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course Phy 241
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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30cm/3 s= 10 cm/s avg velocity therefore final velocity is 20 cm/s then the acceleration is equal to 20 cm/s /3s = 6.67 cm/s^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=30cm , dt= 3 s, v0= 0
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These three quantities all appear in one of the two equations above. Which is it?
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ds= (vf+v0)/2 *dt
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There are four quantities in that equation. What is the fourth?
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The final velocity vf
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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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ds=(vf+v0)/2 *dt
2ds=(vf+v0)*dt
2ds/dt-v0=vf
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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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2(30cm)/3s -0cm/s =20cm/s
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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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I know vf which is 20cm/s and dt=2s and ds=100cm
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@& For the second ramp that would be v0.*@
Apply the definitions of velocity and acceleration to figure out the other two quantities for this interval.
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vf=2ds/dt -v0, vf=29100cm)/2s -20cm/s = 100cm/s -20cm/s = 80cm/s as a final velocity so the average is 80cm/s-20cm/s = 60cm/s then average is 30cm/s so the acceleration is 80cm/s /2 s= 40 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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well answered this question in the previous question
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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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it went around 15 cm in about 1.5 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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ds=(v0+vf)/2*dt plug in what I know, 15cm=(v0+0)/2*1.5s
v0=15cm(2)/1.5s-0=20 cm/s
average velocity = 20cm/s+0cm/s/2 =10cm/s
a= 10cm/s/1.5s=6.67cm/s^2
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@& Acceleration is not ave vel / time interval.
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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=ma
F= 10g *6.67cm/s^2=66.7 g*cm/s^2
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How much work did this force do as the ball coasted to rest?
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work=66.7g*cm/s^2*15cm=1000.5g*cm^2/s^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=1/2m*v^2, KE= ½(10 grams)(20cm/s)^2= 2000g*cm^2/s^2
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@& Note that the work you've calculated is not equal to the change in KE. Since the force you calculated was m * a, that was the net force. The work by the net force must be equal to the change in KE.
The problem goes back to your calculation of acceleration.
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`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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left
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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 tension was greater on the side with the car that had the greater mass the right side
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Just after release, on which side did the car move away from the strap?
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the side with the larger car which was 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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the tension on the right side was greater because the car was larger when we held the two cars at equal lengths away from the strap. while the lighter car on the left had less tension acting on it.
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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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we was able to get the magnet 2cm away from the car and it traveled 22 cm
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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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I used the older acceleration which was 6.67cm/s^2, then one cm from the initial position means it went from 21cm to 0 cm.
I used ds=(v0+vf)/2*dt, 21cm =(v0+0)/2*1.5s, v0=21cm(2)/1.5s-0=28cm/s
or you could use vf=v0+a dt
0=v0+6.67cm/s^2 *1.8s
v0= 12cm/s
average velocity is 6cm/s
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What was the force on the coasting car?
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the force was anything that was slowing the car down which means it could have been friction and the weight of the car.
F=ma, F=12grams*6.67cm/s^2=80g*cm/s^2
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How much work did this force do on the car?
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work= 80 g*cm/s^2*22cm=1760g*cm^2/s^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=1/2mv^2=1/2*12g(6cm/s)^2 =216g*cm^2/s^2
@& Kinetic energy calculated based on on average velocity is almost never a useful or relevant quantity.
Change in KE is equal and opposite to work done by net force.
According to your calculations the net force would do -1760 g cm^2 / s^2 of work on the car.
However you are using an acceleration based on an incorrect calculation (see previous notes). Using a correct acceleration and the resulting calculated v0, your result for KE would be more consistent with your result for the work done by the net force.*@
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What was the momentum of the car just before it started to slow down?
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momentum=mv=12g(12cm/s)=144g*cm/s
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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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(1cm,80g*cm/s^2) and (22cm,0 g*cm/s^2)
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What therefore was the average slope of the graph between your two labeled points?
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(0-80g*cm/s^2)/ (22cm-1cm)= 80g*cm/s^2/21cm=3.81 g/s^2
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@& The weight of a 1-gram mass would be 980 g cm/s^2.
Are you sure your forces are calculated correctly? I suspect you might be missing the multiplication by the acceleration of gravity.*@
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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A=½(b)(h)=1/2(21cm)(80g*cm/s^2)=840g*cm^2/s^2
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What is the meaning of the rise and the run between your two points.
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rise is the change in force
run is the change in distance
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What is the meaning of the average 'graph altitude' between your two points?
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this the average force applied
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What therefore is the meaning of the slope of your trapezoid?
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average force applied per the length of traveled
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@& A graph of force vs. length would be a graph of the tension in a rubber band chain vs. the length of the chain. That graph doesn't include a distance traveled.*@
What is the meaning of the area of your trapezoid?
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represents how much work was done
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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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I think the slope would be .03L-3.9
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@& F would be .03 L - 3.9.
That's not the slope of the F vs. L graph.
If necessary find F for two different values of L, which would allow you to plot two points on the graph. You could then find the slope between those points.
Having done so it might become clear to you why you didn't need to actually find two points. The slope can be read directly from the equation.*@
What is the derivative of the function F(L) with respect to L and what does this quantity tell you?
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I think the derivative is .03L/dL
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@& The derivative of .03 L - .39, with respect to L, is just .03.*@
What is the area of the graph between L = 130 cm and L = 140 cm?
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area= 1/2 (140cm+130cm)*10cm=1350cm^2
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@& Plug in 130 cm and 140 cm for L, then plot the two F vs. L points on the graph.
The region defined by those points is a trapezoid. You need to find its area, and interpret it.*@
What is the area of the graph between L = 140 cm and L = 150 cm?
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area=1/2(140cm+150cm)*10cm=1450cm^2
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What do these two areas tell you about the rubber band?
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the two areas show uniform resistance
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What is the area of the graph between L = 130 cm and L = 150 cm?
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area=1/2(130cm+150cm)*10cm=1400cm^2
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@& You need to recalculate and reinterpret all these areas.*@
`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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w(omega)=2pi rad/1.6 cycles/seconds=3.93 rad* sec/cycles
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@& Right idea, but it's
) 2 pi rad / cycle ) / (1.6 sec / cycle) = 3.93 rad / sec.*@
If omega = sqrt( k / m ), then what is the value of k for the rubber band chain?
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3.93rad*sec/cycles=sqrt(k/.06kg), (3.92 rad*sec/cycles )^2*.6kg=k ,k=9.26 kg*rad^2*sec^2/cycles^2
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@& Good.*@
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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total displacement would be 4 cm
then 9.26 kg*rad^2*sec^2/cycles^2*4cm=37.04 cm*kg*rad^2*sec^2/cycles
you can find the area by a=1/2*4cm*37.04 cm*kg*rad^2*sec^2/cycles=74.08 cm^2*kg*rad^2*sec^2/cycles
******the units are very strange I might have done something wrong
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@& Units need some work, but the idea is correct.
However from x = 0 to x = 2 cm the displacement is 2 cm.
And the force at the 2 cm position is the max force. The force at the 0 cm position is the min force.
To get the work you multiply ave force by displacement.*@
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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½(8cm)( 74.08 cm^2*kg*rad^2*sec^2/cycles)=298.32 cm^3*kg*rad^2*sec^2/cycles
^^^^^^^^^^^^^^^^^^yet again im unsure about the units
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What is the expression for the total energy if the amplitude is A centimeters?
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area=1/2(A)(k*A)
I dont think this is correct because work does correspond with area
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@& This is right.
1/2 k A * A can also be written 1/2 k A^2. Either way is fine right now.*@
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