Acceleration of Gravity (also under Concepts > Motion)

In response to a question about the acceleration of gravity:

We dropped marbles from above our heads and did the 8-count to estimate the time of fall to the floor. 

One typical result:  The ball required .5 seconds to fall 2 meters from rest.

From this information we easily reason out the acceleration.  Knowing `ds and `dt we find vAve, then from v0 and vAve we find vf; from v0 and vf we get `dv and from `dv and `dt we get a.  The results are as follows:

Average velocity is 2 m / (.5 sec) = 4 m/s (based on the definition of average velocity)

Initial velocity is zero, and assuming uniform acceleration we conclude that final velocity is 8 m/s (based on the linearity of the v vs. t graph when acceleration is uniform).

Change in velocity is (8 m/s - 0 m/s) = 8 m/s, so acceleration is 8 m/s / (.5 s) = 16 m/s^2.

Alternatively we can write down the four equations of uniformly accelerated motion, circle the symbols for the know quantities, and solve the the unknown quantity in any equation containing the three known quantities.

The third equation can be solved for a.

Starting with

`ds = v0 `dt + 1/2 a `dt^2, subtract v0 `dt from both sides to get

1/2 a `dt^2 = `ds - v0 `dt, then multiply both sides by 2 to get

a `dt^2 = 2 ( `ds - v0 `dt), and finally divide both sides by `dt^2 to get

a = (2 `ds - v0 `dt) / `dt^2.  Plug in known values:

a = (2 * 2 m - 0 * (.5 sec) ) / (.5 sec)^2 = 4 m / (.5 sec^2) = 16 m/s^2.

More accurate experiments show that the acceleration of gravity in the vicinity of the surface of the Earth, at sea level, is 9.8 m/s^2.

Near the surface of the Earth, any freely falling object accelerates downward at about 9.8 m/s^2.

For nonprecise calculations we can approximate this as 10 m/s^2.  If necessary we can later compensate for the 2% approximation error.

Steel ball off ramp to floor (also Systems > Ball on Incline)

We rolled a steel ball down an inclined ramp and onto a level ramp, from the end of which it fell to the floor 1 meter below. 

• The vertical motion of a freely falling object is independent of the horizontal motion, and is characterized by constant downward acceleration of 9.8 m/s^2.
• The horizontal motion of the object is characterized by 0 acceleration (gravity is the only force acting, and it acts only in the vertical direction).

In this case, when the ball left the ramp it was moving only in the horizontal direction, so its initial vertical velocity was zero.  Thus for the vertical motion, choosing the downward direction as positive, we have v0 = 0, `ds = 1 meter, a = 10 m/s^2 (using the approximation that 9.8 is about 10).  We easily analyze this motion (write out the eqns of unif accel motion, figure out which contain the three known quantities, etc.).

The 4th equation gives us vf = +- sqrt( v0^2 - 2 a `ds). 

When the quantities are plugged in we obtain vf =  +- 4.4 m/s. 

The final velocity is clearly downward, hence positive.

So the average velocity is (0 + 4.4 m/s ) / 2 = 2.2 m/s.

The time of fall is therefore `ds / vAve = 1 m / (2.2 m/s) = .45 sec.

Had we used a = 9.8 m/s^2 the time of fall would have rounded to .43 sec.

Recall that we have used the idea that a ball rolling off the end of a ramp on the tabletop falls to the floor in about .4 seconds.  This is consistent with the experiment in which you time the ball down the ramp, the down the ramp to the floor, using the pendulum timer.  The table is a little less than a meter above the floor, so the above calculation shows that our .4 second approximation is about right.