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course PHY 242
10/17/10; 23:00
Lab-related Questions for 101013Note: Before doing the lab questions you should run through the Sketching Exercise below. That exercise starts with
questions about masses pulled upward by tension and downward by gravity, much along the lines discussed in class. It continues with questions related
to masses on inclines.
In lab you timed the Atwood machine (paperclips on pulley) using your bracket pendulum.
`qx001. What was the length of your pendulum? What would be the period of a pendulum of this length, based on T = .2 sqrt(L)?
11cm
0.66 sec
`qx002. Give the time from release to first, second, third and fourth 'strikes' of the pendulum.
If 1 period = .66 sec=>1/2 a period = .33sec.
The total time for a period=.66sec =>.33+.66*3=2.32sec for four strikes.
`qx003. In your first set of trials there were 3 large clips on each side.
• In the first line give your counts for the first set of trials, separated by commas.
• In the second line give the mean of your counts.
• In the third line give the time interval in seconds which is equivalent to the mean of your counts.
• In the fourth line give the acceleration corresponding to the time interval just reported.
• Starting in the fifth line give an explanation of the results you gave in the third and fourth lines.
13,14,13,14
The number of times the pendulum would have struck the bracket if it had a greater elasticity.
8.6 s
2*50/13.168s[v_ave*2]/13.17=
1.34 cm/s^2= a_ave
I ave my counts then subtracted 1 and multiplied that number by the time interval for one period I then counted the time between the release and the
first strike which = .5 * period; then I added these times to get the total time over which the clips were falling.
To get the acceleration I found the v ave = `dt/`ds. Then I multiplied by two b.c. the object started from rest. This gave `dv; if we `dv/`dt = a
ave.
`qx004. In your second set of trials there were still 3 large clips on each side, but there was a small clip on the side which ascended in the first
set.
• In the first line give your counts for this set of trials, separated by commas.
• In the second line give the mean of your counts.
• In the third line give the time interval in seconds which is equivalent to the mean of your counts.
• In the fourth line give the acceleration corresponding to the time interval just reported.
• You don't need to include an explanation, since the procedure is identical to that of the preceding questions, which you explained in answering
that question. Just make sure your results make sense.
4,5,5,5,5
4.8 ave count
2.85 sec
12.29 cm/s^2
`qx005. In the third set of trials a second small clip was added to the already heavy side.
• In the first line give your counts for this set of trials, separated by commas.
• In the second line give the mean of your counts.
• In the third line give the time interval in seconds which is equivalent to the mean of your counts.
• In the fourth line give the acceleration corresponding to the time interval just reported.
3,3,3,3,3
3
1.66 sec
36.36 cm/s^2
`qx006. If there was a fourth set of trials, report as before:
• In the first line give your counts for this set of trials, separated by commas.
• In the second line give the mean of your counts.
• In the third line give the time interval in seconds which is equivalent to the mean of your counts.
• In the fourth line give the acceleration corresponding to the time interval just reported.
25,27,26,28,24
26
17.9
.35 cm/s^2
Was this last one for the Atwood machine?
`qx007. For the trial with the greatest acceleration, sketch a force diagram showing, to scale, the tension and gravitational forces acting on the
clips on the descending side of the system.
• Which vector was longer?
• By what percent was it longer?
• What is the net force on these clips as a percent of the gravitational force?
Gravitational F on the side of the 3L and 2 Sm clips is much greater.
33 percent larger.
= 0.14 or 14 percent. [F_net / total mass]
[this is if we assess 2Sm clips = 1L clip on the diagram]
if the tension force was 33% greater than the gravitational force, then the acceleration would be 33% of the acceleration of gravity, over 300 cm/s^2.
If the net force was greater than the gravitational force then the acceleration would be greater than 980 cm/s^2.
`qx008. For the trial with the greatest acceleration, sketch a force diagram showing, to scale, the tension and gravitational forces acting on the
clips on the ascending side of the system.
• Which vector was longer?
• By what percent was it longer?
• What is the net force on these clips as a percent of the gravitational force?
The Tension due to the other side was longer
33 percent
=14 percent
????The tension due to the other side of a pulley system is equal and opposite. If we analyze only one side as though it is independently suspended
then the T = mg. So I do not understand the point; I think I missed something?????
If T = m g then the net force on the mass m is 0 and the acceleration is zero.
`q009. At what average rate does the acceleration of the system change with respect to the number of small paperclips?
As the net Weight is doubled so is the acceleration. Why?—As F_net doubles the acceleration of the system nearly doubles so the relation is
one-to-one. Ex. F(1small clip)/m(6L +1Sm). Vs 2Sm/(6L+2Sm) the relation is nearly double.
If friction is negligible then this is the case. However this still doesnt answer the question. What would be the definition of that average rate? How then would you calculated it from your data?
`q010. How much acceleration do we tend to be gaining, per added paperclip?
In our lab about 15cm/s^2 /Sm p.c.= the added acceleration of the system.
Note that this question is actually the same as the preceding question. Be sure you understand why.
`q011. The unbalance in the gravitational forces with each new paperclip is of course significant. It is this unbalance that causes the differences in
the system's acceleration.
The total mass of the system does increase slightly with each added small paperclip, but for the moment let's assume that the resulting change in the
total mass of the system isn't significant.
• What percent of the acceleration of gravity do we get from each added small clip?
• How is this related to the mass of a single clip as a percent of the system's total mass?
• What is your conclusion about the ratio of the mass of a large clip to the mass of a small clip?
F(1 Sm Clip)/m(total)*g.
The percentage * g = the F_net
F_net / g = percentage
to answer the first question, what percent is your 15 cm/s^2 result of the acceleration of gravity?
`q012. This question can be challenging. Don't let yourself get too bogged down on it:
In the preceding you drew conclusions based on the assumption that the changes in the system's total mass due to adding up to a few small paperclips
was insignificant. It is perfectly possible that uncertainties in measuring the time intervals were large enough to obscure the effect of the changes
in the total mass.
However refine your answers to the preceding question to take account of the change in total system mass.
(One possible approach: assume that the requested ratio is r and symbolically solve for the acceleration a in terms of the number N of added small
clips, sketch a graph showing the predicted shape of your a vs. N curve, and see what value of r best matches this graph with a graph of your observed
a vs. N).
R*g=F_net, where R = (T clips-equal # of clips on both sides)/T clips
After manually graphing and approximating the function as linear then the `da/`dclip = r = abt 17.5 cm/s^2/clip
This function defiantly works assuming all paper clips are the same weight. If we then graph the function we can evaluate for any change in weight and
find the mass of one Sm Paper Clip in relation to the L Paper Clip.
&#This looks good. Let me know if you have any questions. &#