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Items falling in liquids

  1. Feb 1, 2016 #1
    Hi, it's my first post.

    When a golf ball and a bowling ball get dropped from the same height. They will both hit the same hit the ground at the same time.

    But when you drop a ball with radius R and another ball eith radius 2R inside a cylinder with oil then the ball with the biggest radius will fall faster. Why doesn't the balls in the oil follow the same rules as the balls in the air and fall with the same speed? Aren't both air and oil a viscous fluid?
  2. jcsd
  3. Feb 1, 2016 #2
    The viscosity of air is several orders of magnitude lower than that of oil.
  4. Feb 1, 2016 #3

    Suraj M

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    Do you think this is actually true?
    There is another condition that you are missing for them to reach the ground at the same time
    Watch this
  5. Feb 1, 2016 #4

    Andy Resnick

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    I wonder if this is sufficient to determine the fluid viscosity.... potentially interesting undergrad problem.
  6. Feb 1, 2016 #5
    There is are commercially available devices out there called Falling Ball Viscometers that measure the viscosity of high viscosity fluids this way.
  7. Feb 2, 2016 #6

    My statement is actually true. Just watch:

    or watch

    Drag is based on shape, not on mass
  8. Feb 2, 2016 #7
    True, but why that doesn't change the fact that in air a heavier ball with bigger radius hits the floor at the same time as any other ball.
    But when in oil why does the ball with bigger R hit the floor first??

    This is my big question

    Both fluids, but why seemingly following different rules
    Last edited: Feb 2, 2016
  9. Feb 2, 2016 #8
    Can you explain the difference? I saw this video and I couldn't explain why the balls with bigger R went faster in oil. But in air all balls fall with the same velocity??

    watch from 0:49

  10. Feb 2, 2016 #9
    Because in the case of the balls falling through air, the viscous drag is negligible, and for the balls falling through a liquid, the viscous drag is very significant. Have you at least tried to model this using Newton's 2nd law? Assume that the drag is proportional to the velocity, but that the constant of proportionality is much higher for the oil (ko) than for the air (ka). Let's see your equations.
  11. Feb 3, 2016 #10

    Andy Resnick

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    For whatever reason, the videos don't display on my browser.
  12. Feb 3, 2016 #11

    Andy Resnick

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    I spent a few minutes calculating v(t) and h, I'm not sure it's a well-posed problem- even when I constrain the sphere densities to be equal. I got:

    h = K1* r2*T/μ - K2* (r2/μ)2 (1-exp(-K3*(μ/r2)T)), where T is the time to fall a distance h, K1, K2, K3 are constants (density, density difference, g, etc.). Given 2 different radii, T and h are the same should (hopefully) give a single value of μ. Maybe if I have a few minutes today I can make further progress.
    Last edited: Feb 3, 2016
  13. Feb 3, 2016 #12

    Suraj M

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    No OP these videos are to prove that the acceleration due to gravity is equal on all objects irrespective of their mass, which is true but the total acceleration would be equal in a vacuum
    In the case of the video the height isn't enough for you to notice the difference in velocities
    You might want to look into "terminal velocity" and its dependence on radius and density of objects, it'll make you understand why it's so easily noticeable in a more viscous liquid rather than just air

    Just to make it a little more clear
    Think of this
    If any 2 objects were actually to fall to the earth at the same time if dropped simultaneously then they're final velocity as they reach the ground should be equal, take a practical scenario
    If a man was to jump of a plane at the height of the clouds I'm sure it's not difficult to assume his final velocity of he were to reach the ground would definitely be more than the velocity of a rain drop coming from the same altitude.
    Now here it is obvious that they don't reach the ground at the same time, this effect is more pronounced in a more viscous fluid.
  14. Feb 3, 2016 #13


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    There are always two forces acting on the balls. The force due to gravity points down, and the force due to drag points in the direction opposite the motion, which is up in this case. So, the acceleration of the ball (or any object) will be the result of the net force on the object. In other words, for a ball of mass ##m_{\mathrm{b}}## with gravitational acceleration ##g##, the ball's acceleration ##a_{\mathrm{b}}## is
    [tex]m_{\mathrm{b}} a_{\mathrm{b}} = \sum F = F_{\mathrm{drag}} - m_{\mathrm{b}} g.[/tex]
    If you drop two balls (or any shape) in a perfect vacuum, then ##F_{\mathrm{drag}} = 0## and ##a_{\mathrm{b}} = -g## independent of mass or shape or anything else. However, if you add in a fluid, ##F_{\mathrm{drag}}## is no longer zero. In general, ##F_{\mathrm{drag}}## depends on the size and shape of an object as well as its velocity and the viscosity of the fluid.

    Now, in air, the viscosity is so low that, even though the drag is not zero, it is much, much less than the weight of the balls in the case of a shape like a sphere moving at the sort of speeds we are talking about here. In other words, ##F_{\mathrm{drag}} \ll m_{\mathrm{b}}g## and you can effectively treat drag as zero for these kinds of experiments. If you had sensitive enough equipment, and could guarantee two balls were released at the same time, then they would not hit the ground at exactly the same time. They just hit so close together as to appear simultaneous. This only works for something like a sphere that is low in drag. If you dropped a sphere and a sheet of paper, it wouldn't work (or two pieces of paper) because the drag on the paper is so much higher relative to the weight of the paper.

    In your video with the balls in syrup, the larger ball falls faster because, while the viscous drag increases as the ball gets larger, the mass of the ball (and thus the force on it due to gravity) increases faster than the drag. That means that the acceleration gets larger as the ball gets larger.
  15. Feb 3, 2016 #14
    Let's look at the terminal velocities of balls of different diameters made out out the same material (i.e., with the same density) dropping through a viscous oil, and then air.

    For a high viscosity oil,

    Mass of ball = ##\frac{\rho \pi D^3} {6}##, where ρ is the density and D is the diameter.

    Weight of ball = ##\frac{\rho g \pi D^3} {6}##

    Buoyant (upward) force on ball = ##\frac{\rho_o g \pi D^3} {6}##, where ρo is the density of the oil

    Drag force (upward) force on ball = ##3\pi \mu D v##, where μ is the viscosity of the oil and v is the (downward) velocity of the ball. So if we do a force balance on the ball, we get:$$\frac{\rho g \pi D^3} {6}-\frac{\rho_o g \pi D^3} {6}-3\pi\mu D v=0$$
    So, if we solve for the terminal velocity v, we obtain:
    $$v=\frac{ g D^2} {18\mu}(\rho-\rho_o)$$
    So a ball of larger diameter will have a higher downward terminal velocity.

    Unfortunately, for a low viscosity fluid like air, the drag force will not be given by the same equation because the air flow is turbulent. For turbulent drag, the drag force is given by ##\frac{1}{2}\rho_ov^2C_D\frac{\pi D^2}{4}##, where CD is the so-called drag coefficient (which, at high values of the so-called Reynolds number) is close to a constant value. If we substitute this into our force balance equation, we obtain:

    $$\frac{\rho g \pi D^3} {6}-\frac{\rho_o g \pi D^3} {6}-\frac{1}{2}\rho_ov^2C_D\frac{\pi D^2}{4}=0$$

    So, for air, $$v =\sqrt{\frac{4gD}{3C_D}\frac{(\rho -\rho_o)}{\rho_o}}$$
    So, even for air, the velocity of the larger diameter ball will eventually be higher (terminal velocity) than that of the smaller diameter ball. Of course, for balls falling in air, the balls start out by falling close to the same velocity at the beginning until the air drag starts kicking in. However, after air drag starts to play a role, the smaller ball will no longer be accelerating downward as fast at the larger ball. However, typically, the balls will not reach their terminal velocities before they hit the ground unless they fall for a very large distance.

    Here's some more important information about the drag coefficient. The drag coefficient is, in general, a function of the Reynolds number for the flow:
    $$C_D=C_D(Re)$$where $$Re=\frac{\rho_o v D}{\mu}$$
    Note that the Reynolds number depends on the diameter of the ball and the fluid viscosity. For very low Reynolds numbers characteristic of high viscosity oils, the drag coefficient approaches the following (in so called Stokes flow):

    $$C_D=\frac{24}{Re}=\frac{24\mu}{\rho v D}$$

    If we substitute this into our general equation for the drag force, we obtain $$F=\frac{1}{2}\rho_ov^2C_D\frac{\pi D^2}{4}=3\pi\ \mu vD$$
    This is the equation we used previously for the drag force for the oil.

    At very high Reynolds numbers characteristic of very low viscosity fluids like air, the drag coefficient becomes a very weak function of the Reynolds number, and is nearly constant over large ranges of velocity.

    So, in conclusion, even though high viscosity oils and low viscosity air have very different drag coefficients, we still find that the terminal velocities of higher diameter balls are greater than lower diameter balls for both.
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