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course Phy 201
Phy 110914The link below is downloadable and/or playable. A recent version of Windows Media Player appears to run it fine. QuickTime should run it well. Let me know.
../../class_notes_current/pendulum_in_changing_water_depth/pendulum_in_changing_water_depth-iPhone.m4v
This doesn't appear to be downloadable to Windows. If you're running a Mac or IPad give it a try and let me know. Also might work on a phone.
../../class_notes_current/pendulum_in_changing_water_depth/pendulum_in_changing_water_depth-iPhone-cell.3gp
The v vs. t trapezoid with 'graph altitudes' v0 and vf, and width `dt, has the following properties and interpretations:
The slope is rise / run = (vf - v0) / `dt = `dv / `dt.
By the definition of rate the quantity `dv / `dt is the average rate of change of velocity with respect to clock time.
This is by definition the average acceleration on the interval represented by the trapezoid.
Thus aAve = (vf - v0) / `dt.
This can be rearranged to the form vf = v0 + aAve * `dt.
(University Physics only: If v is a function of t, then in the limit as `dt -> 0 the slope of a series of trapezoids whose intervals all contain t = t_0 approaches the derivative v ' (t_0); in general then v ' (t) is the function representing the acceleration as a function of t).
The area is average altitude * width = (vf + v0) / 2 * `dt.
Altitudes represent velocities, so the average altitude (vf + v0) / 2 represents average velocity. Multiplying this by time interval `dt yields average velocity * time interval, which is `ds / `dt * `dt = `ds, the displacement corresponding to the interval.
Thus `ds = (v0 + vf) / 2 * `dt.
Assume uniform acceleration, so that aAve can be replaced by just plain a.
The equations we obtain from the above are thus
vf = v0 + a `dt
`ds = (vf + v0) / 2 * `dt
These equations involve the five variables v0, vf, a, `ds and `dt. These five quantities can be represented on a v vs. t trapezoid as the altitudes v0, and vf, slope a, area `ds and width `dt.
Each equation contains four of the five variables.
If for one of these equations we know the values of three of the four variables, we can solve for the value of the fourth.
There are ten possible ways to select three of the five variables, and for eight of the ten possible selections, one of the equations above will contain all three.
These eight cases can all be reasoned out from the definitions, and/or by drawing trapezoids and labeling the given information then figuring out what isn't known. Being able to use the equations does not substitute for knowing the definitions, but it's important to know how to use the equations.
There are two cases where we can't use one of these two equations to solve, and in these two cases we can't directly use the definitions or the trapezoids to easily reason out the results. These cases occur when we know only v0, a and ds, or when we know only vf, a and `ds. For these cases we obtain two additional equations. (University Physics students have already seen these equations, and have also seen how to derive them using calculus).
Next week when we apply F_net = m a to these equations, we will see where the definitions of kinetic energy, work, impulse and momentum come from. (University Physics students have already seen this).
`q001. If you roll a ball down a 30-cm ramp, from rest, and it requires 3 seconds to travel the length of the ramp, what are its average velocity, final velocity and acceleration?
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Ave Velocity is 30 cm / 3 seconds which is 10 cm per secon
Final Velocity is 20 cm per second
The aceleration is going to be
a=Vf- V0/t
which is 20/3 cm/sec^2
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For this interval which three of the quantities v0, vf, `dt, `ds and a are you given?
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ds, dt, v0
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These three quantities all appear in one of the two equations above. Which is it?
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vf = v0 + a 'dt
@& The equations are
`ds = (vf + v0) / 2 * `dt
and
vf = v0 + a `dt.
v0, `ds and `dt all appear in the first of these equations.
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There are four quantities in that equation. What is the fourth?
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acceleration
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Solve that equation for the fourth variable, in terms of the three known variables. Do this symbolically. Don't substitute and numbers until you have a symbolic solution. For example, if you were to solve the first equation for `dt (not something you would do with the present example), you would get `dt = (vf - v0) / a. Include a brief explanation of the algebra steps you used to solve the equation. Don't worry at this point if the algebra gives you a little trouble; the algebra in the General College Physics course isn't that bad, and we can remedy it if necessary. (University Physics students won't have any trouble with the algebra).
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vf = v0 +a 'dt
vf - v0 = a 'dt
vf-v0/ dt = a
a= vf- v0/dt
So, a = 20cm/sec - 0cm/sec /('3 sec)
a= 20cm/sec /3 sec or 6.7 cm/sec^2
@& vf isn't one of the given quantities, so this result doesn't follow from the values of the original three variables.
Of course you can easily reason out vf from the given information, and this is therefore a valid calculation of a.
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@& The solution for vf in terms of the equations:
The equation
`ds = (vf + v0) / 2 * `dt
is solved for vf by first multiplying both sides by 2 to get
2 `ds = (vf + v0) * `dt
then dividing both sides by `dt to get
2 `ds / `dt = vf + v0
and finally subtracting v0 from both sides to get
vf = 2 `ds / `dt - v0.
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Now substitute the values of the three known quantities into your rearranged equation, and simplify. Include units. Again, most likely not everyone at this point will be able to do this with complete success, but with practice this won't be difficult.
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I did that in the question above....
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Now suppose that the ball rolls off the end of the first ramp, right onto a second ramp of length 100 cm. If the ball requires 2 seconds to roll down this ramp, then for this new interval, which of the quantities v0, vf, `dt, a and `ds do you know? Note in particular that the initial velocity on this ramp is not zero, so for this interval v0 is not zero.
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V0 a 'ds 'dt
@& 6.7 cm/s^2 came from the preceding analysis, which was for a different ramp. So it doesn't apply here, and you don't yet know a.
What you know about the motion on this ramp:
The ball ends the first ramp at 20 cm/s, so this will be the initial velocity on the second. So for this interval v0 = 20 cm/s.
The ramp is 100 cm long so `ds = 100 cm.
The time down the ramp is 2 seconds so `dt = 2 sec.
Thus you know v0, `ds and `dt.*@
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Apply the definitions of velocity and acceleration to figure out the other two quantities for this interval.
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Well, if V0 is going to be what the final was before which is 20 cm /sec we can find the avg which will give us the vf as well.
If a is constant and it is 6.7 cm/sec^2
We have Vf = V0 +a ('dt)
Vf = 20 cm/ sec + 6.7 cm/ sec^2 (2 sec)
Vf= 26.7 cm/sec^2 (2 sec)
Vf= 53.4 cm/ sec
For Vave we take the Vf and V0 add and divide by two
20 cm/sec + 53.4 cm/sec= 73.4 cm/sec / 2 = 36.7 cm/sec for Vave
@& This would be good if your value of a was valid for this ramp.
It isn't, though. You'll understand the following:
ave velocity = ave roc of position wrt clock time = change in position / change in clock time = 100 cm / (2 sec) = 50 cm/s.
ave accel = ave roc of velocity wrt clock time = change in velocity / change in clock time.
To get the change in velocity, knowing the initial velocity, we have to find the final velocity. Initial velocity is 20 cm/s, average velocity is 50 cm/s. What do we average with 20 to get 50? The answer is 80, since (80 + 20) / 2 = 50. So the final velocity is 80 cm/s.
Now we can find the change in velocity, which is vf - v0 = 80 cm/s - 20 cm/s = 60 cm/s.
Knowing this we can find the average acceleration
aAve = `dv / `dt = 60 cm/s / (2 s) = 30 cm/s^2.
*@
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Now figure out which of the two equations can be applied to your new information. Solve that equation for that quantity, in the manner used previously, substitute your known values, and see what you get.
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I did that in the problem above
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`q002. When you coasted the toy car to rest along the tabletop, how far did it travel after your finger lost contact with it, and how long did it take to come to rest?
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60 cm in 3 seconds
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Using the definitions of velocity and acceleration, find the car's initial velocity and acceleration, assuming that acceleration to be constant.
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If the car moved 60 cm in the seconds, its initial velocity is 60cm/3sec which is 20cm/sec
V0= 0
Vave= 20cm/sec
Vf= 40cm/sec
So that:
a ave= vf-V0/ 'dt
a= 40cm/sec / 3 sec
a= 13.3 cm/ sec^2
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Assuming the mass of the car to be 10 grams, how much force was required to produce the acceleration you observed?
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F net= m*a
10g * 13.3 cm/sec^2
133 g * cm/sec^2
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How much work did this force do as the ball coasted to rest?
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work is force * distance so
133 g * cm/sec^2 * 60 cm
7980 g * cm^2/sec^2
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What was the initial kinetic energy of the car, i.e., the kinetic energy at the instant it lost contact with your finger?
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KE= 1/2 m v^2
KE= 5 g (20cm/sec) = 100 g * cm/sec
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@& You need to square that velocity.
When you do your result will be equal, or very close to, the value you got for the work in the preceding step. This should reinforce the idea that the work by the net force is equal to the change in the KE.*@
`q003. When you set up the system with the four rubber band chains, two toy cars, the strap and the axel:
As seen by someone facing the strap from the side to which the cars were attached, was the more massive car on the left or the right side of the strap?
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The right side
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While the strap was being held stationary, on which side were the rubber band tensions greater?
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The side where the more massive car was on
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Just after release, on which side did the car move away from the strap?
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the right side
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Just after release, on which side(s) was the tension in the chain connecting the car to the strap less than the tension in the chain pulling 'down' on the car? Explain your thinking.
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During the release, gravity takes over and pulls the car down towards the earth. So, when the car was just released the chain relaxes for a small amount of time which makes the tension on the chain that is connected to the axel have less tension when it is initially released.
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`q004. Assume that the mass of the car and magnet was 12 grams.
How close did you get the magnet to the car, and how far did the car then travel before coming to rest?
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1 cm from the car and it took 21 cm to come to rest in 1.5 sec
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Assuming that the car's acceleration was the same as when you coasted it across the table, how fast was it going when it started to slow down? (This is actually a complicated situation, since you don't know where the car was when the magnet's force became negligible, so just assume that this occurred about a centimeter from the car's initial position).
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V0 = 0
V ave = 21/1.5= 14 cm/sec
Vf= 28 cm/sec
a= 28 cm/sec - 0 /dt
28/1.5 sec= 18.7 cm/sec^2
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@& The car ended up at rest. When it started to slow down the velocity was 28 cm/s, so for that interval vf = 0 and v0 = 28 cm/s.
The magnitude of the acceleration is the same either way, so that's OK.*@
What was the force on the coasting car?
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F net= 12g * 18.7 cm/sec^2 = 224 g* cm/sec^2
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How much work did this force do on the car?
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force * ds
224 g * cm/sec^2 * 21 cm= 4704 g * cm^2/sec^2
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How much kinetic energy do you conclude the car gained from the interaction of the two magnets?
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KE= 1/2 m v^2
KE= 6g (14 cm/sec)^2
KE= 1176 g* cm^2/sec^2
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@& Use the car's initial velocity to calculate its KE.
It's worth remembering that we just about never base a KE calculation on the average velocity.
If you base this on the 28 cm/s result, you'll get just about the same thing you did for the work, which will confirm that the work done by the net force is equal to the change in KE.*@
What was the momentum of the car just before it started to slow down?
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momentum = m * V
p= 12 g * 28cm/sec = 336 g * cm/sec
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University Physics Only:
`q005. What were the coordinates of the two points of the F vs. L graph you sketched and labeled?
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What therefore was the average slope of the graph between your two labeled points?
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What would be the area beneath your graph, between the two labeled points? You may assume that the graph between these points is a straight line.
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What is the meaning of the rise and the run between your two points.
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What is the meaning of the average 'graph altitude' between your two points?
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What therefore is the meaning of the slope of your trapezoid?
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What is the meaning of the area of your trapezoid?
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`q006. If the graph of F(L) vs. L, for a certain rubber band, is a straight line with equation F(L) = .03 Newtons / cm * (L - 130 cm), then:
What is the slope of the graph and what does this quantity tell you?
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What is the derivative of the function F(L) with respect to L and what does this quantity tell you?
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What is the area of the graph between L = 130 cm and L = 140 cm?
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What is the area of the graph between L = 140 cm and L = 150 cm?
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What do these two areas tell you about the rubber band?
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What is the area of the graph between L = 130 cm and L = 150 cm?
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`q007. For the bottle on the rubber band chain, the mass is about 0.6 kg and the frequency of its oscillations is about 1.6 cycles / second. A cycle is 2 pi radians.
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What is the angular frequency omega of the oscillation, in radians / second?
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If omega = sqrt( k / m ), then what is the value of k for the rubber band chain?
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If x is the position of the bottle with respect to its equilibrium posiiton, then the net force on the bottle at position x has magnitude || F_net || = k || x ||. The relative signs of F_net and x are important and relevant, but we won't worry about that right now. What is the area under the graph of || F || vs. || x || between x = 0 and x = 2 centimeters?
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The above is the total energy stored in this oscillation when its amplitude is 2 centimeters. What would be the total energy if the amplitude was 4 centimeters?
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What is the expression for the total energy if the amplitude is A centimeters?
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"
@& You've got a few errors in detail, so be sure to check out my notes.
However you're doing really well here. You'll understand my notes.*@