Coin and Feather

Figure 1

Figure 1

Description:

This apparatus simultaneously drops a feather and a coin into an evacuated cylinder to demonstrate the independence of mass and gravitational acceleration. It is a good idea to drop the coin and the feather by hand in front of the audience and then follow this procedure.

Materials:

  1. Vacuum pump connected to large plexiglass cylinder with two electromagnetic holders (for coin and feather) attached to the cylinder cap.
  2. DC power supply set at 12 V (Model PI-9596 power supply is good)
  3. 2x banana cables (red/black)

Demo:

Figure 2

Figure 2

  1. Lift the cap off of the tube and place the feather and the coin into their respective holders (Figure 2).
  2. Place coin and feather assembly on top of the tube.
  3. To evacuate the cylinder, close the vacuum valve (Figure 3C) and turn on the pump (Figure 3A).
  4. When the gauge shows about 30mm/Hg, stop the pump.
  5. Use the push-button on the base of the tube to release the coin and the feather.
  6. Open the vacuum valve to let air in.
  7. To retrieve the coin and feather from the
    Figure 3

    Figure 3

    bottom of the vacuum tube, wait until the tube has once again filled with air. Make sure to hold the top piece of the tube that holds the coin and feather while you open the tube so it does not fall off and break. Once you are ready to open the tube, unscrew the thumb screw that is at the base of the tube. Once you swing it open, you can carefully retrieve the coin and feather. There is no need to open it up all the way, just far enough so that you can reach your hand inside. Once you’ve retrieved the coin and feather, screw the tube onto the base once again and replace the coin and feather in the top of the tube (Figure 2).

Explanation:

We can explain why in the vacuum the coin and the feather fall at the same acceleration by using Newton’s second law of motion, F = ma.

Here we need to distinguish between two quantities, the acceleration due to gravity, which is constant (mostly, at sea level); and the force acting on the object due to gravity. We will show that the forces due to gravity are different with math, but you can also tell for yourself with a simple experiment.

If we hold a feather in our hand it requires less force to hold it in equilibrium. Remember that to hold an object in equilibrium, the forces must be balanced. You can then tell the relative force of gravity just by seeing what force is required to keep it still (how heavy it is!). A lead ball (or coin) has a much greater force due to gravity, and you have to exert more force to keep it in equilibrium with your hand (it weighs more).

The differences in the force due to gravity are because the masses are different, and according to F = ma, a greater mass leads to a greater force.

Take a 1kg brick and a 1 gram feather under the acceleration of gravity (9.8 m/s2). FgB is the force of gravity acting on the brick, and FgF is the force of gravity acting on the feather.

FgB = mB a = (1kg)(9.8m/s2) = 9.8 N

FgF = mF a = (0.001 kg)(9.8m/s2) = 0.0098 N

This shows us that the gravitational force acting on each object is different, but the force of gravity is not the only thing governing the motions of objects in free fall. Inertia plays an equally important part.

Inertia can be explained by Newton’s first law of motion: an object at rest will stay at rest unless acted on by an external force; an object in motion will continue in motion at constant speed and direction unless acted on by an external force. The greater mass an object has, the greater its tendency to resist a force acting on it. You can see this with another simple experiment: place two bricks, one of cement and the other of styrofoam on a table and push them. The styrofoam brick will begin to move with a much smaller force than is needed to move the cement brick. This shows how mass affects inertia.*

This concept of inertia is displayed in the vacuum tube, and explains why even though the coin has a greater force of gravity acting on it compared to the feather, they fall with the same acceleration. Because the feather is lighter than the coin, it has less inertia and therefore has a smaller tendency to oppose a change in motion. This means that less force is required to start moving it. Conversely, because the coin is heavier than the feather, it has a greater tendency to oppose any force applied to it, and it needs a greater force to begin moving from rest. This means that a coin and a feather will fall with the same acceleration because they are opposing the change in motion (from rest to falling with gravity) to different degrees, proportional to their masses. The small Fg on the feather is enough to start it moving, where the coin needs a substantially larger Fg to get it moving.

Now it is obvious to wonder why the vacuum tube is necessary to demonstrate this property of objects. All objects that fall in the atmosphere are affected to a certain degree by the force of air resistance, which opposes all direction of motion. The vacuum tube creates a vacuum so there is no air for air resistance to play a part. The two most important contributing factors to air resistance are the cross-sectional area of the object and the speed of the object. An increase in either of these will lead to an increase in the air resistance acting to oppose the direction of motion.

Terminal velocity is the speed at which an object is falling such that the force of air resistance (FAR) opposing the direction of motion is equal to the force of gravity acting on an object. Because at this point the forces are balanced, the object will stop accelerating and will maintain a constant velocity (VT). For a more massive object which has a greater Fg, it will require a greater FAR to balance the forces to stop the object from accelerating. Because air resistance is proportional to velocity, it will need to accelerate to a greater velocity to achieve the FAR necessary to balance the forces. Equally, for a light object, there is a very small Fg. This means that there will need to be a very small FAR acting in the opposite direction to stop the acceleration. A small air resistance force can be achieved at very low speeds (especially if there is a large cross-sectional area), so it will reach terminal velocity shortly after it begins accelerating. This is shown in the following image with the example of an elephant and a feather.

elephant-and-feather

(http://www.physicsclassroom.com/mmedia/newtlaws/efar.cfm)

When there is no vacuum tube to remove the air (and therefore the air resistance), the feather has a very large air resistance acting on it (opposing its direction of motion) due to its relatively large cross-sectional area. Because the feather has a large cross-sectional area, and little mass (and therefore a very small force of gravity acting on it), the force of air resistance that is required for it to reach terminal velocity is small and quickly achieved. Since the terminal velocity is very small, the feather falls slowly. While the feather reaches its terminal velocity quickly and maintains a slow velocity throughout its descent, the coin (which does not have a large cross-sectional area, especially proportional to its mass) continues to accelerate and does not noticeably reach terminal velocity in the short descent it makes.

It’s important to remember that the initial accelerations of the coin and the feather are the same regardless of whether they fall in air or in a vacuum, but in air the feather quickly reaches its terminal velocity and then stops accelerating, when the coin doesn’t and continues to accelerate down.

It is important to note here that the styrofoam brick will be easier to move than the cement even if the bricks are on wheels or on a frictionless surface, though undoubtedly the force of friction opposing the direction of motion of a more massive object will be greater than the friction force acting on a less massive object.

Notes:

  • Works well but can only be seen from the first few rows without a video camera
  • Cannot be taken off of Science Hill
  • Needs to be tipped going through most doorways
  • Make sure latch for screws are very tight or else vacuum won’t work as it should
  • Works best at max vacuum pressure (27-30 mm/Hg on gauge)

Written by Sophia Sholtz