course 201
11/22 9 pm
Class 091118Think about the following:
Imagine you're in a car going counterclockwise around a circular path. Is the center of the circle to your right, your left, behind you, in front of you, or none of these? If you slam on your brakes, it what direction do you observe that coffee cup on the seat next to you move? Relative to the circle, in what direction does it move? In what direction does the road surface push the car?
Experiment
Balance a 31-cm ramp (which you should measure accurately) on the edge of a domino, with a stack of dominoes at one end of the ramp. Observe the distance between the balancing point and the nearest edge of the stack.
Repeat of stacks of 1, 2, 3, 4, 5, ... dominoes.
`q001.
For each number of dominoes, give your raw data:
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Ramp length 30.3 cm
1 domino, 10.5 cm
2 dominos, 9 cm
3 dominos, 7.5 cm
4 dominos, 6.8 cm
5 dominos, 5.8 cm
6 dominos, 5 cm
7 dominos, 4.5 cm
8 dominos, 4 cm
9 dominos, 3.5 cm
10 dominos, 3.3 cm
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For each stack, find the distance from the fulcrum (i.e., the domino on which the system is balanced) to the center of the ramp.
The width of a domino is 2.5 cm, length is 5 cm.
Assuming each domino has a mass of 20 grams, find the torque exerted by the stack.
Assuming the full weight of the ramp acts downward from the center of the ramp, use the fact that net torque is zero to find its mass.
For each number of dominoes, give the distance of the middle of the ramp from the fulcrum, then the mass you calculated for the ramp. Use one line for each number of dominoes.
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TauNet=0, TauNet=mstack(g)(-rstack)+mramp(g)(rramp)
For one domino:
0=.2 kg(9.8 kgm/s^2)(-.1175 m)+(.303 m-.1175 m)(mramp)(9.8 kgm/s^2)
.2303 kg^2 m^2/s^2=1.82 kgm^2/s^2 (mramp)
Mramp=.127 kg
1, 11.75 cm, .127 kg
2 , 10.25 cm, .204 kg
3 , 8.75 cm, .244 kg
4 , 8.05 cm, .289 kg
5 , 7.05 cm, .303 kg
6 , 6.25 cm, .312 kg
7 , 5.75 cm, .328 kg
8 ,5.25 cm, .335 kg
9 ,4.75 cm, .335 kg
10 , 4.55 cm, .353 kg
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Explain in detail how you calculated the mass of the ramp for the stack containing 3 dominoes
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TauNet=0, TauNet=mstack(g)(-rstack)+mramp(g)(rramp)
So 0=(.6 kg)(9.8 kgm/s^2)(-.0875 m)+mramp(9.8kgm/s^2)(.303 m-.0875 m)
0=-.5145 kg^2m^2/s^2 +mramp(2.1119 kgm^2/s^2)
. 5145 kg^2m^2/s^2=mramp(2.1119 kgm^2/s^2)
Mramp=.244 kg
Well done, but the distance from fulcrum to the center of mass of the rod is only about 5 cm, not (.303 m - 0.875 m), from the balancing point. The balancing point is 10 cm from one end (7.5 cm + 2.5 cm = 10 cm), hence 20 cm from the other and 5 cm from the center of the rod.
Your calculation should end up with ramp mass in the neighborhood of 6 times that of a domino, or around .12 kg.
The masses that result from your data range from about .12 kg to .9 kg, decreasing as you go down the column (.12 kg for the 1-domino data, .09 kg for the 10-domino data). Given the width of the domino you used to balance on, these results are good and consistent. I would expect the least error for the first five measurements, with a ramp mass between .11 kg and .12 kg.
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What are the sources of uncertainty in this experiment? For which stack do you think the uncertainty in your determination of the mass was least, and what is the basis for your answer?
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The exact weight of the domino stack, as not all of the dominos were from the same set
I feel like my stack of 8 dominos may have the closest to expected value for the weight of the ramp because it is most consistent with the values above and below it. At that point we were not moving the ramp much with each additional domino, so little weight was added to the side of the ramp.
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Geosychronous satellite
A geosychronous satellite orbits the Earth once per day. The period of the orbit is therefore 24 hours.
We want to find the radius r of this orbit.
Recall that by setting centripetal force equal to gravitational force we find the velocity of a circular orbit of radius r is
v = sqrt( G * M_earth / r).
The velocity of an orbit is also v = `ds / `dt, where `ds is the distance around the orbit and `dt the time requires to complete an orbit. The circumference of the orbit is 2 pi r so
v = 2 pi r / period.
We therefore have two equations in the unknowns v and r. We can easily solve the equations simultaneously to determine r. Setting our two expressions for v equal, we eliminate v and obtain the equation
sqrt( G * M_earth / r) = 2 pi r / period
which we easily solve for r, obtaining
r = ( G * M_earth * period^2 / (4 pi^2) ) ^ (1/3).
`q002. What are the radius and velocity of a geosychronous orbit?
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Velocity= sqrt(G*mearth/r)
Radius=2*pi*r
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Don't do the calculation, but explain how you would use your information to calculate the total mechanical energy (total mechanical energy = PE + KE) of the orbit.
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In orbit there are no Nonconservative forces, so `dKE=-`dPE
KE=.5m*v^2
You have only one orbit so there is no `dKE or `dPE involved. PE = - g M m / r^2.
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Getting used to negative PE
Let's assume that everyone in this discussion has the same mass.
If you are at the bottom of the well and I am at the opening to the well, you will easily come to the conclusion that my PE is greater than yours, since you would have to do work against gravity to get to my position.
You might choose to think that your PE is zero and that mine is positive.
I might choose to think that my PE, since I'm on the surface, is zero and yours is negative.
Either of us would be right, but without a common reference point we would come to very different conclusions about PE.
For example a guy halfway down the well would have negative PE according to my reference point, and positive according to yours. We might agree on how much PE he would lose if he fell to your position, or how much he would gain if he climbed up to mine, but if we want to describe his PE in terms of a single number, we won't be able to do it.
Suppose we agree to use my perspective: PE is zero at the surface of the Earth.
Then the guy halfway down has negative PE, and you have even more negative PE. Both PE's are negative, but yours has the greater magnitude. At the same time your PE is less than his; since your negative number has greater magnitude, it's further 'below zero' than his, so it's less.
Just as we agreed to use the surface of the Earth as the reference point for the example of the well, we need to agree on a reference point for orbital PE. In this case our reference point is 'far, far away'. In fact it's the limit of 'far, far away', an infinite distance.
If we agree to measure PE relative to infinity, then we use the formula PE = - G M m / r. This formula takes care of the fact that g keeps changing (which makes m g h irrelevant since you don't know what value to use for g).
Using this formula everything has negative potential energy. The smaller the value of r, the greater the magnitude of the PE, so things closer to the center of the Earth have a negative PE which less than (but of greater magnitude than) the PE they would have if they were further.
This is analogous to the example of the well. If you're on the surface of the Earth you now have a big negative PE; if you're in orbit halfway to the Moon you have a negative PE but not as big as the one you had on the surface; and to go from a bigger negative PE to a smaller negative PE, your PE has to increase. Put another way, your PE at the surface is 'way below zero'; halfway to the Moon it's still below zero, but not as far below, and to get there your PE must increase.
`q003. Earth's mass is about 6 * 10^24 kg. The Moon's mass is about 1/60 times the mass of the Earth. The two are separated by about 370 000 km.
What is the gravitational force exerted on the Moon by the Earth?
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Fgrav=G*mearth*mmoon/r^2
Fgrav=6.67*10^-11 Nm^2/kg^2*(6*10^24 kg)*(1*10^23 kg)/(3.7*10^8 m)^2
Fgrav=1.08*10^21 N
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What therefore is the acceleration of the Moon toward the Earth?
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Force=mass*acceleration
1.08*10^21 N=(1*10^23 kg)*a
A=1.1 *1044 m/s^2
You made an error with your exponents. The solution to your equation would be a = 1.08 * 10^21 N / (1 * 10^23 kg) = 1.08 * 10^-2 m/s^2, about .01 m/s^2.
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What is the acceleration of the Earth toward the Moon?
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F=m*a
1.08*10^21 N=(6*10^24 kg)*a
A=1.8 *10^44 m/s^2
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What is the radius of a circle if a point traveling around the circle once every 28 days has the acceleration you calculated in the preceding exercise?
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I am unsure of how to solve this one
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If the Moon were to move 10 000 km closer to the Earth, what average force would act on it during this interval?
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Fgrav=G*mearth*mmoon/r^2
Fgrav=6.67*10^-11 Nm^2/kg^2*(6*10^24 kg)*(1*10^23 kg)/(3.6*10^8 m)^2
Fgrav=1.11*10^21 N
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How much work would gravity do on the Moon during this interval?
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Work=F*`ds=(9.8 kgm/s^2)(10000 km)=9.8 *10^7 kgm^2/s^2
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By approximately how much would its gravitational PE change as a result?
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PE=-work
PE=-9.8 *10^7 kgm^2/s^2
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By how much would its PE relative to infinity change?
^^^^
PE=-G*M*m/r
PE=-(6.67*10^-11 Nm^2/kg^2)(6*10^24 kg)(1*10^23 kg)/3.6 *10^8 m
PE=-1.11 *10^45 Nm
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How much energy would it take to move the Moon to a very great distance from the Earth?
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Im not sure which formulas I should use for this problem
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What is the Moon's moment of inertia in its orbit around the Earth?
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Moi=sum(mr^2)
Moi=(1*10^23 kg)(3.7 *10^8 m)^2
Moi=3.7 *10^39 kgm^2
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What are the angular velocity and angular momentum of the Moon's orbit?
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Velocity=circumference/period=2*pi*3.7*10^8 m/28 days= 2pi*3.7*10^8 m/2419200 seconds= 7.17*10^12 m/sec
momentum=moi*v= 3.7 *10^39 kgm^2*7.17*10^12 m/sec=2.65 *10^52 kgm^3/sec
Should I have used the omega=`dtheta/`ds formula here? I wasnt sure
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What is the KE of the Moon's orbit?
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KE=.5m*v^2=.5*m*v^2=.5*1*10^23 kg*(7.17*10^12 m/s)^2=2.57 kgm^2/s^2
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`q004. What is the velocity of a satellite in a low-Earth orbit at a distance of 7000 km from the center of the Earth?
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V=sqrt(G*mearth/r)
V=sqrt(6.67*10^-11 Nm^2/kg^2 *6*10^24 kg/7*10^6 m)=sqrt(5.72*10^19 N m/kg)=7561179046 m/s
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What are the KE and PE of this orbit?
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KE=.5mv^2=.5(6*10^24 kg)* 7561179046 m/s^2 =1.7 *10^44 kg m^2/s^2
Since no nonconservative forces, KE=-PE
PE=-1.7 *10^44 kg m^2/s^2
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If a satellite changes from a circular orbit at 7000 km from the center of the Earth, to a circular orbit at 7001 km, what is `dKE between the orbits and what is `dPE?
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V= V=sqrt(6.67*10^-11 Nm^2/kg^2 *6*10^24 kg/7001000 m)=7560.6 kg m/s
KE=.5m*v^2=.5(6*10^24 kg)( 7560.6 kg m/s)^2=1.7 *10^32 kgm^2/s^2
PE=-1.7 *10^32 kgm^2/s^2
`dKE=1.7 *10^32 kgm^2/s^2- 1.7 *10^44 kg m^2/s^2=-1.7 *10^44 kgm^2/s^2
`dPE=-1.7 *10^32 kgm^2/s^2- -1.7 *10^44 kg m^2/s^2=1.7 *10^44 kgm^2/s^2
These figures dont seem right, did I do something wrong here?
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How much work must have therefore be done on the system by nonconservative forces?
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Nonconservative force=`dKE=`dPE=-1.7 *10^44 kgm^2/s^2+1.7 *10^44 kgm^2/s^2=0
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`q005. The Moon's orbit around the Earth is actually elliptical, with its closest approach occurring around 360 000 km and its furthest distance around 380 000 km.
No significant nonconservative forces act on the Moon in its orbit.
Does the Moon gain or lose KE as it moves from the 380 000 km distance to the 360 000 km distance?
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It gains KE as it gets closer to the Earth
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How much KE does it gain or lose? (hint: you should start by calculating the Moon's PE at both points; you should not start by assuming that the Moon moves between two circular orbits with radii 380 000 km and 360 000 km)
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V38=sqrt(G*mearth/r)=sqrt(6.67*10^-11 N m^2/kg^2*6*10^24 kg/3.8*10^8 m)=1.0*10^11 m/s
KE38=.5*m*v^2=.5*6*10^24 kg*1*10^11 m/s^2=3.2 *10^46 kg m^2/s^2
V36= sqrt(G*mearth/r)=sqrt(6.67*10^-11 N m^2/kg^2*6*10^24 kg/3.6*10^8 m)=1.1*10^11 m/s
KE36= 5*m*v^2=.5*6*10^24 kg*1.1*10^11 m/s^2=3.3 *10^46 kg m^2/s^2
`dKE=KE36-KE38=3.3 *10^46 kg m^2/s^2-3.2 *10^46 kg m^2/s^2=1*10^45 kg m^2/s^2
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`q006. An object moves around a circle with radius r = 100 cm at velocity v = 50 cm/s and centripetal acceleration a_cent = 25 cm/s^2.
Sketch the radial, velocity and centripetal acceleration vectors corresponding to angular position theta = 45 degrees. Sketch the x component of the radial vector.
Sketch auxiliary axes to determine the directions of the velocity and centripetal acceleration vectors, and using the auxiliary axes sketch the x components of each of these vectors.
Describe your sketch in detail:
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There is a circle with angle of 45 degrees extending 100 cm to meet its circumference. At this point two vectors originate: the acceleration vector which follows the radial vector back towards the center of the circle, and the velocity vector which extends tangent to the circle.
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Estimate the x component of each vector. Give your estimates below.
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Rx about 80 cm
Ax about -18 cm/s^2
Vx about -45 cm/s
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Using sines and/or cosines as appropriate, calculate the x component of each vector.
Give your calculations below
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Rx=100 cm(cos 45)=70.7 cm
Ax=25 cm/^2(cos 225)= -17.7 cm/s^2
Vx=50 cm/s (cos 135)=35.4 cm/s
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Repeat for angles theta = 90 deg, 135 deg, 180 deg, 225 deg, 270 deg, 315 deg and 360 deg.
List the x components of the radial vector for angles 45 deg, 90 deg, 135 deg, 180 deg, 225 deg, 270 deg, 315 deg and 360 deg.
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Rx 45=100 cm(cos45)=70.7 cm
Rx 90=100 cm(cos90)=0
Rx 135=100 cm(cos 135)=-70.7 cm
Rx 180= 100 cm(cos180)=-100 cm
Rx 225=100 cm(cos225)=-70.7 cm
Rx 270=100 cm(cos 270)=0
Rx 315= 100 cm(cos315)=70.7 cm
Rx360= 100 cm(cos360)= 100 cm
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List the x components of the velocity vector for angles 45 deg, 90 deg, 135 deg, 180 deg, 225 deg, 270 deg, 315 deg and 360 deg.
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Vx45=50 cm/s(cos135)=-35.4 cm/s
Vx90=50 cm/s(cos180)=-50 cm/s
Vx135=50 cm/s(cos225)=-35.4 cm/s
Vx180=50 cm/s(cos270)=0
Vx225=50 cm/s(cos315)=35.4 cm/s
Vx270=50 cm/s(cos360)=50 cm/s
Vx315=50 cm/s(cos45)=35.4 cm/s
Vx360=50 cm/s(cos90)=0
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List the x components of the centripetal acceleration vector for angles 45 deg, 90 deg, 135 deg, 180 deg, 225 deg, 270 deg, 315 deg and 360 deg.
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ax45=25 cm/s^2(cos225)=-17.7 cm/s^2
ax90=25 cm/s^2(cos270)=0
ax135=25 cm/s^2(cos315)=17.7 cm/s^2
ax180=25 cm/s^2(cos360)=25 cm/s^2
ax225=25 cm/s^2(cos45)=17.7 cm/s^2
ax270=25 cm/s^2(cos90)=0
ax315=25 cm/s^2(cos135)=-17.7 cm/s^2
ax360=25 cm/s^2(cos180)=-25 cm/s^2
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Can the x component of the position vector have the same sign as the x component of the velocity vector?
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Yes it can
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Can the x component of the position vector have the same sign as the x component of the centripetal acceleration vector?
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No, except in the cases where they are both zero
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Can the x component of the centripetal acceleration vector have the same sign as the x component of the velocity vector?
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yes
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Homework:
Your label for this assignment:
ic_class_091118
Copy and paste this label into the form.
Answer the questions posed above.
You have already seen most of the ideas in the qa's and Introductory Problem Set mentioned below. If you work through these documents as assigned, you will get plenty of practice and should develop good expertise with these concepts.
No new qa's are assigned. Try to catch up.
You should familiarize yourself with text chapters:
chapter 5 on circular motion an gravitation
chapter 8 on rotational motion
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_28.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_31.htm
See my notes, and see the file I sent you by email. Couldn't append it; it contains graphics.