class 091207

Class 091207

Suggested approach to problems:

First list the names, symbols, units and values of the quantities you are given. This helps get your thinking into the context of the situation. Be sure you're using the right words, the right symbols, the right units. The better you do with this, the more clearly you are likely to think about the situation.

Unless it's obvious don't even think about what the question asks you to figure out. Think first about what you can do with the information you have, rather than confining your thinking to a specific goal. Ask yourself what you can figure out from the given information. Use sketches, diagrams, words, etc..

Once you know what you can do with the information, it's time to focus yourself on the goal. This is the time to start thinking about what the question wants you to figure out.

Orbital Mechanics

`q001. Answer the following:

What quantities are associated with a satellite in circular orbit around a planet?

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Given the mass, velocity and orbital radius of a satellite in a circular orbit, what other quantities can we find, and what specific reasoning process do we use?

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If we give a quick impulse to a satellite in circular orbit, in the direction of its motion, what can we predict will happen to the shape of the orbit, and to the kinetic, potential and total energies?

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Suppose a satellite is in an elliptical orbit. What difference does it make to the shape of the new orbit if a given impulse is delivered at the perigee vs. at the apogee? What difference does it make to the total energy of the new orbit?

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Which has greater total energy, a circular orbit or an elliptical orbit which remains inside the circular orbit except at two 'extreme points' where it just touches the circular orbit?

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Test Problems

The following worked problems are from Physics 121 Test 2 and Physics 121 final exam.

Problem Number 2

During a 6.2 -second time interval, a disk is observed to accelerate from 9.699 radians/second to 12.39 radians/second.

If the net torque on the disk is known to have been 6 meter Newtons, what must have been the moment of inertia of the disk?

The first sentence tells us the following:

given time interval is `dt = 6.2 sec
initial angular velocity is omega_0 = 9.7 rad/s
final angular velocity is omega_f = 12.4 rad/s.

The second sentence also provides information, but we can determine a whole lot from the three quantities given in the first sentence so we stop here and sketch out what we know we can find:

We're given time interval and initial and final angular velocities. These constitute three of the five variables (`dt, `dTheta, omega_0, omega_f and alpha) in terms of which we can analyze uniform acceleration, so we know we can find alpha and `dTheta.
As it turns out we can easily use the definitions to reason out these quantities, and in the process we will also reason out omega_Ave and `dOmega.
We could alternatively use equations, which are valid but unnecessary in this case.

Now the second sentence gives us net torque, which is tau_net = 6 m N. What can we now find?

We know we can find alpha, `dTheta, `dOmega and omega_ave from the information in the first sentence alone.
What do we know about net torque?

Net torque is tau_net = I * alpha.

The torque of a force exerted at a point is tau = r X F

tau_net = sum(tau_i) (we get net torque by adding all the individual torques)

What is relevant to the current situation?

We know alpha and tau_net so we can easily find I.

Now would be a good time to ask what the problem wants us to find.

The problem asks us to find I. So we proceed as follows:

From omega_0, omega_f and `dt we find the acceleration alpha. We get alpha = (omega_f - omega_0) / `dt.

From tau_net = I * alpha we get I = tau_net / alpha.

Evaluating alpha we have alpha = (12.4 rad/s - 9.7 rad/s) / (6.2 s) = (2.7 rad/s) / (6.2 s) = .43 rad/s^2.

Evaluating I we get I = 6 m N / (.43 rad/s^2) = 14 kg m^2.

note: be sure you do the units calculations correctly; these are illustrated in detail in earlier class notes.

Are there aspects of this problem that you could usefully visualize using graphs?

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A graph of omega vs. t could be useful in determining the angular acceleration and angular displacement, which are respectively the slope of and area beneath the omega vs. t trapezoid.****

`What definitions apply to this situation, and what are the names and symbols for the quantities are involved?

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The definitions of angular velocity and angular acceleration can be used to infer the angular acceleration and angular displacement.

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`Does this problem involve any of the formulas you have been advised to memorize? What relevant formulas result from definitions and basic laws?

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tau_net = I * alpha is the rotational analog to F_net = m a. It is important that you know both forms of Newton's Second Law.

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`What situations seem similar to this, but perhaps not exactly the same? What concepts, ideas and procedures have a similar 'feel'?

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The motion 'feels' a lot like linear motion; the behavior of torque and moment of inertia are analogous to the behavior of net force and mass.

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`What is the nature of the reasoning you use when working through this problem?

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Direct reasoning from definitions is recommended, where it works, in solving problems where acceleration or angular acceleration is uniform.

It is recommended that you use the strategy of listing what is known and what can be determined from what is known.

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`How have the quantities and relationships used in working out this problem been directly experienced in experiments and other hands-on activities?

Problem Number 3

A tower rises from its base at the surface of the Earth, a distance of 6400 km from the center, and rises an altitude of 900 kilometers above the surface. An individual of mass 87 kg wishes to climb the tower.

How much would the person weigh at the bottom and at the top of this tower?
If the average weight of the person during the climb was equal to the average of the two weights just found, and if the person was capable of an average sustained power output of .99 watt/kg for 8 hours per day, then how many days would be required for the climb?

We are given the mass of the individual, his initial distance from the center of the Earth and the altitude of the tower. Before reading further figure out what you can do with this information:

You can determine the person's distance from the center at the bottom and the top of the tower.
Knowing the two distances you can use Newton's Law of Universal Gravitation to find the gravitational force on the individual at each point.
Knowing the two distances you can find the gravitational PE relative to infinity at both points.
Knowing the two distances you could find the corresponding orbital velocity at both points. However it isn't practical to orbit at the surface of the Earth, so only one orbital velocity could potentially be useful.
You know the acceleration of gravity at the surface, so you could find the gravitational force on the person at the bottom of the tower, without having to resort to Universal Gravitation.
Using inverse square proportionality of gravitational acceleration with distance from center, we could find the acceleration at the top of the tower and use this to find the weight of the person at that position.

Now looking at the first question, we see that we're asked for the weight of the person at the top and bottom of the tower. We can easily answer:

We easily find the weight at the bottom, using either Universal Gravitation or the known acceleration of gravity at the surface.
We use Universal Gravitation (or proportionality) to find the weight at the top.

We are given a power output of .99 watts per kg of body mass, sustainable for 8 hours.

A watt is a Joule/second.
Body mass is 87 kg so power output is .99 watts / kg * 87 kg = 86 watts.
This is 86 Joules / second, sustainable for 8 hours so available daily energy is 86 J / sec * 8 hr * 3600 sec / hr = 86 J / s * 28800 sec = 2 400 000 J, approx. (about 2/3 of a kilowatt-hour)

We need to figure out how much energy is required to climb the tower.

We can average the gravitational force at top and bottom, then multiply by the distance up the tower, in order to get the approximate PE change.
We could calculate PE relative to infinity (using - G M m / r) at both points and subtract the two results. This would give us the accurate change in PE.

This problem is solved in the Introductory Problem Set.

Are there aspects of this problem that you could usefully visualize using graphs?

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A graph of gravitational force vs. position on the tower could be useful. The work is the area beneath the graph.

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`Are there things about this problem that you could usefully represent using pictures and/or diagrams?

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A picture of the tower with gravitational force vectors decreasing with height might be very helpful.

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`What definitions apply to this situation, and what are the names and symbols for the quantities are involved?

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work = ave force * displacement

`dPE is equal and opposite to work done by conservative force

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`Does this problem involve any of the formulas you have been advised to memorize? What relevant formulas result from definitions and basic laws?

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You probably need Newton's Law of Universal Gravitation.

You need to know that weight is force exerted by gravity.

You might need the expression - G M m / r^2 for potential energy relative to infinity.

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`What situations seem similar to this, but perhaps not exactly the same? What concepts, ideas and procedures have a similar 'feel'?

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The work here is done by a changing force. In this sense there is a similarity to the work done when stretching an elastic object.

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Problem Number 8

What is the mass of a planet about which a small satellite orbits with a period of 52 minutes at a distance of 20000 km from the planet's center?

Let's first ignore the fact that we're asked for the mass, and figure out what we can do with the given information.

We are given the time required for a single orbit and the distance from the center of the planet, which is the radius of the orbit.

From the radius we can find the circumference of the orbit.
From the circumference and the time required we can find the orbital speed.

Now let's think about the question of mass.

We know that the gravitational force on the satellite is G M m / r^2, its PE relative to infinite separation is - G M m / r, and its orbital velocity is v = sqrt(G M / r).
We don't know the mass of the satellite, though we could divide the gravitational force by its unknown mass m to get its centripetal acceleration.
We might notice that v = sqrt( G M / r) doesn't involve the mass of the satellite. Looking a little more closely we see that v and r are known, as is the value of the universal constant G, leaving only M as unknown.
So we solve v = sqrt( G M / r) for M, obtaining M = v^2 r / G.

121 final

Problem Number 4

As an Earth satellite of mass `mass kg, originally in a circular orbit of radius 7.05 * 10 ^ 6 m, increases its orbital radius to 7.17 * 10 ^ 6 m, what are the KE and PE changes and the ratio of PE to KE change?

The mass, designated by `mass, is not given. The symbol `mass should have been represented as a number. You are free to substitute any number you wish. You could choose 1 kg, a good choice because it simplifies the arithmetic. You are also free to regard `mass as an unknown quantity, and represent your solution in terms of the symbol `mass.

We are given the two orbital radii. For each we could calculate any of the following in terms of the mass `mass:

Gravitational force is G M * `mass / r^2
PE relative to infinity is -G M * `mass / r
KE is 1/2 G M * `mass / r, being positive and half the magnitude of the PE.
v = sqrt( G M / r), the orbital velocity at radius r.

We are asked to find PE and KE changes.

We easily find PE for each radius.
We easily find KE for each radius.
We easily subtract PE at the first radius from PE at the second to obtain our (positive) change in PE.
We easily subtract KE at the first radius from KE at the second to obtain our (negative) change in KE.
Whatever means we used to find KE we find that its magnitude is half that of the change in PE.

Problem Number 6

An ideal spring has restoring force constant 950 Newtons/meter. An unknown mass on the spring is observed to complete 47.616 cycles every minute. What is the unknown mass, in kilograms?

We are given the force constant k = 950 Newtons / meter.

We are given the number of cycles completed every minute.

From this we can easily determine the time required to complete a cycle (i.e., the period of the motion), or the number of cycles completed per second (i.e., the frequency of the motion).
Knowing that 1 cycle corresponds to 1 revolution about the reference circle, we can use the fact that a revolution about the reference circle corresponds to an angular displacement of 2 pi radians, along with either the number of cycles per second or the time required to complete a cycle, to determine the angular velocity omega of the reference-circle point. This angular velocity is also called 'angular frequency' when referring to a model of simple harmonic motion.

At this point we know the angular frequency omega, and the force constant k. This should call your attention to the formula omega = sqrt( k / m ).

Knowing k and omega, we can easily solve to find m.
We get

m = omega^2 / k

We easily substitute for omega and k, obtaining mass m. Be sure the units work out.

`Does this problem involve any of the formulas you have been advised to memorize? What relevant formulas result from definitions and basic laws?

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This problem requires the formula omega = sqrt( k / m ), which is to be memorized.

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Problem Number 7

If 37 J of net work is done on a system while 17 J of work are done on the system by nonconservative forces, then by how much do the PE and the KE of the system change?

This requires only knowledge of the basic work-kinetic energy theorem and the definition of `dPE:

37 J is the work done by the net force, so since `dKE = `dW_net_on we have `dKE = 37 J.
`dPE is equal and opposite to the work done on a system by nonconservative forces, so `dPE = - 17 J.

`What definitions apply to this situation, and what are the names and symbols for the quantities are involved?

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We use the definition of `dPE as equal and opposite to the work done on a system by conservative forces.

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`Does this problem involve any of the formulas you have been advised to memorize? What relevant formulas result from definitions and basic laws?

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We use the basic work-KE theorem `dW_net_on = `dKE.

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Problem Number 13

An object of mass 4 Kg is acted upon by an unknown force F for `dt = 6 seconds. Its velocity is observed to change during this time from 9 m/s to -1017 m/s.

Use the Impulse-Momentum Theorem to determine the average force exerted on the object.
Verify your results using your knowledge of uniformly accelerated motion.

We are given

`dt = 6 sec
v0 = 9 m/s
vf = -1017 m/s
m = 4 kg

We can determine a number of quantities from this information:

From m, v0 and vf we can determine initial and final KE, and initial and final momentum.
So we can find change in KE, and we can find change in momentum.
The change in KE is equal to the work done by the net force.
The change in momentum is equal to the impulse of the net force.
We can also find the acceleration and displacement of the object, reasoning directly from v0, vf and `dt.

We are then asked to use the impulse-momentum theorem:

impulse = change in momentum
impulse is F_net_ave * `dt
change in momentum is `dp, the change in mv, easily calculated as noted above
thus F_net_ave * `dt = `dp, and F_net_ave = `dp / `dt

We are asked to also use uniformly accelerated motion:

From v0, vf and `dt we find acceleration a.
We apply Newton's Second Law F_net = m a, with the known mass, to get F_net. Our result should agree with the preceding result obtained using the impulse-momentum theorem.

Problem Number 14

An object moves around a circle of radius 3 meters, making a revolution every 4.7 seconds. Assume that its angular velocity is constant. Starting at t = 0, when its angular position is 0 radians, what are the x and y coordinates of its position after t seconds (express your answer in terms of the parameter t)?

We are given the radius and time required for 1 revolution around a circle:

r = 3 m
T = 4.7 sec

We can get of the solution in two different lines of reasoning:

One line of reasoning:

It's important to visualize the motion of the object. Let's think for a minute about how the position of the object changes:

It will take 1/4 * 4.7 sec = 1.175 sec to reach the 90 degree or pi / 2 position; 1/2 * 4.7 sec = 2.35 sec to get to the 180 deg or pi position, and 4.7 sec to get to the 2 pi position.

After 1 second it will have traveled through 1 / 4.7 revolutions, after 2 second it have traveled through 2 / 4.7 seconds, after 4.7 sec it will have traveled through 4.7 / 4.7 revolutions = 1 revolution.

The angular positions corresponding to 1 sec and two sec are 1 / 4.7 * 2 pi rad = 1.34 radians, 2 / 4.7 sec * 2 pi rad = 2.67 rad.

After t seconds it will have traveled through t / T revolutions, so its angular position will be

theta = t / T revolutions = t / T * 2 pi radians.

The radial vector has magnitude 3 m, matching the radius of the circle, so the x and y components of position after t seconds are

x = 3 m * cos(t / T * 2 pi rad) = 3 m * cos( (2 pi / T) * t)

y = 3 m * sin(t / T * 2 pi rad) = 3 m * sin((2 pi / T) * t)

Since 2 pi rad / T = 6.28 rad / (4.7 sec) = 1.34 rad / sec, approx., we have

x = 3 m * cos(1.34 rad/sec * t)

y = 3 m * sin(1.34 rad/sec * t)

Another line of reasoning:

Given the period T = 4.7 sec we can find the angular frequency omega. It takes 4.7 sec to go around the circle, which corresponds to an angular displacement of 2 pi radians. So the angular velocity of the object is 2 pi rad / (4.7 sec) = 1.34 rad / sec. Formally

omega = dTheta / `dt = 2 pi rad / (4.7 sec) = 1.34 rad / sec.

After t seconds the angular position of the object is

theta = omega * t = 1.34 rad/sec * t.

Once more the radial vector has magnitude 3 m, matching the radius of the circle, so the x and y components of position after t seconds are

x = 3 m * cos(1.34 rad/sec * t)

y = 3 m * sin(1.34 rad/sec * t)

Note on a bunch of confusing formulas and why they probably aren't worth memorizing:

The following formulas all say things that are obvious if you know how to think about motion on the reference circle:

omega = 2 pi / T, where T stands for the time required to complete a revolution (i.e., the period of the motion) says that angular velocity of a point moving around a circle is 2 pi radians divided by the time required to complete a revolution.

omega = 2 pi f, where f stands for frequency, says that the angular velocity of a point moving around a circle is 2 pi radians multiplied by the number of times the point goes around the circle in a second.

f = 1 / T says that the number of times you go around a circle in a second is the reciprocal of the time required to go around the circle (e.g., if it takes .1 second to complete a revolution, then you complete 1 / .1 = 10 revolutions in a second).

T = 2 pi / omega says that if you divide 2 pi radians by the angular velocity you get the time required to complete a revolution (i.e., if you divide angular displacement by angular velocity you get the time interval).

f = omega / (2 pi) says that if you divide the number of radians per second by the 2 pi radians around the circle, you get the number of times the object goes around the circle every second. For example at 6 pi radians / sec you go around the circle 3 times in a second.

Most of these formulas include 2 pi, mostly in the numerator though in some cases it's in the denominator. One doesn't have 2 pi at all. T is in the numerator. f is in the denominator except when it isn't. Most people can't keep the formulas straight without understanding their meanings; if the meanings are understood then there's no need to memorize them.

However all these formulas can be reduced to two:

omega = 2 pi / T

f = 1 / T.

All the others are just combinations and/or rearrangements of these two. If you need to memorize something, memorize these two. But by all means be sure you understand motion on a circle so you know what they mean.

Problem Number 15

What is the KE change of an object of mass 7.9 kg as its gravitational PE increases by 65 J if at the same time it also does 49 J of work against the net nongravitational force?

This requires application of energy conservation:

Gravitational PE increases by 65 J. `dPE is equal and opposite to the work done by the gravitational force on the object, so the gravitational force does -65 J of work on the object.

The object does 49 J of work against other forces, so the other forces do -49 J of work on it.

The sum of the gravitational and nongravitational forces is the net force (every force is one or the other).

Thus the work done on the object by the net force is -65 J + (-49 J) = -114 J.

That is, `dW_net_on = -114 J.

Since `dW_net_on is equal and opposite to the change in KE, it follows that `dW_net_on = -114 J.

Homework:

Your label for this assignment:

ic_class_09207

Copy and paste this label into the form.

Answer the questions posed above.

Consider asking questions, etc., about the test problems sent out last weekend.