Physics I 070829
Basic situation: A ball rolls 60 cm down a ramp in 2.5 seconds.
Analysis:
We want to find at least the initial, average and final velocities of the ball between its release and reaching the end of the ramp.
The ball begins its acceleration down the ramp at the instant it is released from rest, at which instant its velocity is 0. We say that the initial velocity of the ball is zero. When the ball was released various clocks in the lab and around the world read different times; it's possible that you might have determined the time using a clock that didn't read zero at the instant of release. So, depending on what clock was used, the clock time at release might or might not have been zero. However since you started your timing by releasing the pendulum at the same instant as the ball, you will probably regard this as the 'start' of your 'clock' and you will say that the initial clock time is zero.
Under these assumptions our graph of velocity vs. clock time would start with a point at (0, 0), indicating that the initial event (the release of the ball) occurs at clock time 0 and velocity 0.
The average velocity of the ball is
- ave vel = 60 cm / (2.5 sec) = 24 cm / sec.
The net force accelerating the ball down the ramp is the resultant of the normal force and the gravitational force acting on the ball. (Don't worry about it yet, but note that this is actually an ideal case and is really an oversimplification of the actual system we're observing, but we don't yet have the machinery to deal with all aspects of the situation).
The picture that represents the forces acting on the ball is the same all the way down the ramp. So the net force acting on the ball is uniform.
From this it follows that the velocity vs. clock time graph of the ball is a straight line.
Assuming that this straight line begins at the origin, it will consist of a straight line segment extending up and the the right into the first quadrant, and will terminate when it reaches the point where t = 2.5 seconds. The average velocity of 24 cm/s will occur at the midpoint of this segment, and the final velocity at the end of the segment. It should be clear from the geometry of the graph that, since the initial velocity is 0 and the line is straight, the final velocity is double the average velocity, or 48 cm/s.
In the above analysis we used the idea of average velocity, which we defined as displacement / change in clock time.
We are going to make another equivalent definition of average velocity, this time within the context of the rate of change concept.
First we define rate of change:
Definition: The average rate of change of A with respect to B is
The average velocity is now defined as a rate of change between two events:
For the ball on the ramp the first event is the release of the ball at the top and the second is the arrival of the ball at the lower end of the ramp.
The definition of average rate tells us how to calculate average rate of change of A with respect to B.
We need here to calculate average rate of change of position with respect to clock time.
Comparing the phrase
'average rate of change of A with respect to B' with the phrase
'average rate of change of position with respect to clock time'
we see that the two phrases are identical, is only we let 'A' stand for 'position' and let 'B' stand for 'clock time'.
'Plugging' into the definition of average rate the word 'position' in for 'A' and the phrase 'clock time' for 'B' we get
Definition of average rate of change of position with respect to clock time: The average rate of change of position with respect to clock time is
Between the two events (release and reaching the end of the ramp) the position of the ball changed by 60 cm and the clock time changed by 2.5 seconds. So we have
This is the average velocity.
Now we apply the definition of average rate of change to a different question:
Average velocity is the average rate of change of position with respect to clock time.
New Question: What is the average rate of change of velocity with respect to clock time between these two events?
Remember, just as before:
Definition: The average rate of change of A with respect to B is
We compare the phrases
average rate of change of velocity with respect to clock time and
average rate of change of A with respect to B
and see that in this case 'A' should stand for 'velocity' and 'B' should stand for 'clock time'. So, rewriting the 'average rate' definition and making this substitution, we obtain the following:
Definition: The average rate of change of velocity with respect to clock time is
For the situation in which the ball accelerates thru 60 cm in 2.5 sec, from rest and with a linear v vs. t graph, we have seen that the velocity starts at 0 and ends at 48 cm/s. Thus we have
average rate of change of velocity with respect to clock time is
Note how we get cm/s^2. In our calculation (cm / s) / s is cm/s divided by s. To divide by a quantity you multiply by its reciprocal, so (cm/s) / s means (cm/s) * (1/s). To multiply fractions you multiply the numerators and multiply the denominators. So you get (cm * 1) / (s * s) = cm / s^2.
Repeat the preceding experiment using the steel ball.
Report in the following format:
Additional report: If you roll the ball down the ramp and off the edge, allowing it to fall to the floor, what is its horizontal range (relative to a straight drop from the edge of the ramp)?