#$&*
course phy 201
9/23 7:00 pm
`q001. For your graphs of rubber band length vs. chain length:Find chain lengths for four different representative lengths of rubber band 1.
****
40 cm: 13cm; 45 cm: 14cm; 50 cm: 15.5 cm;
#$&*
Find the lengths of rubber band 2 for those four chain lengths.
****
40 cm: 16 cm; 45 cm: 16.5 cm; 50 cm:
#$&*
Graph the lengths of rubber band 2 vs. the lengths of rubber band 1.
****
The graph is increasing at a steady rate. However, the first rubber band chain does not increase as fast as the second rubber band chain, so the graph eventually has a vast gulf between the x and y axis points.
#$&*
`q002. Reason out the answer to the following questions, based on the definition of average velocity as average rate of change of position with respect to clock time and average acceleration as average rate of change of velocity with respect to clock time, and the fact that for uniform acceleration the velocity vs. clock time graph is a straight line. Show explicitly and in detail how you are using the definitions and the graph.
A ball accelerates uniformly from rest, traveling 100 centimeters in 10 seconds. What are its final velocity and its acceleration?
****
initial velocity: 0 cm/s; average velocity: 10 cm/s; final velocity: 20 cm/s; average acceleration: 2 cm/s^2
#$&*
A ball accelerates uniformly, with its velocity decreasing from 20 cm/s to 10 cm/s as it travels 90 cm. How long does this take, and what is its acceleration?
****
change in velocity: 10 cm/s; average velocity: 15 cm/s; time taken: 6 seconds. Average acceleration: -1.67 cm/s^2
#$&*
A ball accelerates uniformly through a displacement of 60 centimeters, during which its average velocity is 20 cm/second and its final velocity is 30 cm/second. How long does it take and what is its acceleration?
****
initial velocity: 10 cm/s; change in velocity: 20 cm/s; time taken: 3 seconds; Average acceleration: 6.67 cm/s^2
#$&*
A ball accelerates from 10 cm/s to 30 cm/s, accelerating at a uniform 6 cm / s^2. How long does this take and how far does the ball travel?
****
change in velocity: 20cm/s; time taken: 3.33 seconds; average velocity: 20 cm/s; distance travelled: 66.6 cm.
#$&*
`q003. Give your data for the rotating strap, along with a brief explanation of how you obtained your data.
****
Counter-Clockwise
3.54 seconds: 900 degrees
4.30 seconds: 1080 degrees
2.39 seconds: 450 degrees
2.99 seconds: 720 degrees
5.14 seconds: 1350 degrees
Clockwise
4.75 seconds: 900 degrees
6.89 seconds: 1530 degrees
12.2 seconds: 3030 degrees
12.93 seconds: 4320 degrees
9.14 seconds: 1860 degrees
#$&*
Find the average angular velocity of the strap for each trial.
****
Units in degrees/second
254.23
251.16
188.28
240.80
262.64
189.47
222.06
248.36
334.10
196.93
#$&*
Find the angular acceleration of the strap for each trial, assuming that the angular acceleration is constant.
****
units in degrees/second^2
71.81
58.40
78.78
80.53
51.09
39.88
32.22
20.35
25.83
21.54
#$&*
@&
You appear to have divided the average angular velocity by the time interval. This does not give you the average angular acceleration.
*@
Which appears to have the greater angular acceleration, the unloaded strap or the strap loaded with magnets? (University Physics students might not have observed the strap loaded with magnets).
****
no data
#$&*
For the unloaded strap, construct a graph of angular acceleration vs. average angular velocity, and describe any trend of this graph.
****
#$&*
`q004. Once more consider the domino on the balanced metal track. If you balance the system with the domino near the end of the track, you find that you can move it some distance without upsetting the equilibrium. If you then move the domino twice as close to center of the track and rebalance it, do you think you would be able to move the domino further without upsetting the equilibrium, or the same distance, or would a lesser distance suffice to upset the equilibrium?
****
you would be able to move it less distance
#$&*
"
@&
Good, but you need to revise your average angular accelerations.
You should be able to do so easily and quickly, once you understand your error. I do recommend that you connect the process to the definitions. These quantities are otherwise very difficult to keep straight.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
*@
Explain how you would find the area and width of a 'graph trapezoid', given its two altitudes and its slope.
****
Add the alltitudes and divide by 2 to get the rise and average height, average height * width = area. Subtract the altitudes(y2-y1), and divide by the width = slope
#$&*
@&
You could do that given the two altitudes and the width.
However given the two altitudes and the slope you would do things in a different order. To start, two altitudes give you the rise, the slope is rise / run, so you can get the run. That's the width, and from there you're home free.
*@