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(c) Terminal velocity  (balancing forces - rolling down hill and objects falling downwards under gravity in fluids)

(c)(i) Terminal velocity and a trolley - running board experiment

If you set up a running board and let an object like a trolley roll down it, the 'diluted' effect of gravity gets the object to roll down the incline. At first it speeds up as the force of gravity exceeds the friction of the trolley movement.

The force of gravity is constant, but as the speed of the trolley increases, so does the effect of friction e.g. the wheels and axle, wheel contact with running board.

Eventually the resistive forces of friction balance the diluted gravity effect and the trolley continues moving at a constant speed.

This final maximum speed is called the terminal velocity.

You can actually demonstrate this with a ticker tape timer experiment (results illustrated above).

 You attach a long piece of ticker tape to the trolley and the tape passes through a little 'ticker tape' machine that stamps a dot onto the tape every fraction of a second.

The distance between the dots gives you the relative speed.

At the start the dots are close together at the lowest speeds, but as the trolley speeds up the distance between the dots increases.

Eventually the dots are evenly spaced at their maximum distance apart showing the terminal velocity of the trolley was reached (first at time T).

I presume you can do a similar experiment with a series of light gates to show the same effect, not as much fun though!

The above example illustrate the idea of terminal velocity, but next we consider objects falling down in a fluid due to the force of gravity (air or a liquid).

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(c)(ii) The terminal velocity of a small sphere falling down through a liquid

Apparatus

You can demonstrate the effect of resistive forces in a fluid using the experiment illustrated in the diagram above.

Theory: When an object falls through a fluid there are three forces to take into consideration.

W↓ is the weight of the ball bearing due to gravity (Weight = mass x gravity, W = mg)

U↑ is the upthrust experienced by any object in a fluid. See section 6. Pressure in liquid fluids

F↑ is the resistive force of friction between the ball and the fluid.

W and U are constant and so W-U is a constant and the resultant force in the downward direction.

Initially F is zero at the point where the ball enters the fluid, but, as it descends and speeds up F increases and when F equals W-U the ball bearing descends with uniform velocity - its maximum velocity = terminal velocity.

Method:

A large glass tube, sealed at one end with a rubber bung. is filled with a viscous liquid e.g. oil or glycerine (the latter I think is the best and easier to clean out). You need at least a 50 cm depth of liquid.

The glass tube is marked with suitable depth intervals e.g. every 10 cm.

Small steel ball bearings are carefully dropped down a thinner glass tube to make the entry into the fluid as smooth as possible.

The time it takes for the tiny ball to fall between the distance markers is timed.

You need an accurate stopwatch and convert the cm or mm of h into m.

(vertical) velocity (m/s) = d (m) ÷ t (s)

You should find the steel ball falls relatively slowly at first and then attains a maximum velocity - the terminal velocity when the force of the weight of the object is balanced by the resistive forces of friction as the surface of the ball interacts with the liquid.

The 'theoretical results' are shown in the velocity-time graph below.

The more viscous the liquid, the smaller the terminal velocity as the friction forces are increasing.

At first when the object starts to fall the accelerating force due to gravity, W↓, is greater than the frictional force slowing it down.

You can tell this from the steep positive gradient at the start where you get the biggest acceleration.

As the speed increases, friction (force F↑, drag effect) increases and this reduces the acceleration.

But, as time goes on, the acceleration decreases (gradient decreasing) because the value of F↑ is increasing.

When all the forces balanced (W-U) = F, the resultant force is zero and the graph becomes horizontal (at time T) and the horizontal value is the maximum speed or terminal velocity.

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(d) The complex behaviour of a falling parachutist

This is an interesting case because it involves an acceleration, a deceleration and two terminal velocities!

1.    2.    3.  

These diagrams show the relative magnitude of the two forces acting on a descending parachutist.

Note the relative size and direction of the arrows. The weight force F2 is constant.

1. When the parachutist opens the parachute, there is an immediate large drag force due to big increase in air resistance (air friction).

2. As the parachutist slows down (decelerates) the air resistance (F1) reduces and continues to do so until F1 = F2.

3. When drag force F1 equals the weight force F2, the parachutist is descending at a steady speed - a terminal velocity.

Remember the drag force on an object due to friction (air resistance), increases with speed i.e. as more air brushes over the surface of the object in the same time.

We now look at the whole sequence of events after the parachutist has jumped out of the aircraft

speed/velocity-time graph for a parachute descent

  weight due to gravity W↓ and the drag effect due to air resistance D↑

(I'm ignoring the minor effect of upthrust U↑ that applies to any fluid - gas or liquid)

The force vectors operating in the jump

(1) The parachutist jumps out of the aircraft and immediately accelerates due to gravity (weight force W↓), but the parachute is not opened yet, so the parachutist is in free fall!

Between (1) to (3) acceleration is taking place, but although W↓ is constant, the drag force D↑ due to air friction (air resistance) is increasing, so the acceleration is decreasing (gradient decreasing).

At (3) the terminal velocity is first reached when the drag force D↑ = weight W of the parachutist.

 Its apparently great to do a spot of sky diving first at ~120 mph (~54 m/s) with arms and legs stretched out (increased surface area) and ~200 mph (~89 m/s) if body compact (minimum surface area).

In the graph I've assumed its one or the other!

Under the influence of the Earth's gravitational field this first terminal velocity depends on the shape of your body, but its pretty fast!!!.

Between (3) to (5) the descent continues at the (1st) constant terminal velocity because the resultant force is zero.

At (5) the parachutist opens the parachute to increase air resistance. The wide area brushing through the air produces a massive increase in air resistance (friction), so the drag effect D↑ massively increases, and is far greater than W, so rapid deceleration immediately takes place.

Between (5) to (7) the deceleration continues and as the parachutist decreases in speed, so the drag forces decreases too.

At (7) the drag force once again equals the weight force and the parachutist attains a 2nd, somewhat slower, terminal velocity.

Between (7) and (8) the parachutist descends far more slowly with constant speed to land safely at ~15 mph, again the resultant force is zero.

 

Terminal velocity experiments

You do simple experiments with toy parachutes attached to a weight to determine their terminal velocities.

You can vary the mass attached to a parachute of constant surface area - will it fall faster the greater the mass?

For the same weight of an attached mass, does it fall more slowly with a larger sized area parachute?

There is scope for investigating factors affecting the drag effect.

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(e) Other falling object situations and thought described and explained!

The earliest theories on falling objects suggested that the heavier the object, the faster it falls.

We now know that all objects fall at the same speed if only the force of the local gravitational field is acting on the them.

Some 'actual' and 'thought' experiments.

You need to consider the W (weigh, depends on local gravitational field strength), U (upthrust) and F (friction, drag) forces described in the ball bearing - liquid filled tube experiment and the four objects illustrated above.

(1) The hammer and the feather experiment on the moon

When the first astronauts landed on the moon they carried a beautifully simple experiment.

One astronaut held up a feather and a hammer and let them both fall simultaneously.

Both objects (i) fell to the ground in the same time experiencing the same gravitational acceleration and (ii) they seemed to fall to the ground more slowly than if they had been dropped on Earth.

(Weight = mass x gravity, W = mg)

(i) Both objects experience the same gravitational acceleration of 1.62 m/s², BUT the Moon has no atmosphere so there is no air resistance. With no drag effect due to friction both objects, whatever their mass or shape (surface area) are free to fall with maximum acceleration due to the Moon's gravity.

(ii) Acceleration on Earth due to the gravitational field is 9.80 m/s². The Moon has a much smaller mass and gravity is much weaker and only produces an acceleration of 1.62 m/s². Therefore objects on the Moon will fall much more slowly than on Earth (ignoring air resistance).

On Earth: Hammer weight = 0.20 x 9.80 = 1.96 N  and  feather weight = 0.002 x 9.80 = 0.0196 N

On the Moon:  Hammer weight = 0.20 x 1.62 = 0.32 N  and  feather weight = 0.002 x 1.62 = 0.0032 N

 

(2) The hammer and the feather experiment on Earth

 If you drop the feather and hammer simultaneously on Earth, the feather takes much longer to fall to the ground because it has a relative greater surface area/mass ratio greatly increasing drag effect.

If it wasn't for air resistance, no drag effect can happen and all objects would fall with the same acceleration.

You can do a laboratory experiment with two objects (one 'light' like a feather and one heavy like a lead weight) by allowing them to fall vertically in a large tube from which the air has been pumped out to create a vacuum.

It seems uncanny when they both hit the bottom of the tube at the same time.

 

(3) Dropping two spheres of equal surface area, but different mass

If you simultaneously dropped the dense pool ball and light hollow plastic ball with the same radius (giving same surface area) the pool ball will hit the ground before the lighter hollow plastic ball.

This is because the greater weight of the pool ball will overcome the drag effect of air resistance more than the less 'weighty' plastic ball and the pool ball will appear to accelerate at a greater rate.

The surface area is the same, and if both surfaces are smooth, the friction effect should be the same.

On the Moon you would not notice any difference, see (a) for the argument.

 

(4) The general behaviour of a falling object on Earth

 If you drop any object (safely!) from a tall building on Earth, it will always achieve a terminal velocity due to the friction effect of air resistance.

The terminal velocity s reached when the downward weight force equals the air resistance friction force.

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Motion and associated forces notes index (including Newton's Laws of Motion)

1. Speed and velocity - the relationship between distance and time, distance-time graphs gcse physics

2. Acceleration, velocity-time graph interpretation and calculations, problem solving gcse physics revision notes

3. Acceleration, friction, drag effects and terminal velocity experiments gcse physics revision notes

4. Newton's First, Second and Third Laws of Motion, inertia and F = ma calculations gcse physics revision notes

5. Reaction times stopping distances and example calculations gcse physics revision notes

6. Elastic and non-elastic collisions, momentum calculations and Newton's 2nd law of motion gcse physics notes