course Phy 201
Fractional cycles of a pendulumRegard 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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2cm, positive,50% (. 4seconds)
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3/4 cycle
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3cm,positive,75% (.6seconds)
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2/3 cycle
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2.5cm,positive,63.5% (.5seconds)
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5/4 cycle
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10cm,neigitve,100% (.10seconds)
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7/8 cycle
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3.5cm,positive,85% (.7seconds)
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.6 cycle
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3cm,positive,75%
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After 1/4 cycle the pendulum would be at equilibrium, at x = 0. It would be at its max velocity (100%), moving in the negative direction.
After 1/2 cycle the pendulum would be at the opposite extreme point, at x = -4 cm, and its velocity would be zero.
After 3/4 cycle the pendulum would be back at equilibrium, at x = 0. It would be at its max velocity (100%), moving in the positive direction.
After 1 full cycle the pendulum would be at the original extreme point, at x = +4 cm, and its velocity would be zero.
After 5/4 cycle the pendulum would again be at equilibrium, at x = 0. It would be at its max velocity (100%), moving in the negative direction.
etc.
2/3 cycle is between 1/2 cycle and 3/4 cycle, so the pendulum would be between x = -4 and x = 0, moving with a velocity between zero and 100% of maximum, in the positive direction.
Other estimates can be made accordingly. The acceleration isn't uniform so we're unlikely at this point of the course to arrive at accurate estimates.
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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16cm length string, the height at which was dropped was 87.5 cm, we did not have a coin so we used a metal ball that was on the table so if you take are info and divide it by half we would be in the right range.
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Show how to start with your raw data and reason out the acceleration of the falling coin, assuming constant acceleration:
First we got the average velocity 87.5/.3=291.66
Then to find acceleration 583.33/.3=1944.4cm/s
Units are cm/s^2.
The pendulum went through 3/4 of a cycle, which would have taken 3/4 * .8 seconds = .6 seconds. This would reduce your result to around 500 cm/s^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? .3
How long after the light in the room reflects off the pendulum does it strike your eye? Within a hunderdth of a second.
Is either delay significant compared to other sources of uncertainty in this experiment? Yes.
It would take sound less than 1/100 second to get to your ear, and light about a billionth of a second (a nanosecond) to get to your eye.
1/100 second would not be significant with this observation, but if we had very accurate means of measuring time (as we will) it could become 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:
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On my first trial my time was 2.5 full oscilations which would be 5 ticks and the ball travled in the horizontal direction. The ball we 22.5 cm up the ramp. Out of 2.5 full oscilations the ball only took 1 oscilation to come down the ramp which would be 2 ticks. *Each Tick will repersent .5seconds.
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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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22.5*2=45cm the length of the total ramp.
Average Velocity =45cm/2.5s=18cm/s
Final Velocity =18*2=36cm/s the reason we can double it is because the initial velocity was 0.
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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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Vf ramp-ball’s final velocity=36cm/s
Displacement ‘ds x projectile=the horizontal displacement of the ball leaving the ramp and striking the floor.
Well I know that the distance from the ramp to the floor is 90cm.
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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What I know is that
Vo-0
A-1000cm/s^2
‘ds-90cm
Vf=squr. V0^2+2a ‘ds
Vf=+or- 0+2(1000cm/s^2)(90)
Vf+or- 424.26cm/s
‘ds=90cm/212.13=.42seconds
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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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36cm/s / .42sec=85.7cm./s^2
85.7 cm/s^2 is not an answer to the question of 'how far'.
The answer to this question depends only on the definition of average velocity.
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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?
90cm/85.7cm/s^2=1.05 cm/sec^2
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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.
Determine the initial velocity and acceleration of the ball for the interval between the first and second event.
Initial velocity-0
Acceleration-20cm/s
You don't show how you get this. Your units aren't correct, though your number 20 is reasonable. However I can't tell whether you're reasoning this correctly.
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Determine the final velocity and acceleration of the ball for the interval between the second and third event.
Final Velocity-30
Acceleration-30cm/s
same comment as on the preceding
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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.
Greater for the second event because we timed both going up and coming down and it on took two ticks for it to come down. The reason I fill this way is because its losing acceleration going up a hill and gaining it when coming down
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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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Yes I fill it is accurate enough. On the way down.
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