#$&*
course Phy 241
9/5 9
Questions: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?
ANSWER
I completed this experiment several times and my data seemed to always be, 8 for my count at 3 sec. Therefore time in seconds between collisions was (3/8) = .375 sec. From magnet to top of bead was a little less than 5 inches, when calculated for the mid-section of the bead it was exactly 5 inches or 12.7 cm.
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?
ANSWER
Period of my pendulum was approx. 1.3167 oscillations per sec. I got this approx. by using my data which was where I observed my pendulum oscillating for total of 39/40 times in a 30 sec time interval. I then divided 39.5 by the 30 sec to get 1.3167 approximately. With units this is 39.5 cycles / 30 sec = 1.3 cycles / sec, which is the frequency, not the period.
Note that I was only capable of performing this observation one time due to time restrictions, so I had nothing to compare it to. It took a little less than 4 times the amount of time to perform an oscillation when compared to the ’hits’.
your results for this and the preceding are very nicely consistent; calculate the period for the present problem and you might see what I mean.
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?
ANSWER
While observering the steel ball rolling down the ramp while being timed with my pendulum, I gathered the following data. With one domino I counted between 5 and 6 ‘hits’. While with two dominoes I observed 4 ‘hits’. With three dominoes I counted 3 ‘hits’.
With this data I calculated that for a ramp of 30 cm with one domino, the steel ball traveled with an average velocity of approx. 5.4545 cm/s. With two dominoes the steel ball traveled with an average velocity of approx. 7.5 cm/s. With three dominoes the steel ball traveled with an average velocity of approx. 10 cm/s.
4 'hits' at .375 sec between 'hits' would be 1.5 sec, which would give you an average velocity of 20 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?
ANSWER
While observering the steel ball rolling down the ramp and then falling to the floor, I gathered the following data. With one domino I counted between 7 ‘hits’. While with two dominoes I observed 5 ‘hits’. With three dominoes I counted 4 ‘hits’.
With this data it clearly shows that there was time added to the previous data collected, which most would attribute to the time needed for the ball to fall that extra distance to the floor. When I really examined the data collected I noticed a direct correlation between the smaller the velocity at which the ball came with off the ramp, the more ‘hits’ that were added to the total number of ‘hits’ before the ball hit the floor. So it is feasible to say that the velocity had a direct effect on time required to reach the floor. When you think about it only makes sense, but that was only observation I had regarding time required 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.
ANSWER
I believe it would have had to due to the fact that every experiment observed the ball started at a velocity of zero. !!!!OR!!!! Maybe the added time is not coming from the velocity its simply making the ball travel a further distance with every domineo added and this increased distance that must be travled is what is accounting for the added time?????
I'm not sure you were here for this one. We'll be repeating it, and you'll understand. The time of fall is pretty much the same for each.
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.
ANSWER
1. v0 -has least amount of velocity, due to starting from rest.
2. v mid t - is next due to the fact that the ball is gaining velocity going down the ramp so that it will take more time to travel the same distance when looking at velocity at earlier t’s. Meaning it may take 3 or 4 times as long to travel the same distance the first 1/3 of the ramp compared to the last 1/3. So knowing there will be a lot more time spent on the starting side of the ramp, we know that the mid t it will be centered closer to the starting point(which has the least velocity) with respect to all other aspects we are measuring.
3. v mid x – is next because it’s position will come later than the v mid t point(Because ball is constantly gaining velocity, later position means greater velocity), but after the vAve(average velocity). Again because the ball is gaining velocity at a greater and greater rate the velocity at the center of the ramp will be less than the vAve(!!!!!!!!Unless the velocity is growing at a constant rate, then vAve and v mid x would be equal!!!!!!)
4. vAve – I explained why I believe this value should be greater than v mid x. It is easy to see that vAve is equal to (total displacement/total time taken) and `dv is simplily change in velocity in initial velocity and final velocity. Because the initial velocity is zero then `dv will be equal to final velocity. This in our case, with velocity always increasing must be greater than average velocity.
5. `dv – I am putting it next to last because in every other situtation except for when starting with a neg. velocity or a velocity of zero, then when working with a situation like ours where velocity is increasing positively in an increasing manner. Then the final velocity will always be larger.
6. vf – which is really equivalent with `dv is what I picked as the greatest of all which is the final velocity.
Explain your reasoning.
Good thinking. We'll be designing an experiment to test this.
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 .
I am not sure about 2 ramps we only did one, but I’m guessing that the first ramp is higher than the second so that velocity is growing at a faster rate during first first ramp. But does ball lose velocity once it reaches second ramp???
If so….
1. v0
2. v mid t
3. v mid x
4. vAve
5. `dv
6. vf
Place the same quantities in order assuming that v0 is relatively large.
??????But does the ball keep gaining velocity or is it slowed down???????
both ramps are sloped in the same direction, as demonstrated, so the ball continues speeding up
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?
vAve will get larger as well as final velocity if velocity is being gained and not lost. Assuming that the final velocity can only be so great, `dv will be smaller the larger v0 gets. If the ball is going to gain the same amount of velocity on the ramp regardless of starting velocity then mid-points should be unchanged.
8. If a ball requires 1.2 seconds to travel 30 cm down the ramp from rest:
* What is its average velocity? (total displacement/`dt) = 30cm/1.2sec = 25 cm/sec
* What is its final velocity?[d/t = (vi + vf)/2
2d/t = vi + vf
(2d/t)-vi = vf, no plug in our info
vf = ((2*30)/1.2) – 0 = 60 cm/1.2 sec = 50 cm/sec
* What is the average rate of change of its velocity?
Ave Acceleration = `dv/t = 50 cm/s/1.2 = 41.6667
50 cm/s/(1.2 2) = 41.6667 cm/s^2. You need the units.
Be sure to explain your reasoning.
"
&#Good responses. See my notes and let me know if you have questions. &#
#$&*