course 201
9-1 12 am
Count your pendulum for a minute. xxxx
Notebook length pendulum had 53 cycles in 60 seconds
Time the ball as it moves down 10 feet of ramp, using a pendulum. Repeat for 9 feet, then for 6 feet, then for 4 feet.
10 ft completed – 5.25 cycles
9 ft completed- 5 cycles
6 ft completed- 3.5 cycles
4 ft completed – 3 cycles
Graph position vs. clock time for the ball rolling down the ramp, with clock time in units of half-cycles.
• Sketch the graph pretty carefully and sketch a smooth curve that you believe represents the actual position vs. clock time behavior of the ball.
• The speed of the ball is constantly changing, so your graph will not contain any straight lines.
• There is some uncertainty in your timing, so while your curve will probably come close to your data points, it shouldn't be expected to actually pass through any of them.
• In other words, your curve should represent the actual behavior of the ball, as best you can infer it from the data points, but you shouldn't go out of your way to make your curve actually go through any of the points.
Based on your graph estimate the following:
• The time required to travel down the first one-foot ramp.
• .8 cycles of my pendulum
• The distance that would be traveled during each half-cycle of your pendulum (i.e., the distance from the start to the end of the first half-cycle, the distance from start to end of the second half-cycle, etc.).
• 0-.5 cycles: about 50% of the first ramp or about 15 cm
• .5-1 cycles: a little past the first ramp, or about 17 cm
• 1-1.5 cycles: nearly all of the second ramp, or about 26 cm
• 1.5-2 cycles: over half the third ramp, or about 20 cm
• 2-2.5 cycles: a quarter down the fourth ramp, or about 18 cm
• 2.5-3 cycles: to the fifth ramp, or about 21 cm
• 3-3.5 cycles: nearly all the fifth ramp, or 28 cm
• 3.5-4 cycles: nearly all the sixth ramp, or 30 cm
• 4-4.5 cycles: a quarter through the seventh ramp, or 40 cm
• 4.5-5 cycles: to the ninth ramp, or 52 cm
• 5-5.5 cycles: to the tenth ramp, or 30 cm
My values do not quite follow what my logic tells me, which would be that the centimeters would increase as the ball progressed and gained speed, but I derived them from the graph I drew as best as I could.
• The average rate of change of position with respect to clock time on the first, the third, the fifth, the seventh and the ninth 1-foot ramps.
• The first: 30 cm/2 half cycle= 15 cm/half cycle
• The third: 30 cm/1.5 half cycles= 20 cm/half cycle
• The fifth: 30 cm/1.25 half cycles= 24 cm/half cycle
• The seventh: 30 cm/2.25 half cycles= 13 cm/half cycle
• The ninth: 30 cm/.5 half cycles= 60 cm/ halfcycle
Again there are inconsistencies in my data that I attribute to my hand drawn graph.
Trapezoids
Sketch a y vs. x coordinate system. The y axis is vertical (up and down the page), the x axis horizontal (left and right across the page).
You have a 'graph trapezoid' on your desk.
• A 'graph trapezoid' has the property that one of its sides is perpendicular to two other sides.
Orient your trapezoid so that its base rests somewhere on the x axis. (The base is the side which is perpendicular to two other sides; not every trapezoid has a base in this sense, but a 'graph trapezoid' does).
Estimate the two 'graph altitudes' of your trapezoid, its 'altitude' and its 'base'. You can use any unit with which you are comfortable to make your estimate (e.g., centimeters, inches, feet, kilometers, nanometers, pounds, liters, gallons, kilograms, slugs, cubic feet, miles per hour; whatever you think works best for you is fine, though length units are probably most appropriate to this exercise). (The 'graph altitudes' are the sides which are parallel to the vertical axis when the base rests on the horizontal axis).
Make on fold in the trapezoid so that if you tear the paper along the fold, the two pieces can be reassembled to make a rectangle.
Answer the two questions below, and in your answers explain your reasoning by giving the estimated dimensions, and a complete description of what you did. Your explanation show how you proceeded from your estimates to your results.
• What is the 'graph slope' associated with your trapezoid? The graph slope of my trapezoid would be 0.4. To arrive at this answer, I first had to measure the sides and base of my trapezoid. The sides (which supplied the altitudes) measured 11.5 cm and 8.5 cm while the base measured 7.5 cm. Slope formula is rise/run, so (11.5-8.5)/7.5 will give the total change in the rise (y value) divided by the total change in run (the x value). Divided out, this gives 3/7.5=.4
• What would be the dimensions of this rectangle?
• The dimensions of the area rectangle can be found by using the initial base length, 7.5, but finding the midpoint between the two altitudes and using it as the new height. This gives the dimensions of 10 cm by 7.5 cm. This is done because if a line were to be drawn parallel with the base crossing through that midpoint on the line of slope, it would create a triangle cut off of the top of the trapezoid that could orient itself to fit perfectly in the gap needed to form a perfect rectangle.
Definitions of average velocity and average acceleration:
These are the central definitions for the first part of your course. Everything you do in analyzing motion should come back to these definitions:
• The average velocity of an object on an interval is its average rate of change of position with respect to clock time on that interval.
• The average acceleration of an object on an interval is its average rate of change of position with respect to clock time on that interval.
Analyzing the motion of the Lego racer:
We estimated that the Lego racer traveled 60 cm in 1.5 seconds to rest as it traveled in the direction opposite our chosen positive direction, then 30 cm in 1.2 seconds to rest as it traveled in our chosen positive direction.
Applying the definition of average velocity to the second motion:
• By the definition, we are finding average rate of change of position with respect to clock time.
• The A quantity is the position of the racer.
• The B quantity is the clock time.
• The average rate is by definition of average rate equal to (change in A) / (change in B).
• Having identified the A and B quantities we find that
average velocity = average rate of change of position with respect to clock time = (change in position) / (change in clock time).
• According to our information, the change in position is +30 cm and the change in clock time is +1.2 seconds.
• Thus our average velocity is
average velocity = average rate of change of position with respect to clock time = (change in position) / (change in clock time) = (+ 30 cm) / (+1.2 s) = +25 cm / sec.
Find the average velocity for the first motion, using similar steps to connect your result with the definition of average velocity.
If we sketch a graph of velocity vs. clock time for the second motion:
• we know that the velocity ended up at zero
• we know that the cart was moving in the positive direction as it slowed to rest
• if we assume that the graph is a straight line, we conclude that the line decreases toward a point on the horizontal axis during the 1.2 second interval (you should have a sketch of the graph in your notes)
• we know that the average velocity is 25 cm / s; since the final velocity is zero we conclude that the initial velocity is greater than 25 cm / s; and since we expect the average velocity to occur at the middle of the time interval we conclude that the initial velocity was 50 cm/s
Our graph therefore forms a trapezoid with base 1.2 seconds, and altitudes 50 cm/s and 0 cm/s (in this case the trapezoid is in fact a triangle). We could find the trapezoid's associated slope and area.
• The slope is rise / run. The rise is the change in velocity. Velocity changes from 50 cm/s to 0, so the change in velocity is
change in velocity = final velocity - initial velocity = 0 cm/s - 50 cm/s = - 50 cm/s.
• The run is 1.2 seconds. So the slope is
slope = rise / run = - 50 cm/s / (1.2 s) = -42 (cm / s) / (s) = -42 (cm / s) * ( 1 / s) = -42 cm / s^2.
• Since the rise represents change in velocity and the run represents change in clock time, our calculation gives us (change in velocity) / (change in clock time).
This is the form of an average rate of change. Recalling the definition of average rate of change, we see that velocity is the A quantity, clock time the B quantity, so that this is the average rate of change of velocity with respect to clock time.
This is the definition of acceleration.
The slope of this graph represents the acceleration of the car.
• Note that our reasoning requires that the v vs. t graph be a straight line. Otherwise we could not have concluded that the initial velocity is 50 cm/s.
What is the slope of the graph of the first motion (the distance was 30 cm and required 1.5 seconds)?
• Answer by duplicating the reasoning used above.
• Assuming you meant the distance to be 60 cm, as it truly was in our first motion:
• Slope is rise/run. Rise corresponds to the change in velocity. Knowing that our average velocity was -40 cm/sec, our starting velocity can be assumed to be twice that, or -80 cm/sec. To find the change in velocity, we subtract -80 cm/sec, the starting velocity, from 0 cm/sec, the ending velocity, to get 80 cm/sec. Run is given as 1.5 seconds. Therefore m=rise/run=80 cm/sec /1.5 sec=53 cm/sec^2
What is the area of the graph trapezoid corresponding to the first motion, and what does this area represent?
• Answer by identifying all the quantities you use to find the area, and as best you can reason out the meaning of your result. Reason in detail similar to that used above, though the reasoning process will be different for the question of area.
• The graph trapezoid in this case is conveniently a triangle, which has an area of .5*base*height. In this case, the height is negative as it pertains to a negative velocity totaling -80 cm/sec. The base is given as 1.5 seconds. Area would then be a=.5*1.5 sec*-80 cm/sec=-60 cm. This value tells us the distance traveled over time- the original -60 cm we measured but did not input into this graph trapezoid!
Excellent work. Great insight recognizing how the 60 cm distance came 'full circle' in that last set of calculations.