very large one.
https://oll.libertyfund.org/titles/galilei-dialogues-concerning-two-new-sciences
(To find the passage on the page, search the page for something unique, I used "larger outstrips the smaller".) This passage is at the core of problem 2.36 in John Taylor's Classical Mechanics. Two questions are asked in the problem
(a) Given that the density of iron is about 8 g/cm3, find the terminal speeds of the two iron balls.
(b) Given that a cubit is about 2 feet, use Equation (2.58) to find the time for the heavier ball to land and then the position of the lighter ball at that time. How far apart are they?
Equation (2.58) is calculated within the model that the drag force is proportional to the square of the speed,$$y=\frac{v_{ter}^2}{g}\ln \left[\cosh \left(\frac{gt}{v_{ter}} \right)\right].$$ Enough information is given in the book to derive the equation for a sphere falling through air$$v_{ter}=\left( \frac{mg^3\pi^2\rho^2}{6^2\gamma^3} \right)^{1/6}$$where $\rho$ is the average density of the object and $\gamma=0.25~N\cdot s^2/m^4$, the latter calculated as the answer to problem 2.4 using the simplifying assumption that all the air mass that the ball encounters per unit time is accelerated to the speed of the ball.

The answers I calculated are (drum roll)
(a) $v_{ter} = 88.5 ~m/s$ (1-pound ball); $v_{ter} = 191 ~m/s$ (100-pound ball).
(b) $\Delta t=36~ms$, $\Delta y=1.22~m.$

The difference is about 2 cubits, not two finger-breadths which would be noticeable to Galileo. I suspect this difference is overestimated because of all the simplifying assumptions that went into the calculation.

Last edited: Feb 11, 2019
18. Feb 11, 2019

### liroj

Since I'm no good at math, I used the online calculators and did some calculated "experiments".
https://www.omnicalculator.com/physics/free-fall-air-resistance
https://keisan.casio.com/exec/system/1224830797

For example,
A body with mass of 1kg, air resistance of 0,24kg/m, falling from 56m (about the height of the tower of Pisa) takes 9,2s
A body of 2kg, in the same conditions takes 6,8s
A body of 10kg takes 4,2s.

1kg body has a terminal velocity of 6,39 m/s and reaches it in 2.5 seconds. After 1s it's speed is 5,82 m/s
2kg body has terminal velocity of 9,04 m/s and reaches it in 3.8s. After 1s it's speed is 7,19 m/s
10kg body has terminal velocity of 20,2 m/s. (takes 7.3s to reach it, so it doesn't reach it in the fall from 56m) After 1s it's speed is 9,1 m/s.

Is there something wrong with those calculators of with my way of thinking or is this OK as an inductive way to prove the point? (besides all the things not covered with the calculations - increasing air density and gravity as the fall progresses and god knows what else....)

19. Feb 12, 2019

### PeterO

I saw a video a colleage made where a bowling ball and a rubber beach ball of approx the same diameter were dropped from a 4th floor apartment window - co-incidently 9.8m high; not that that mattered - which showed the bowling ball win the race by about 80 cm. The result was not unexpected.
Mathematics could , no doubt, have predicted this, but who needs mathematics when you can observe what happens.

20. Feb 12, 2019

### PeterO

Suppose you assumed their acceleration was the same until they reached one half of the lighter ball's terminal velocity (we can consider later whether that assumption is a good one)
In that case both balls would be travelling at the same speed.
Having the same speed and shape, the drag force on each would be equal.
The acceleration of the balls from then then would be dependent on the NET force acting - weight force minus drag force.
For the light ball, the drag force may be as much as 30% of the weight force, so the acceleration would be significantly reduced.
For the heavy ball, the drag force may only be 3% of the weight force, making only a slight reduction in acceleration.
From that time, the heavy ball will be accelerating at a greater rate than the light ball.
To me that suggests that once the balls began to fall (and gain some speed) the heavier ball was accelerating at a greater rate, and that original assumption was not a good one.