course phy 121
9/29 7 pmI am unsure about three questions that I have marked with three question marks (???)
I also would like to know where I messed up with the last set of acceleration questions. I think that my munbers are correct, and I knew what signs (positive or negetive) were approate but I could not get the negetive displacement to work out maybe you can see my problem.
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
x = -4 cm, speed 0 %
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• 3/4 cycle
x = 0 cm, speed 100 %
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• 2/3 cycle
x = -2 cm, speed 50 %
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• 5/4 cycle
x = 2 cm, speed 50 %
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• 7/8 cycle
x = 3 cm, speed 25 %
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• .6 cycle
x = 1 cm, speed 75 %
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Good. It turns out that the acceleration isn't actually uniform (more force and therefore more acceleration the further you are from equilibrium), so at the instant halfway between release and equilibrium the velocity isn't exactly halfway between zero and max velocity.
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:
Pendulum is 8 cm long, displacement is 85 cm, ¾ of a cycle
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Show how to start with your raw data and reason out the acceleration of the falling coin, assuming constant acceleration:
Period = squr(length) * .2
Period = squr(8) * .2
Period = 2.82 * .2
= .57 sec
¾ * .57 s = .4275s
85 cm / .4275s =198.83 cm/s this is the ave vel since the v0 was 0 if we mult this # by 2 we have vf
397.66 cm/s / (.4275 s) = 930.201 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?
It takes relatively 0 sec to reach my ear and my eye so these delays are not significant.
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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:
Trial Distance Time (10 cm pendulum)
1 25 cm 3 cyc
2 22 cm 3 cyc
3 20 cm 3 ¼ cyc
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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?
The initial velocity is 0, the average vel is 30 cm/ 1.36= 22.05, final vel is 44.1
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Moving at vf_ramp, how long would it take the ball to travel through displacement `ds_x_projectile?
44.1 cm/s / 25 cm = 1.7 s
cm/s / cm is 1/s or s^-1, not s
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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?
???
I can't tell what your question is. You know v0 and a. Presumably you measured `ds_y_projectile. So you can solve the vertical motion.
I don't know if you don't have this displacement, or if you don't know what to do with it (or both).
Just in case, `ds_y_projectile is about 90 cm for the tables in the lab.
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In the time interval you just calculated, how far would the ball travel if moving at velocity v_f_ramp?
???
I don't know whether you are confused because you didn't get the time interval in the preceding, or for some other reason; I assume the former
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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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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. Up ramp is positive
Determine the initial velocity and acceleration of the ball for the interval between the first and second event.
29cm/1.5s = 19.3 cm/s is the ave accel final vel is 0, so v0 is double that, v0 = 38.6 cm/s
Ave rate of change of vel wrt clock time is -38.6 cm/s/ 1.5 s = -25.73 cm/s^2
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Determine the final velocity and acceleration of the ball for the interval between the second and third event.
-29 cm/ 2.25s = -12.89 cm/s is the ave vel, v0 is 0, therefore the vf is doubled that, -25.77 cm/s/ 2.25 s = -11.45 cm/s^2 I thought that this number should me positive. ???
the acceleration is down the ramp in both cases, so it wouldn't change sign
looks like you got this right
The displacement is neg so the velocity is neg because you can not have a neg time. The vel that is 0 is the v0, and it is final vel is neg. -vf – v0= a neg number. I do not know what I did wrong.
you appear to have done it right
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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.
I think that the acceleration for the first and second event was greater because we bumped the ball with something causing the ball to be forced up the ramp
that force was done before the timed interval started (or should have been, since the interval was to have been a uniform-acceleration interval)
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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?
Yes, I found that the acceleration was greater in the first and second event.
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Good. See my notes and feel free to follow up with additional questions.