1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Falling Bodies - Does mass play a role?

  1. Sep 28, 2011 #1
    Me and a few friends of mine were discussing if mass plays a role in the speed of a free falling body. They argued that equal bodies (same volume, same geometry) with different weights (mass) would fall at different velocities. I defended that mass has no part on the matter, but only air resistance. Meaning that if two equal balls (same radius), one made of iron another made of cork, were dropped from an airplane, both would reach the ground at the same time. I am aware that in vacuum as long as both bodies are dropped from the same height they will hit the ground at the same time, regardless the shape or weight. Since there's no air resistance. Was I right or wrong? Is air resistance affected by mass or just by the contact surface (geometry)?

    I searched the forum for similar threads but couldn't find a straight answer.

    Thanks
     
  2. jcsd
  3. Sep 28, 2011 #2
    What's important is the ratio of mass to drag. If mass is large compared to drag, as in a vacuum, then both will hit at the same time. If mass is small compared to drag then they will not.

    You know how fine sawdust will float in the air, but a solid piece of wood wont? A dust particle and a 2x4 are made of the same material so why does the 2x4 fall so much faster then the dust particle?

    Answer: Imagine 2 cubes. 1 is 1 meter per side, the other is .1 meters per side...

    1 meter cube has a volume of 1 cubic meter and a surface area of 6 square meters. ratio of 1:6

    the 1 cm cube has a volume of 0.001 cubic meters and a surface area of 0.06. ratio of 1:60

    So the smaller cube has far more surface area in relation to it's volume. Assuming constant density it also has more surface area per mass. Surface are is proportional to drag so it will have more drag per unit mass. If the particle is very small, like a dust particle the drag will be the dominant force and the particle will waft about with the air currents for a while before eventually settling to the floor.
     
  4. Sep 28, 2011 #3
    So according to your explanation I was wrong. Even though in the situation I depicted the surface area is the same because both balls have the same radius the mass/drag ratio is different. The iron ball has a higher density and therefore a higher mass/ratio so it will reach the ground first, correct? Is this visible also for smaller heights?
     
  5. Sep 28, 2011 #4

    BobG

    User Avatar
    Science Advisor
    Homework Helper

    Yes, although the short time periods involved might be hard to deal with.

    One of the best ways to measure the difference is to set up an apparatus where the ball and a pendulum will be released at the same time. Put some paper along the pendulum. You can put carbon paper on top of the paper so the carbon paper will leave a mark where the pendulum struck the ball or cover the ball in chalk so it leaves a mark on the paper, etc.

    You might have to vary the length of the pendulum so the ball doesn't hit the ground before the pendulum has time to strike it.

    Just measure the difference between the dots.
     
  6. Sep 28, 2011 #5
    For most experiments that you might do, such as wood vs. iron balls the difference in fall times will be smaller than the margin of error. To make the difference detectable you need to design your experiment so that the difference is large and/or the margin of error is small.
     
  7. Sep 29, 2011 #6
    In a realistic sense, and ignoring air resistance, all objects fall at the same speed and accelerate the same amount.

    However, I believe that a heavier mass, will have a slightly higher acceleration, but negligible.

    The force of gravity is expressed by F = GMm/(r^2)

    F = gravitational force, G = gravitational constant, M = attracting body's mass(like Earth), m = falling body's mass(like an apple), and r = radius from the center of mass.

    Recall that F = ma, so a = F/m

    Divide the gravitation force by m, so you get a = GM/(r^2)

    Therefore all objects are attracted to an object at the same rate, determined only by the attracting body's mass and the distance from the center of mass.

    However, Newton also says that for every force, there is an equal and opposite reactive force. So as the Earth pulls on the apple, the apple pulls on the Earth.

    F(apple) = -F(Earth)

    So if you find the gravitational acceleration of the Earth to the apple, you divide the M:
    a = Gm/(r^2)

    Therefore, the Earth accelerates to the apple.

    Now, the Earth has an enormous mass compared to anything on it. I think the mass in kilograms is in the area of 10^31 or so. An apple, however, is less than 1 kg. The acceleration(relative to mass, the Earth's gravitational pull is only 9.81 m/(s^2)) is so tiny because G is tiny, about 6.67 x 10^-11. If an object as massive as the Earth only produces 9.81 m/(s^2), how much would the apple produce? The amount would be negligible, even if the falling mass weighed 10^5 tons. This is because the Earth's acceleration is so much larger that an difference is too tiny to measure.

    In short, a heavier mass does technically fall slightly faster. However, in reality, this difference is very very very very very very very very very tiny, to the point of no effect.
     
  8. Sep 29, 2011 #7

    DaveC426913

    User Avatar
    Gold Member

    Technically it does not fall faster; it falls at the same rate, but Earth rises to meet it, resulting in a shorter duration before impact.
     
  9. Oct 1, 2011 #8
    Well yes, the earth's gravitational pull is identical. When he said falling, I figured that would include both pulls, like if Earth and Venus were suddenly next to each other and pulled each other into a collision. In that case, the acceleration including both bodies(g[itex]_{e}[/itex]+g[itex]_{v}[/itex]) should be higher, right? However, if you only include the gravitational attraction of Earth, then the falling body would fall the same regardless of mass.

    I suppose I should have specified that. Most people only take the Earth's acceleration into account for everything, since the field exerted by everyday objects is too small for much of anything, meaning there is no noticable difference of falling. I don't know if current technology can see any difference between ordinary everyday objects.
     
  10. Oct 1, 2011 #9

    epenguin

    User Avatar
    Homework Helper
    Gold Member

    First get a leaning tower...
     
  11. Oct 2, 2011 #10
    Compare a slowly falling balloon, with a solid iron ball of the same size...
     
  12. Oct 2, 2011 #11

    DaveC426913

    User Avatar
    Gold Member

    Did not realize this question was as simple as it is.

    No. The iron ball will fall much faster.
     
  13. Oct 2, 2011 #12

    Ken G

    User Avatar
    Gold Member

  14. Oct 2, 2011 #13
    as the density of iron is more than cork, mass of iron ball will be more than that of cork ball. there are three forces acting on falling ball: weight, upthrust and viscous force. same upthrust acts in both bodies. so (weight - upthrust ) of iron is larger than of cork. and hence a larger viscous force is req to balance larger (weight - upthrust). by stokes law viscous force is directly proportional to velocity. hence there will be larger terminal velocity(velocity at equilibrium) for iron ball.

    so it will fall faster.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Falling Bodies - Does mass play a role?
  1. Falling body question (Replies: 13)

  2. Falling body equation (Replies: 2)

Loading...