course PHY 201
20091005 0330
Class 090930Horizontal range of a pendulum let loose at equilibrium
Hold a washer pendulum stationary at its equilibrium point and let it drop to the floor. Mark or measure
the position at which it
strikes the floor.
Now, holding the pendulum so its equilibrium point is the same as before, pull it back and release it so
that it swings back toward
equilibrium. The instant it reaches its equilibrium point, let go of the string and let it fall to the
floor. Record the distance
the pendulum was pulled back from its equilibrium position, its length, its height above the floor, and the
position at which it
strikes the floor.
Repeat for a few trials.
Report your raw data below:
****
Trial 1 - Pullback = 15 cm, Length Travelled = 47 cm
T2 - Pb = 20 cm, Lt = 60 cm
T3 - Pb = 10 cm, Lt = 36 cm
Length of pendulum for all trials - 25 cm
Vertical Displacement - 89.5 cm
&&&&
Starting with the event of letting go and ending with the event of first contact with the floor, we assume
that the washer is in free
fall, the only force acting on it being the force of gravity (air resistance on the washer and on the thread
will actually be
present, but will be insignificant compared to our uncertainties in measurement). We will analyze the data
to determine its velocity
at equilibrium.
If you hold the pendulum string in a fixed point, you can move the washer around a circle whose radius is
equal to the length of the
pendulum. So as long as you are holding the string in a fixed position the pendulum will move along an arc
of a circle.
* As it swings back toward equilibrium, the direction of its motion is mostly horizontal but it also
descends, so its motion has
a downward vertical component.
* As it approaches equilibrium it moves faster and faster, but due to the shape of the circle it
descends more and more slowly.
* After it passes equilibrium it continues for a time moving in the horizontal direction as it moves
upward in the vertical
direction.
At equilibrium the washer is at its lowest point, neither rising nor falling, so at that instant it is
moving entirely in the
horizontal direction.
* If you let go of the string before the washer reaches equilibrium, then its fall will begin with a
downward vertical velocity.
* If you let go after equilibrium, then the initial vertical of its fall will be upward.
* If you let go exactly at the equilibrium point, then the initial velocity of its fall will be entirely
horizontal and its
initial vertical velocity will be zero.
* Of course you can't let go exactly at the instant the washer reaches the equilibrium position, and the
initial velocity of the
fall won't in fact be entirely vertical. The initial position of the washer won't be exactly at the
equilibrium point, either.
* Since in our analysis we will assume that the initial event occurs at the equilibrium point, with zero
vertical velocity, there
are multiple sources of uncertainty in this experiment.
Using your raw data show how you find the following, for the interval between the event of letting go of the
string and the event of
the washer's first contact with the floor.
* The time required to fall your observed vertical distance, starting with initial vertical velocity
zero and accelerating
downward at 980 cm/s^2.
* The displacement of the washer in the horizontal direction.
* The horizontal velocity of the washer.
For each of your trials, report the pendulum length, the pullback distance, and the horizontal velocity of
the falling washer. Use
one line to report the results of each trial:
****
Trial 1:
Pendulum Length - 25 cm, Pullback distance - 15 cm
Horizontal Velocity is a constant cm/s, acceleration is 0 cm/s^2, displacment is 47 cm
Vertical vf = 418.8 cm/s (found using vf = +/- sqrt(v0^2 + 2a`ds) , `dt = `ds/v_ave = 89.5cm/209.4cm/s =
.427 seconds
This horizontal time interval is the same as the vertical time interval so we can no find the average
horizontal velocity:
V0_hor = 2`ds -`dtvf / `dt = 2(47cm) - (.427sec*0cm/s) / .427sec = 220.14 cm/s
Horizontal velocity is constant, so vf = v0 = vAve. vf is not zero.
The average horizontal velocity is 110.07 cm/s
Trial 2:
Pendulum Length - 25 cm, Pullback distance - 20 cm
vf_ver = 418.8 cm/s
v0_hor = 281.03 cm/s
v_ave_hor = 140.52 cm/s
Trial 3:
Pendulum Length - 25 cm, Pullback distance - 10 cm
v0_hor = 168.62 cm/s
v_ave_hor = 84.309 cm/s
&&&&
You need do this part for only one of your trials:
Suppose the pendulum was released a little early, so that the magnitude of its initial vertical velocity was
10% of the horizontal
speed you just calculated. (If you didn't get that part you can assume a horizontal speed of 60 cm/s)
* How long would it then take to reach the floor? (you know the initial vertical velocity, vertical
displacement and vertical
acceleration)
****
Using 1st Trial:
v0_ver = .10 * 220.14 cm/s = 22.014 cm/s
knowns: v0_ver - 22.014 cm/s, `ds - 89.5 cm, ave_a - 980 cm/s^2
Using the 3rd equation - `ds = v0`dt + (a(`dt^2))/2 ---> `dt = (-v0 + sqrt (v0^2+2a`ds))/a = (-22.014cm +
(sqrt (22.014^2cm + 2
(980cm/s^2)89.5cm))) / (980 cm/s^2) = .405 seconds
To double check work, will do the long process:
vf = +/- sqrt(v0^2+2a`ds) = +/- sqrt(22.014^2cm/s + 2(980cm/s^2)89.5cm = 419 cm/s
Using 2nd equation - a_ave = `dv/`dt --> `dt = `dv/a_ave = (419 cm/s - 22 cm/s) / 980 cm/s^2 = .405 seconds
&&&&
Estimate the vertical 'drop' as the pendulum swings to equilibrium
You need do this part for only one of your trials:
Draw a circle on your paper. The diameter of your circle should be at least half the length of your paper.
Sketch the pendulum, hanging at equilibrium, as follows:
* The center of the circle represents the point at which you held the pendulum.
* The radius of the circle represents the length of your pendulum.
* The 'lowest' point on the circle, vertically below the center, will represent the center of the
washer.
* Sketch the washer, centered at this 'lowest' point.
* Sketch the string, which will extend from the center of the circle to the washer.
Now sketch the pendulum at its 'pullback' position (the 'held' end of the string will still be at the center
of the circle). Keep
your sketch reasonably to scale.
Presumably you have observed the pullback and the length of the pendulum. What is the pullback of the
pendulum as a percent of its
length? (e.g., if the length is 16 cm and the pullback 4 cm then pullback is 25% of the length).
****
Using 3rd Trial:
Pendulum length = 25 cm
Pullback Length = 10 cm
The pullback is 40% the length of the pendulum
&&&&
Since the radius of the circle represents the length of the pendulum, the number you just gave is also the
pullback as a percent of
the radius of the circle.
When the washer was pulled back in the horizontal, it was also raised in the vertical direction. Estimate
the distance it was raised
as a percent of the distance it was pulled back.
****
Pendulum Length - 25cm
The vertical displacment is about 1.85 cm
&&&&
What therefore is your estimate of the distance the washer was raised as a percent of the pendulum's length?
****
7.14%
&&&&
You measured the length. Based on your preceding estimate, how much was it raised?
****
1.85cm
&&&&
If a coin was dropped from rest, and allowed to fall a distance equal to your previous result, how fast
would it be going at the end
of its fall?
****
acceleration = 980 cm/s^2, `ds = 1.85 cm, v0 = 0 cm/s
vf = sqrt(v0^2 + 2a`ds) = 60.2 cm/s, down being the positive direction
&&&&"
Good, except for that vf = 0 assumption (see my note; easy to fix).