Tuesday, April 20, 2010

A study of uniformly accelerated motion

A motion is said to be uniformly accelerated/decelerated if during equal time-intervals, we have equal increments/decrements of speed that is the acceleration is a positive/negative constant. For example that can be the case for a free falling object, but beware that this would be only an ideal approximation of its motion. It is only valid when we can neglect the air friction compared to the weight of the object.

Therefore it would be an inadequate model for a feather falling on earth which has a very small weight or mass but offers a large resistance to the air. Instead it's a good approximation for a stone falling on earth or a feather on the moon whose atmosfere is extremely rarefied. And it's an exact surprising law only in vacuum when there are no other forces involved other than gravitation.

The constant value of gravitational acceleration was found for the first time by the great ancient Italian physicist Galileo Galilei. He had the smart idea of using inclined planes in order to delay the motions by reducing the gravitational force (and hence the proportional acceleration). By making the rolling balls and ramps very smooth, decreased obtained acceleration is also constant and no other appreciable friction is introduced. Galileo didn't have a wristwatch for time measurements, instead he used a very simple water clock which is not very precise. That is the reason why he needed to slow down the falling motion.

Here's a mathematical derivation of the constant acceleration motion equations applied to something that is left to fall. For more general formulas see also here.

Ingredients: Newton's second law, mathematical general definitions of acceleration and speed

Results: how position and speed change over time in such a motion


f = m a

f = m dv / dt

f dt = m dv

For simplicity when we integrate this, we can set the starting time to 0, but this is unnecessary.
To be fussy the upper integration limit should be named differently from the integration variable t, but as long as we don't confuse things we can avoid introducing another name for the time variable in the result.

t           v(t)
∫  f dt = m   ∫ dv
0           v(0)

Let's suppose that f is constant respect to the time t, as gravitational force is. Really is not, but this approximation is as much good as the distances covered are small compared to the earth radius. That's about 6371 km and we're not launching a space shuttle here ;-)

Taking f outside of the integral sign we get:

f t = m [ v(t) - v(0) ]
v(t) = v(0) + a t

This tells us how speed changes. It increases linearly over time. Acceleration a here plays the role of a proportionality constant. If the moving object starts from rest, of course v(0)=0.

If we now mix in the definition of speed (velocity modulus):

v(t) = ds / dt

set the origin of our coordinate system so that s(0)=0:

t           s(t)
∫ v(t) dt =   ∫ ds
0          s(0)=0

and insert the previous found expression for v(t), we finally have:

s(t) = v(0) t + a t^2 / 2

E.g. if you let fall a high density object like a hammer from rest, the distance it will travel over time grows quadratically as:

s(t) = g t^2 / 2

Here a = g, the gravitational acceleration. In all honesty is not constant, not even throughout earth surface, because it depends on latitude too. A reasonably approximate mean value also found by Galileo is 9.8 m/s^2. After one second a hammer or a piece of wood will cover about 4.8 metres. Mass counts for very little if the object density is not too low so that floating effects and air friction can be ignored. More on this here.

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