Freefall
SPH 3U
Introduction
An object near the surface of the earth will fall due to the force of gravity created by the Earth.
The object will accelerate towards the centre of the earth at 9.8 m/s2 (980 cm/s2). This acceleration will
be reached if there is no friction. As previously discussed, friction is a force which opposes motion and
it can not be eliminated from this lab. Scientist have performed experiments in a vacuum, a
frictionless environment to confirm that a feather and bowling ball fall at the same rate - 9.8 m/s2 (980
cm/s2)[towards the centre of the earth].
The total frictional forces acting on an object in freefall depends on the objects shape and the
speed of the object. As the speed increases for a given shape, the force of friction also increases.
When someone goes parachuting, the frictional forces will eventually offset the force of gravity and the
jumper will fall to the ground with a constant velocity. This is referred to as the objects terminal
velocity. Once the chute opens, the shape of the object is changed and the frictional forces will be
greater than the force of gravity creating an upward acceleration. It is unlikely that the jumper will
travel upwards because as they slow down the frictional forces change and a new terminal velocity is
reached.
The object you drop should only accelerate at 9.8 m/s2 (980 cm/s2) for a brief moment at the
very beginning of its freefall before frictional forces lower its acceleration. As much as possible, try
and get as many tangents as possible from the beginning of the position time graph.
Graphing provides us with a very visual tool to examine a relationship. Unfortunately, graphing is a
very tedious and delicate art. In theory, drawing tangents to determine instantaneous velocity should
produce an accurate account of the objects motion. However, in practice, the uncertainty around
drawing lines of best fit and tangents creates too much error. Students within the same group will
produce completely different results even though they use the same method and data. This raises
serious questions for scientists concerned with reproducibility.
Averaging techniques eliminate the uncertainty associated with lines of best fit and tangents.
Although averaging values will never produce a true value, it succeeds where the tangent method fails
in creating results that are reproducible. It must be stressed however that these values are only
averages and often averages can be misleading.
Recall that friction cannot be eliminated from this lab and that a value of 9.8 m/s2 should only
exist at the beginning of an object's freefall. The non-graphing method we will use will allow us to
produce many estimates of the rate of acceleration for many different intervals
Purpose:
1) To determine if the acceleration due to gravity is 9.8 m/s2 (980 cm/s2).
2) To compare 3 different methods in obtaining an acceleration value from the same
position time data.
Materials
small mass
ticker timer
ticker tape
carbon disc
tape
Procedure
Drop a small dense object and use the ticker timer to record its position and time during the fall. The
ticker timer generates dots at 60 Hz. (60 dots/second) Since the object will only accelerate at 9.8 m/s2
at the beginning of the fall, we will record its position every 3 dots or 0.05 s.Observations
Time (s)
0
Position (cm)
Time (s)
Position (cm)
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
Method 1 – Tangents 9 (generate 2 graphs)
Plot the position time data and determine the instantaneous velocity at 5 different locations by drawing
tangents. Show your calculations on the back of the graph and record these slopes in the table below.
Time (s)
Velocity (cm/s)
(where the
tangent touches
the curve)
Plot these points (t,v) on a velocity-time graph. If the acceleration is uniform, the dots will suggest a
straight line of best fit. Determine the slope of this line.
Questions to consider
Is (0,0) a point on your velocity time graph? Was it moving at your reference point?
Was your acceleration constant? How can you tell?
How will you determine precision with a graph and only 1 trial?
Method 2 – Averages from averages
Using the same data used in method 1, complete the following table.
Time
(s)
Position
(cm)
Average velocity
Average Acceleration
vavg = (d2-d1)/(t2-t1)
aavg = (v2-v1)/(t2-t1)
Time
(s)
Position
(cm)
Average velocity
Average Acceleration
vavg = (d2-d1)/(t2-t1)
aavg = (v2-v1)/(t2-t1)
tavg (s)
Vavg (cm/s)
tavg (s)
aavg (cm/s2)
0
0.025
0.05
0.05
0.1
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
Questions to consider
Was your acceleration constant? How can you tell?
How will you determine accuracy with so many values calculated?
How will you determine precision with a graph and only 1 trial?Method 3 – Assuming that v1 = 0
when (t,d) = (0,0)
If one can assume that v1 = 0 then the displacement equation for uniform acceleration can be simplifed.
The calculated accelerations apply for the entire time interval used. (The average accelerations apply
only to the specific time interval used for the calculation.)
Complete the following table
Time (s)
Position (cm)
Acceleration
Estimate assuming a
zero initial velocity
a = 2Δd/Δt2
(cm/s2)
Time (s)
Position (cm)
Acceleration
Estimate assuming a
zero initial velocity
a = 2Δd/Δt2
(cm/s2)
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
Questions to consider
Was your acceleration constant? How can you tell?
How will you determine accuracy with so many values calculated?
What is different between these values calculated and the ones used from the average's table?
Lab Report
Will be discussed in class tomorrow.