course 201
9/30 10 am
Class 090928The following conventions will allow your instructor to quickly locate your answers and separate them from the rest of any submitted document, which will significantly increase the quality of the instructor's feedback to you and to other students.
When answering these questions, give your answer to a question after the **** and before the &&&&.
When doing qa's, place your confidence ratings and self-assessment ratings on the same line as the prompt.
If you don't follow these guidelines you may well be asked to edit your document to make the changes before I can respond to it.
Thanks.
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Fractional cycles of a pendulum
Regard the equilibrium position of a pendulum as the origin of the x axis. To the right of equilibrium x values are positive, and to the left of equilibrium x values are negative.
Suppose you release a pendulum of length 16 cm from rest, at position x = 4 cm.
Estimate its position in cm, its direction of motion (positive or negative) and its speed as a percent of its maximum speed (e.g., speed is 100 % at equilibrium, 0% at release, and somewhere between 0% and 100% at every position between) after each of the following time intervals has elapsed:
• 1/2 cycle
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-4 cm, negative, 0% (it is on the other side of its cycle, slowing to a stop to return back in the positive direction)
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• 3/4 cycle
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0 cm, Positive, 100%
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• 2/3 cycle
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1 cm, positive, 75%
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• 5/4 cycle
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0 cm, negative, 100%
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• 7/8 cycle
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3 cm, positive, 25%
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• .6 cycle
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-3.2 cm, positive, 20%
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Acceleration of Gravity
Drop a coin and release a pendulum at the same instant. Adjust the length of the pendulum so that it travels from release to equilibrium, then to the opposite extreme point and back, reaching equilibrium the second time at the same instant you hear the coin strike the floor. Measure the pendulum.
• Give your raw data below:
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10 cm long pendulum
3/4 cycle for ball to drop from table
87 cm from table to floor
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Show how to start with your raw data and reason out the acceleration of the falling coin, assuming constant acceleration:
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Period of 10 cm pendulum is .2(L)^1/2=.2(10)^1/2=.65 seconds
3/4 of a period passed, so .65seconds/period(3/4 period)= .49 seconds to drop
Velocity=`ds/`dt=87 cm/.49 sec=177.6 cm/sec
Since initial velocity was zero, it can be assumed that `dv is twice the average velocity, or 2(177.6 cm/sec)=355.1 cm/sec
Acceleration=`dv/`dt=355.1 cm/sec / .49 sec= 724.7 cm/sec^2
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There are two delays between the events you are observing and your perceptions:
• How long after the coin strikes the floor do you hear it?
• How long after the light in the room reflects off the pendulum does it strike your eye?
• Is either delay significant compared to other sources of uncertainty in this experiment?
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Both are significant as I first wait to hear the ball drop, and then I immediately look at the pendulum. I lose time in between both events due to the speed of the sound getting to me later than the actual drop, and then my reaction time to catch the position of the pendulum. These delays may account for my calculated acceleration being lower than the actual acceleration of the ball- I found the ball to take more time in seconds than it actually took.
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Introduction to Projectile motion
Time a ball down a ramp, and measure how far it travels in the horizontal direction.
• Give your raw data below:
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Using a stopwatch, we found that for the ball to go down the 30 cm ramp and fall to the floor on a 1 domino width (.5 cm) incline, it took 3.2 seconds
For the ball to fall straight down from the end of the ramp to the floor alone, it took .5 seconds
Measuring the difference in drop points of the two balls, there was a 7.4 cm difference, as in, the ball on the ramp traveled 7.4 cm horizontally further than the ball that was dropped from rest.
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To keep things straight, let's use the following notation in the rest of this analysis:
• `dt_ramp is the time required to travel the length of the ramp starting from rest
• `ds_ramp is the displacement of the ball along the ramp
• vf_ramp is the ball's final velocity on the ramp
• `ds_x_projectile is the horizontal displacement of the ball between leaving the ramp and striking the floor
• `ds_y_projectile is the vertical displacement of the ball between leaving the ramp and striking the floor (for the tables in the lab we may assume that `ds_y_projectile is about 90 cm).
• `dt_projectile is the time interval between leaving the ramp and striking the floor
Answer the following questions:
According to the time `dt_ramp required to travel down the ramp and its length `ds_ramp, what are the average and final velocities on the ramp, assuming uniform acceleration?
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To find `dt ramp, subtract the ball drop time from the ball ramp + drop time. `dt ramp= 3.2 sec - .5 sec=2.7 sec
`ds ramp= 30 cm
V ave=`ds/`dt=30 cm/2.7 sec=11.1 cm/sec
Since v0 ramp=0, it can be inferred that vf ramp=2*Vave=2*11.1cm/sec=22.2 cm/sec
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Moving at vf_ramp, how long would it take the ball to travel through displacement `ds_x_projectile?
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`ds x projectile= 7.4 cm
22.2 cm/sec /7.4 cm= 3 seconds
cm/s / cm is 1/s or s^-1
it doesn't take 3 seconds to move 7.4 cm at 22 cm/s
this is of course easy to correct
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Accelerating at 1000 cm/s^2, how long would it take the ball to fall from rest through displacement `ds_y_projectile?
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`ds y projectile= 90 cm
1000 cm/sec^2 /90 cm= 11.1 sec^2= 3.3 sec
cm/s^2 / cm = 1/s^2
a / `ds isn't a significant quantity; neither is `ds / a.
You know v0, `ds and a. How do you solve this situation? You need to use the equations.
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In the time interval you just calculated, how far would the ball travel if moving at velocity v_f_ramp?
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22.2 cm/sec (3.3 sec)= 73.3 cm
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Accelerating at the rate you calculated in the preceding exercise, how long would it take the ball to fall from rest through displacement equal to `ds_y_projectile?
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22.2 cm/sec /90 cm=.25 seconds
22.2 cm/s is a horizontal velocity; 90 cm is a vertical displacement
the two don't belong together in the same calculation
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Ball up and down ramp
'Poke' a ball (perhaps using your pencil as a 'cue stick') so that it travels partway up a ramp then
back. Observe the clock time and position at three events: the end of the 'poke', when the ball comes
to rest for an instant before rolling back down, and its return to its original position.
Choose your positive direction.
Positive direction is down the ramp, back towards the initial point.
`dt up the ramp= 1.16 seconds
`dt total= 2.96 seconds
29 cm up to vf=0, 29 cm back down (-29 cm, then +29 cm)
Determine the initial velocity and acceleration of the ball for the interval between the first and second event.
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V ave=`ds/`dt=-29 cm/1.16 sec= -25 cm/sec
Since vf=0, v0= 2*Vave=2*-25 cm/sec=-50 cm/sec
Acceleration=`dv/`dt=vf-v0/`dt=-50cm/sec-0cm/sec /1.16 sec=-50cm/sec /1.16 sec=-43.1 cm/sec^2
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Determine the final velocity and acceleration of the ball for the interval between the second and third event.
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`dt down ramp=`dt total-`dt up ramp=2.96 sec-1.16 sec=1.8 seconds
Vave= `ds/`dt=29 cm/1.8 sec=16.1 cm/sec
if `ds was positive in the first calculation it will be negative in this calculation
Since v0=0 cm/sec, it can be assumed that vf=2*Vave=2*16.1 cm/sec=32.2 cm/sec
Acceleration=`dv/`dt=vf-v0/`dt=32.2 cm/sec-0 cm/sec /1.8 sec= 17.9 cm/sec^2
the acceleration is down the ramp both times; your signs imply that it's down in one case, up in the other
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Do you think the acceleration of the ball is greater between the first and second event, or between between the second and third event? Or do you think it is the same for both? Give reasons for your answer.
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I think the acceleration of the ball is greater between the first and second event, even though this acceleration is negative. This makes sense as the initial push was given before the first event, and that forces the ball up the ramp. When it slows to a stop and begins to roll back down, the only contributor to the acceleration is the incline of the ramp. If we had a ramp of a greater incline, it may be enough to increase the acceleration to higher than the initial acceleration up the ramp. But for us, with an incline of only .5 cm, this was not the case.
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Are your data accurate enough to determine on which interval the acceleration is greater? If so, on which interval do you determine it is greater? If not, how accurate do you think your data would need to be to decide this question?
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I feel that yes, our data is accurate enough.
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Homework:
Your label for this assignment:
ic_class_090928
Copy and paste this label into the form.
Report your results from today's class using the Submit Work Form. Answer the questions posed above.
You should complete the Major Quiz by Wednesday, October 7. You are urged to complete it this week. Remember if your grade isn't up to your expectations you can take another.
Note the Major Quiz sample test questions distributed by email last week.
Note the recommendation to work through the q_a_ documents listed below. This is a recommendation, not a requirement. Note that many students (though not all) find that this is the most time-effective way to learn the material. If you choose to do this:
Read each question and work it out, at least in your head. If your answers and procedures agree with the given solution and you are sure you understand, you don't have to insert anything. If your answers don't agree with the given solution, consider inserting a self-critique.
Links are provided below:
• q_a_3
• q_a_4
• q_a_5
• q_a_6
• q_a_7
• q_a_8
You have a few errors, though most of your work is correct. Be sure to see my notes and let me know if you have questions.
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