course phy 201
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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1: 10.5 cm
2: 9 cm
3: 8 cm
4: 7 cm
5: 6 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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1: distance= 10.5 cm mass of ramp= 10.8 grams
2: distance= 9 cm mass= 17.14 grams
3: distance= 8 cm mass= 21.8 grams
4: distance= 7 cm mass= 24.3 grams
5: distance= 6 cm mass= 25 grams
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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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I already knew the distance of the dominoes from the ramp, the mass of the dominoes, the torque, and the distance of the end of the ramp to the center.
First I multipied the r_stack*m_stack: 0= -8cm(60g) + 22cm(m_ramp), 0= -480 + 22cm(m_ramp)
then I added that to both sides of the equation: 480= 22cm(m_ramp)
and finally I divided both sides by 22 cm: 480cmg/22cm = m_ramp
m_ramp= 21.8 grams
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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 sources of uncertaint are the distance between the dominos and the center, and the actual mass of the dominos.
the 1 domino stack had the least amount of uncertainty, and the 5 domino stack had the greatest uncertainty because as the number of dominos you use increases so does the amount of uncertainty.
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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 / (4 pi^2) ).
`q002. What are the radius and velocity of a geosychronous orbit?
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V= srt(G*m_earth/ r)
r= cubedrt(G*m_earth*'dt/ 4(pi)^2)
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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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KE= 1/2 mass*V^2 and PE= -G*M*m/r^2
Just plug V into the KE formula and r into the PE formula then add your answers together.
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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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F_gravity= 2.92 *10^20 N
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What therefore is the acceleration of the Moon toward the Earth?
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F_cent= F_grav
(2.92 * 10^20 N)/(1 * 10^23kg)= a_cent
a_cent= .0029 m/s^2
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What is the acceleration of the Earth toward the Moon?
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I'm not sure
equal but opposite force acts on the Earth; you know the mass of the Earth
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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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r= 2.76 * 10^16 m
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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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centripital acceleration
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How much work would gravity do on the Moon during this interval?
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F_grav= 3.08 * 10^26 N * 10,000,000m=
(I may have made a mistake)
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By approximately how much would its gravitational PE change as a result?
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'dPE= -6,265,633,100 J
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By how much would its PE relative to infinity change?
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PE= -1.1 * 10^32 j
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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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I don't know
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What is the Moon's moment of inertia in its orbit around the Earth?
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I= 1.369 * 10^34 mN
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What are the angular velocity and angular momentum of the Moon's orbit?
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V=sqrt(GM/r)= G(6*10^24kg)/370,000m= 32,888 m/s
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What is the KE of the Moon's orbit?
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KE= 1/2mV^2= .5 * (1*10^23 kg) * (32,888 m/s)= 5.4*10^31 Joules
I got all the way down here and just realized I have been doing my math wrong. I don't have time to fix it so, Sorry.
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I think you've got inconsistencies in your units; can't be sure but you're probably using most of the right relationships.
`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= 7561 m/s
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What are the KE and PE of this orbit?
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ok I don't know how to find the mass so I'm going to make one up: the mass of the satelite is 40 Kg.
KE= 1/2 * 40kg * (7561 m/s)^2= 1.143 bilion Jueles
PE= -GMm/r= (6.67*10^-11)(6*10^24 kg)(40 kg)/(7000000 m)= 2.286 billion Jueles
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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=sqrt[(6.67*10^-11 G)(6*10^24 kg)/(7,001,000 m)] = 7560 m/s
KE_2= 1/2 * 40kg * (7560 m/s)^2=1,143,072,000 jueles
PE_2= (6.67*10^-11)(6*10^24 kg)(40 kg)/(7001000 m)=
'dKE= (1,143,374,420 Jueles) - (1,143,072,000 jueles)= 302,420 Joules
'dPE= (2,286,857,143 Joules) - (2,286,530,496 Joules)= 326,646 Joules
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How much work must have therefore be done on the system by nonconservative forces?
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'dw_nc_on= 302,420 Joules + 326,646 Joules = 629,066 Joules
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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
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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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`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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Its a circle with a line, r= 100 cm, going away from the center at a 45 degree angle. At the end of the line there is the velocity vector that is perpindicular to the first line, V= 50 cm/s, its angle is 135 degrees. The a_cent. is going 25 cm/s^2 in the opposite direction of the first line at an angle of 225 degrees.
From that point I drew another auxiliary axes where the velocity, a_cent., and radial meet. The angles I mentioned earlier for the Velocity and a_cent are all from this point.
Also I have vectors for the radial
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Estimate the x component of each vector. Give your estimates below.
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xcomponent:
radial= 60
velocity= -33
a_cent= -17
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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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calculations:
xcomponent:
radial= 70
velocity= -35
a_cent= -17 (wow right on the money)
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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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xcomponent at 45 degrees:
radial= 70
xcomponent at 90 degrees:
radial= 0
velocity= -50
a_cent= 0
xcomponent at 135 degrees:
radial= -70
velocity= -35
a_cent= 17
xcomponent at 180 degrees:
radial= -100
velocity= 0
a_cent= 25
xcomponent at 225 degrees:
radial= -70
velocity= 35
a_cent= 17
xcomponent at 270 degrees:
radial= 0
velocity= 50
a_cent= 0
xcomponent at 315 degrees:
radial= 70
velocity= 35
a_cent= -17
xcomponent at 360 degrees:
radial= 100
velocity= 0
a_cent= -25
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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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xcomponent at 45 degrees:
velocity= -35
xcomponent at 90 degrees:
velocity= -50
xcomponent at 135 degrees:
velocity= -35
xcomponent at 180 degrees:
velocity= 0
xcomponent at 225 degrees:
velocity= 35
xcomponent at 270 degrees:
velocity= 50
xcomponent at 315 degrees:
velocity= 35
xcomponent at 360 degrees:
velocity= 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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xcomponent at 45 degrees:
a_cent= -17
xcomponent at 90 degrees:
a_cent= 0
xcomponent at 135 degrees:
a_cent= 17
xcomponent at 180 degrees:
a_cent= 25
xcomponent at 225 degrees:
a_cent= 17
xcomponent at 270 degrees:
a_cent= 0
xcomponent at 315 degrees:
a_cent= -17
xcomponent at 360 degrees:
a_cent= -25
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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
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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
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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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No
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Good overall, but be sure to see the notes I sent by email. I used email because the file contains some graphics.