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course Phy 241
September 3 around 9:00am.There is a few questions I had trouble understanding. I put
???
1. What was your count for the pendulum bouncing off the bracket, and how many seconds did this take? What therefore is the time in seconds between collisions with the bracket? What was the length of your pendulum?
The count for the pendulum bouncing off the bracket was approx. 8 beats and took approx. 3 seconds since it took about 12 seconds for 32 beats (12/4=3). The time in seconds between beats with the bracket is approx. 3/8 of a second or 0.375. The length of my pendulum was 9cm, from the top of the non-moving string down to the center of the bead.
2. What was the period of your pendulum when it was swinging freely? Give your data and briefly explain how you used it to find the period. How does your result compare with the time between 'hits' in the first question?
I propped one end of the bracket up with one domino so the bead was not in contact with the bracket. I then released it, allowing it to go side to side. As soon as I released it, I watched the clock and it “swayed” for 50 seconds until it came to a stop. I also did it another time and it “swayed” for 30 seconds and had approx 42 sways.
That would be about .71 seconds per cycle.
3. Give your data for the ball rolling down the ramp, using the bracket pendulum as your timer. Assuming the ball traveled 30 cm each time, what are the resulting average velocities of the ball for each number of dominoes?
Ball rolling down the ramp (onto table): Avg velocities of ball if traveled at 30cm each time:
1 domino: ball took 7 beats on pendulum (2.625 sec) 11.43 cm/s
2 dominos: ball took 5 beats on pendulum (1.875 sec) 16 cm/s
3 dominos: ball took 4 beats on pendulum (1.5 sec) 20 cm/s
4 dominos: ball took 3 beats on pendulum (1.125 sec) 26.66 cm/s
4. How did your results change when you allowed the ball to fall to the floor? What do you conclude about the time required for the ball to fall to the floor?
When I allowed the ball to fall to the floor, my results changed in how many beats on the pendulum it took until it hit the floor. It took the ball more time, therefore had more “beats” on the pendulum.
Ball rolling down the ramp (into the floor):
1 domino: ball took 8 beats on pendulum (3 sec)
2 dominos: ball took 6 beats on pendulum (2.25 sec)
3 dominos: ball took 5 beats on pendulum (1.875 sec)
4 dominos: ball took 4 beats on pendulum (1.5 sec)
So looking at my results from the two different experiments (onto table vs. into the floor), it’s clear that it took the ball more time to fall into the floor since the beats of the pendulum is about 1 second more than the first results. It also took LESS time (seconds) for the ball to hit the floor the more dominos you add.
5. Look at the marks made on the paper during the last class, when the ball rolled off the ramp and onto the paper. Assuming that the ball required the same time to reach the floor in each case (which is nearly but not quite the case), did the ball's end-of-ramp speed increase by more as a result of the second added domino, or as a result of the third? Explain.
???I don’t really understand what this question is asking. However, looking at my marks on my paper, from the 2nd added domino to the 3rd domino, the marks are about the same distance apart. So, I’m thinking the answer to this question would be “neither” but I need help on an explanation. I also know the ball’s end of ramp speed increased more as a result of the 3rd added domino because it was raised at a higher angle and therefore caused the speed to increase.???
If you have marks corresponding to 1, 2 and 3 dominoes, then if the spacing between the 1-domino and 2-domino marks is equal to that between the 2- and 3-domino marks, then if we assume equal 'falling times', the added distances correspond directly to increases in velocity and 'neither' would be an appropriate answer.
6. A ball rolls from rest down a ramp. Place the following in order: v0, vf, vAve, `dv, v_mid_t and v_mid_x, where the quantities describe various aspects of the velocity of the ball. Specifically:
* v0 is the initial velocity,
* vf the final velocity,
* vAve the average velocity,
* `dv the change in velocity,
* v_mid_t the velocity at the halfway time (the clock time halfway between release and the end of the interval) and
* v_mid_x the velocity when the ball is midway between one end of the ramp and the other.
Explain your reasoning.
V0 < v_mid_t < v_mid_x < vf=dv
???I still have vAve to put in. I don’t know where it goes exactly, but I DO know that vAve is going to be between v0 and vf.???
Good so far. I'm going to let everyone think awhile longer about where to put vAve, but we are very likely to either talk about this Wednesday, or to do an experiment to try to measure these different quantities.
7. A ball rolls from one ramp to another, then down the second ramp, as demonstrated in class. Place the following in order, assuming that v0 is relatively small: v0, vf, vAve, `dv, v_mid_t and v_mid_x .
???Even though v0 is relatively small, I’m thinking that the order is going to be the same order as above.???
Place the same quantities in order assuming that v0 is relatively large.
???If v0 is relatively large, I’m thinking that the order will ALSO be the same as above because if v0 is big, the rest are going to be bigger. ???
Which of these quantities will larger when v0 gets larger? Which will get smaller when v0 gets larger? Which will be unchanged if v0 gets larger?
If v0 gets larger, the change in velocity (dv)
Good. That in itself will change the order in which `dv appears.
8. If a ball requires 1.2 seconds to travel 30 cm down the ramp from rest:
* What is its average velocity? 25 cm/s
* What is its final velocity? 30 cm/s
if you average 30 cm/s and 0 cm/s you don't get 25 cm/s. So I'm not sure about that final velocity.
* What is the average rate of change of its velocity? =change in velocity/change in clock time
=30/1.2
=25 cm/s^2" good reasoning, except possibly for the final velocity
&#Good responses. See my notes and let me know if you have questions. &#
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