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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4 cm, positive direction of motion, 100%
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3/4 cycle
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6 cm, positive direction, 75%
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2/3 cycle
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6 cm, positive direction, 75%
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5/4 cycle
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0 cm, negative direction, 50%
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7/8 cycle
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0 cm, negative direction, 50%
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.6 cycle
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5 cm, positive direction, 75%
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That pendulum will never get more than its release distance of 4 cm from equilibrium.
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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Pendulum length = 9cm, Clock Time = Ύ cycles, ds = 88cm
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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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First we have to find out the pendulum period.
Pendulum Period = 0.2 sqrt L, so 0.2 * sqrt 9 = 0.6
Now we find out how long it takes one cycle to complete.
0.6 * Ύ = 0.45 seconds
Now we can find out average rate of change of position with respect to clock time or average velocity.
V = 88 centimeters/0.45 seconds = 195.5 cm/s
Now we can find Vf.
195.5cm/s = (0cm/s + Vf)/2 = 391.1cm/s
Now we can find average acceleration which is change in velocity with respect to clock time.
A = (0cm/s 391.1cm/s)/0.45 seconds = -869.14 cm/s^2
Well we know that average acceleration cant be a negative number so the average acceleration is 869.14 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?
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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Sound travels about 340 meters/second. So we probably hear the coin about 1/10 second after it hits. Light travels at massive speeds as well. So it was probably about 1/10 second as well after the pendulum struck our eye.
For safety's sake keep the pendulum from striking your eye.
Light from the pendulum, assuming a reasonable level of illumination, is not a problem.
Neither of these numbers are significant enough for uncertainty in this experiment.
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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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Pendulum is 10 centimeters long, ramp is 30 centimeters long, ball traveled 28 centimeters, and clock time was 3 cycles.
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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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Well the initial velocity was 0 cm/s because the ball was released from rest.
So now we can figure out the average rate of change of position with respect to clock time or the average velocity.
V = 28 cm/3 cycles = 9.33 cm/cycle
Now we can find Vf.
9.33cm/cycle = (0cm/cycle + Vf)/2 = 18.66cm/cycle
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So far so good.