#$&*
course
8/30 10Questions:
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?
During the exercise of counting to 8 four times with the same beat of the bead, it took 10 seconds to complete it. Therefore the time between each bounce is 0.2 seconds. My pendulum is nine cm long.
That would be 32 counts in 10 seconds, which would be 10 seconds / (32 counts) = 0.3 seconds/count, not 0.2 seconds/count.
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?
When allowing the pendulum to swing freely, it took 36 cycles to reach 30 seconds (clock time). The period of the pendulum swinging is 30/36=0.83. A period is the time it takes to complete one cycle. Compared to my data from the “hits”, one cycle swinging freely is approximately 4 times as large as allowing it to hit the bracket.
The ratio would be lower if you used 0.3 sec, but your reasoning is fine.
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?
1 Domino—6 beats on pendulum
2 Dominoes—4 beats on pendulum
3 Dominoes—3 beats on pendulum
Average Velocities
1 Domino—30 cm/ 6 beats— 6 cm/ beat
2 Dominoes—30 cm/ 4 beats—7.5 cm/ beat
3 Dominoes—30 cm/ 3 beats – 10 cm/ beat
Since a beat is about 0.3 seconds, 6 cm/beat would be 6 cm / (0.3 sec) = 20 cm / sec.
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?
The amount of beats it takes to roll down the ramp and hit the floor is:
1 Domino—7 beats
2 Dominoes—5 beats
3 Dominoes—4 beats
Compared to the amount of beats it takes solely to roll to the end of the ramp, I concluded that it takes approximately 1 beat for the ball to fall to the floor.
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.
The ball’s end of ramp speed increased more by as a result of the third added domino. I concluded this by measuring the distance between the marks on the paper from each ball drop corresponding to the number of dominoes placed under the ramp. The fewer dominoes under the ramp, the greater the distance between drops, but as you add dominoes, the distance becomes smaller indicating that the ball is falling faster.
It would be expected that 'as you add dominoes, the distance becomes smaller' (that's a nice concise statement).
However some students have said that this indicates a progressively smaller increase in velocity for each additional domino.
I'm not taking sides on this one; let's try to get a discussion of this going in Wednesday's class and see who convinces whom.
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— vAve--`dv—--v_mid_x—v_mid_t—vf Initial velocity is the beginning and the smallest. The average velocity is mean. Then the change in the velocity is the difference between initial and final velocity. It may be the same or equal to vAve. The v_mid_x and the v_mid_t are potentially similar to each other. The final velocity is the velocity at which the object reaches the end.
****This question and the following 2, I had issues trying to differentiate between the different velocities and the size of each. While they are all my best effort, I realize they are probably not exactly right.
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 .
v0—vAve--`dv—v_mid_x—v_mid_t—vf
Place the same quantities in order assuming that v0 is relatively large.
v0--`dv—vAve—v_mid_t—v_mid_x
Which of these quantities will larger when v0 gets larger? vAve
Which will get smaller when v0 gets larger? `dv
Which will be unchanged if v0 gets larger? V_mid_t and v_mid_x
8. If a ball requires 1.2 seconds to travel 30 cm down the ramp from rest:
* what is its average velocity?
* What is its final velocity?
* What is the average rate of change of its velocity?
Be sure to explain your reasoning.
Average velocity=30 cm/1.2 s = 20 cm/s ==displacement/elapsed time
Final velocity=30 cm/1.2 s = 20 cm/s ==displacement/elapsed time
(I realize the answer is the same to each, but once again, I’m uncertain)
Average rate of change== (20 cm/s)/1.2 s = 16.7cm/s^2
The change in velocity, in this case 20 cm/s- 0 cm/s divided by the elapsed time
" ""
Good work. Not every answer is correct, but most are and you are doing a good job overall explaining your work.
#$&*