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course Phy 241
12/7 3:15As far as I can tell I owe you one more lab folling this one for 11/17/10. Please let me know if this is incorrect. Thanks in advance
Conservation of Energy on a Ramp:
Use an incline with a very small slope.
Strike the steel ball so it coasts up, comes to rest, then accelerates back down the incline. Let the ball continue to roll off the incline and fall to the floor, and mark the position at which it hits the floor. (Be sure you also mark the straight-drop position).
Also note the position on the incline at which the ball comes to rest before accelerating back down.
At the same time, using the TIMER program, record the time interval from the end of the 'strike' back to the end of the incline.
Repeat for at least two good trials.
Report your data:
using ramp and audacity My trails are as follows
Trail1:
Ramp was 30.5cm long but steel ball rested 2cm from end of ramp. The ht of the ramp was 2cm so the following data represents what I observed
Start time: 2.285sec
Dist travled up ramp: 14cm
Time max dist up was reached: 3.63sec and 3.77sec
Dist travled down ramp: 16cm
Time ball went off end of ramp: 4.568sec
Time floor was struck by ball: 5.034sec
Dist. travled from drop pt on ground: 3.5cm
Trail2
Ramp was 30.5cm long but steel ball rested 2cm from end of ramp. The ht of the ramp was 3cm so the following data represents what I observed
Start time: 10.55sec
Dist travled up ramp: 10.1cm
Time max dist up was reached: 10.97sec and 11.235sec
Dist travled down ramp: 12.1cm
Time ball went off end of ramp: 12.006sec
Time floor was struck by ball: 12.373sec
Dist. travled from drop pt on ground: 6.5cm
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From each of the stopping points on the ramp, as you previously observed them, release the ball from rest and time it down the incline. Record the positions at which it strikes the floor.
Report your data:
16cm mark(1st trail)= 5cm displacement
14cm mark = 7cm displacment
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From your data, infer how long it took the ball to go up the ramp for each trial.
Calculate the ball's acceleration up the ramp, and its acceleration down the ramp.
Report your results, and indicate how they were obtained:
1st trail
Up accel: approx. 13.89cm/s^2
Down accel: approx. 42.47cm/s^2
A ball that travels to or from rest, covering 14 cm in 4 sec, would have average velocity 3.5 cm/s, final velocity 7 cm/s and acceleration about 1.75 cm/s^2. The accelerations corresponding to your data are much lower than those you report.
Your times down are significantly greater than your times up, which would imply an acceleration of a lesser magnitude for the downward motion than for the upward motion.
2nd trail
Up accel: approx. 79.34cm/s^2
Down accel: approx. 34.57cm/s^2
!!!Guess I hit the ball a lot harder and gave a greater v0 for the 2nd trail!!!!!!!!!!
Calulating an approx. of time needed to travel down ramp using `dx to calculate V0 of steel ball as it leaves the ramp(is vf in regards to the ramp’s properties). Then having v0 and vf for the ramp and `ds on the ramp we can calculate V_ave and then `dt. Subtracting this time from the time we observed for the ball to travel up and back down we have an approx. for time needed to travel up the ramp.
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From your two accelerations, you can infer the coefficient of rolling friction. What is your result?
Accel = wt_|| - F_frict, where for small slope wt_|| = wt* slope so …
F_frict = wt_|| - accel(I could easily solve this but I’m not sure what the mass of stellball)
In one direction the net force is wt|| - F_frict; in the other direction it is wt_|| + F_frict. The difference is 2 F_frict.
The acceleration one way is (wt_|| - F_frict) / m, the acceleration the other way is (wt_|| + F_frict) / m.
The difference in accelerations is therefore 2 F_frict / m, so F_frict = 1/2 m * (difference in accelerations).
for example if acceleration up is 12 cm/s^2 and acceleration down is 8 cm/s^2,
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The large ball has diameter 2.5 cm, the small ball diameter 2.0 cm. The ball is made of steel with an approximate density of 8 grams / cm^2.
Find the KE the ball attained while rolling down the ramp with the PE it lost while rolling down. Compare the two:
I am not sure what is being referenced I only have the one steel ball and am not sure if it is 2cm diameter or the 2.5?????????
KE = .5m*v^2 = .5((`pi*2cm*8gr/cm^2)*31cm/s = approx. 779J
KE = .5m*v^2 = .5((`pi*2cm*8gr/cm^2)*37cm/s = approx. 930J
You would use the radius to get the volume, and you would multiply by the squared velocity.
The volume of a sphere is 4/3 pi r^3, not pi * diameter.
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Calculate the energy lost to rolling friction. What do you get?
slope = .066 and .098, wt = [16gr/cm*`pi]*980cm/s^2
So that wt_|| = .066*16gr/cm*`pi*980cm/s^2 = approx. 3.25N for trail 1
=.098*16gr/cm*`pi*980cm/s^2 = approx. 4.83N for trail 1
Accel = wt_|| - F_frict
42.47cm/s^2 = 3.25N – (F_frict/(16gr/cm*`pi))
???I’ve gotten lost in my logic somehow please tell if you see my errors?????
you have a number of errors to correct here
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Can you account for all PE that was lost as the ball rolled down the ramp? Give details.
I’m not sure about all but most of PE is accounted for.
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Rotational dynamics with magnets on strap
Basically, after a little setup and practice, you are going to spin the strap 15 times, using a rubber band chain. The system doesn't behave all that consistently, so don't go to a lot of trouble trying to be very precise. But do take some care to use the length of the rubber band chain for each of your 'real' trials. Setup and data collection should take between 15 and 30 minutes.
You will then answer some questions. The amount of calculation necessary is not extensive, and most of the calculations are of the same type you have done many times already in the course. Do be sure to use units throughout your calculations. The main challenge will be interpretation of the questions in terms of the system.
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You will accelerate the strap using a chain of rubber bands, as demonstrated in class.
Assume for now that the restoring force of your chain is
F = - k x,
where x is the length of the chain in excess of its length when supporting a single suspended domino, and k is
k = (1/2 N / cm) / (number of rubber bands in chain).
For example, suppose you have a chain of 7 rubber bands, which has length 44 cm when supporting the weight of a domino. When the chain is 50 cm long, it exerts a force, which is calculated as follows:
For this chain k = ( 1/2 N / cm ) / 7, since there are 7 rubber bands. We get k = .07 N / cm, approx..
The chain is 6 cm longer than it was when supporting a domino, so x = 6 cm.
The force exerted at this length is therefore F = - k x = -.07 N / cm * 6 cm = -.42 N.
Forces for other lengths can be calculated using F = - k x = -.07 N / cm * x. Just remember that x is the length in excess of the 44 cm basic length.
Loop the last rubber band in your chain around the edge of the strap in such a way that when released the strap rotates in the clockwise direction. Most likely this will cause the nut to spin in such a way that it descends the threaded rod.
Extend the rubber band chain far enough that when the strap is released it will rotate through several revolutions before coming to rest. However make it spin it fast enough to be dangerous, and make sure it's spinning slowly enough that you can count its revolutions.
Find a chain length that accomplishes the above goals.
What is the length?
76cm
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You will use this chain length for every trial in your experiment.
Advice:
For each of the trials below, if the nut does rotate, return it to its original position between trials.
If you wish you can lubricate the system with WD-40 or something similar. This is entirely optional. If you don't have anything like WD-40 handy, don't worry about it. Your instructor doesn't yet know whether this will yield better results, or worse results, than not lubricating the system.
Using this chain length for every trial, do five trials, releasing the system and counting the half-revolutions. It should take several seconds for the system to come to rest, so you should also be able to look at the second hand or digital display of a clock or wristwatch when you release the strap, and when it comes to rest. You should be able to remember the two clock positions and the number of half-revolutions long enough to write them down.
Report your data for these five trials:
12 counts, 6.5sec
14counts, 5.06sec
44counts, 14.8sec
24counts, 8.7sec
32counts, 13.896sec
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Now repeat, but this time with a magnet halfway between the axis and each end of the strap. Again do five trials.
Report your data for these five trials:
19count, 9.9sec
2count, 1.6sec
16count, 6.3sec
12count, 7.5sec
22count, 25sec
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Repeat once more, with the magnets near the ends of the strap. Again do five trials.
Report your data for these five trials:
23.5count, 19sec
18.5count, 16sec
30count, 22sec
22count, 15.8sec
12count, 13sec
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Based on your data:
What is the median number of half-revolutions through which the strap rotated for each set of five trials?
24, 16, 22
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What is the median time required for the strap to come to rest for each set of five trials?
8.7sec, 7.5sec, 16sec
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Based on your median values, what is the average angular velocity of the strap for each set of five trials, calculated in half-revolutions / second?
Using 2*`pi*radius to find circum. and dividing by `dt we get
1st group = [2*pi*15cm]/[24(half rev.)/8.7sec] = approx. 34.16(1/2 rad. per sec)
2nd group = [2*pi*15cm]/[16(half rev.)/7.5sec] = approx. 44.18(1/2 rad. per sec)
3rd group = [2*pi*15cm]/[22(half rev.)/16sec] = approx. 68.54(1/2 rad. per sec)
Angular velocities don't consider the radius; 24 half revolutions is 24 * pi radians, which divided by 8.7 seconds is a bit under 9 rad / sec.
For the second and third groups we get about 7 rad / sec and 4 rad / sec.
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Assuming uniform angular acceleration, what is the initial angular velocity of the strap for each set of five trials, calculated in half-revolutions / second?
1st trail = 34.16(1/2 rad. per sec)*2 = 68.32 (1/2 rad. per sec)
2nd trail = 44.18 (1/2 rad. per sec)*2 = 88.36 (1/2 rad. per sec)
3rd trail = 68.54 (1/2 rad. per sec)*2 = 137.08 (1/2 rad. per sec)
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Assuming uniform angular acceleration, what is the acceleration of the strap for each set of five trials, calculated in half-revolutions / second^2?
1st trail = 68.32 (1/2 rad. per sec)/8.7sec = approx. 7.85 half-revolutions / second^2
2nd trail = 88.36 (1/2 rad. per sec)/ 7.5sec = approx 11.78 half-revolutions / second^2
3rd trail =137.08 (1/2 rad. per sec) /16sec = approx 8.57 half-revolutions / second^2
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The moment of inertia of the strap is about 1/12 m L^2, where L is its length (about 30 cm) and m its mass (about 65 grams). The moment of inertia of a magnet whose center is at distance r from the axis is m r^2, where m is its mass (about 50 grams).
What therefore is the moment of inertia of the strap (plus magnets if present) in each of your trials?
1st trail: I = 1/12*65gr*(30cm)^2 = 162.5 gr*cm^2
2nd trail = 162.5 gr*cm^2 + [50gr*(15/2)^2(because mag where halh way between middle and end an r =15)]
there are two magnets so this would be 162.5 gr*cm^2 + 2 * (50gr*(15/2cm)^2). Similar comment on 3d trial.
= 162.5 gr*cm^2 +2812.5gr*cm^2 = 2975gr*cm^2
3rd trail = 162.5 gr*cm^2 + [50gr*(15)^2] = 162.5gr*cm^2 + 11,250gr*cm^2 = 11,412.5gr*cm^2
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The angular kinetic energy of an object whose moment of inertia is I is given by the formula KE = 1/2 * I * omega^2, where omega is its angular velocity. Since a half-revolution corresponds to pi radians of angular distance, the angular velocity in rad / sec is pi times the angular velocity in half-rev / sec.
What is the angular KE of the strap corresponding to each set of five trials, based on the initial angular velocity you previously calculated?
1st trail: KE = 1/2*162.5 gr*cm^2*[`pi*34.16(1/2 rad. per sec)] = 8,719.49
2nd trail = 1/2*2975gr*cm^2*[`pi*44.18(1/2 rad. per sec) ] = 206,458.4
3rd trail = 1/2*11,412.5gr*cm^2*[`pi*68.54(1/2 rad. per sec)] = 1,228,643.08
????I’m not sure my calculations are correct????????????????
Your angular velocities are not correct. See previous notes.
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What average force was exerted by the rubber band system during each trial? This result should be the same for all trials, if you followed the instruction to use the same chain length for each trial.
I’m having a brain melt down and can’t figure out what I need to do. The force would simply be the tension force which is “quadruple” but at this very moment I cannot remember how to apply this info???
It should look something like F =-kx, where x is pt. in relation to equilibrium or dist. rubber bands are stretched.
My fault here. I didn't ask you to measure the 'basic length' of the chain--the length at which it begins to exert force. You should know how many rubber bands you had in the chain (probably about 8 or 9), so you should be able to find k. If you had be basic length you could calculate how far beyond that length your 76 cm chain was, and use F = - k x to find the force at that position.
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Through what distance do you estimate the rubber band chain exerted its force? Again a single result will probably apply to all trials.
The dist. the rubber band exerts its force is the dist. from where rubber bands are slack to the final position you pull them back to. That is start position of rod before interval is init.
Ex. x = pt where rubber bands are not stretched at all and `dx is pt. where rubber bands are stretched no more but released.
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About how much work did the rubber band chain therefore do during each trial? Once more a single result will probably apply to all trials.
If the force is given in form
F = -kx, then
F_ave = -(1/2)kx, and
`dW = F_ave*`ds = -(1/2)kx*x = -(1/2)kx^2
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What is the average torque on the system during each trial? (recall that torque = force * moment-arm; in this case the moment-arm is the distance from the axis to the point where the rubber band exerted its force)
moment arm = radius and force = -1/2kx
So
torque = -(1/2)kxr
???I’m not so sure this is in the correct form??????
That would be the expression for the average torque; to avoid confusion maybe 12 k * x * r would be better form, but nothing wrong with your answer here.
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Through what fraction of a radian do you estimate the system traveled between release and the instant the rubber band chain went slack?
Not much, maybe 1/8 or 1/5???
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What therefore is the product of average torque and angular displacement, for this interval?
`dKE
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You've made some errors here, and to see how this works out you'll need to correct them and resubmit. That will be worth the effort.
See my notes.
&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions 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.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you. &#
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