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Guide Questions for Physics I.
At the indicated stage of the course you should be able to answer all the associated questions accurately without reference to notes. Preferably you will also be able to state and answer all, or at least most of the questions (or perhaps an equivalent set of questions) without notes.
It is suggested that you periodically copy these questions into a document and insert your answers as indicated, then submit the document for the instructor's review.
These ideas alone are not sufficient to guarantee success in your course, but they do form a core structure, a basis from which to master the skills and concepts that will lead to success. In working through the assignments for the course you should think in terms of these definitions and quantities.
1. What is the definition of rate of change?
Answer:
The definition is as follows:
The rate of change of A with respect to B is (change in A) / (change in B).
2. What is the definition of average velocity?
Answer:
The definition of average velocity is as follows:
average velocity is average rate of change of position with respect to clock time
By the definition of rate of change it follows that
average velocity = (change in position) / (change in clock time).
3. What is the definition of average acceleration?
Answer:
The definition of average acceleration is as follows:
average acceleration is average rate of change of velocity with respect to clock time
By the definition of rate of change it follows that
average acceleration = (change in velocity) / (change in clock time).
4. What are the four equations of uniformly accelerated motion and how do they follow from the definitions of velocity, acceleration, and the linearity of the v vs. t graph?
Answer:
For motion on an interval, where `ds is change in position, `dt is change in velocity, v_0 is the velocity at the beginning of the interval, v_f the velocity at the end of the interval and a stands for acceleration:
`ds = (v_0 + v_f) / 2 * `dt
v_f = v_0 + a * `dt
`ds = v_0 * `dt + .5 * a * `dt^2
vf^2 = v_0^2 + 2a * `ds
The first equation follows from the definition of average velocity and the linearity of the v vs. t graph (which is guaranteed by the uniformity of acceleration). The linearity implies that the average velocity is the average of initial and final velocities. The definition of average velocity implies that vAve = `ds / `dt, so that (vf + v0) / 2 = `ds / `dt, and by simple algebraic rearrangement `ds = (vf + v0) / 2 * `dt.
The second equation follows from the definition of average acceleration. In the case where acceleration is uniform the average acceleration is the acceleration. The definition of average acceleration therefore tells us the a = `dv / `dt; since `dv = vf - v0 this means that a = (vf - v0) / `dt. Simple algebra allows us to rearrange this to the form vf = v0 + a `dt.
The third equation follows from the first two by eliminating vf. This is done simply by substituting vf = v0 + a `dt into the first equation and simplifying.
The fourth equation follows
from the first two by eliminating `dt. If we solve the second equation for
`dt and substitute the result for `dt in the first, and simplify, the resulting
equation can be rearranged to give us the fourth equation.
5. What are the seven quantities in terms of which we understand uniformly accelerated motion and what are the standard units of each?
Answer:
We understand uniformly accelerated motion by understanding the quantities v0, vf, `ds, `dt and a (see #4) as well as `dv and vAve, and by understanding how these quantities are related to one another and to the definitions of average velocity and average acceleration.
The standard unit of `ds is the meter and standard unit of `dt is the second.
By the definitions of velocity
and acceleration vAve, v0, vf, `dv and vAve all have units of meters / seconds,
abbreviated m/s, and a has units of meters / second^2 or m/s^2 (understood as
(meters / second) / second).
6.
What are the five quantities used in the equations of uniformly accelerated
motion and what are the standard units of each?
Answer: The equations are expressed in terms of `ds, v0, vf, `dt and a. The quantities `dv and vAve are essential in applying the definitions of velocity and acceleration in order to obtain the equations, but these two quantities do not appear in the equations.
The units are as given in #5.
7. Given a graph of one
quantity vs. another, how do we calculate the area beneath a straight line
segment on the graph and the slopes of the segment,
and how do we interpret the area and the slope?
Answer:
A graph of quantity A vs. quantity B has the quantity A on the 'vertical' axis and the quantity B on the 'horizontal' axis.
A line segment runs from one point (called its initial point) to another (called its terminal point).
We always ask the following questions:
What does the 'rise' of the graph between the two points represent?
What does the 'run' of the graph between the two points represent?
What does the slope of the graph between the two points therefore represent?
What does the midpoint value of A represent?
How closely does the midpoint value of A represent the average value of A on the interval?
What does the 'run' of the graph between the two points represent?
What therefore does the area of the graph, which is the product of the average value of A and the 'run', represent?
The slope between two points is equal to the rise divided by the run, where the rise is the change in the 'vertical' coordinate and the run is the change in the 'horizontal' coordinate.
The 'rise' is measured in the units of the 'vertical' coordinate A and the 'run' in terms of the 'horizontal' coordinate B.
The units of the slope are therefore equal to the units of the vertical coordinate divided by the units of the horizontal coordinate.
The 'rise' is the change in the quantity A, the 'run' is the change in the quantity B, so slope = rise / run represents (change in A) / (change in B), i.e., the average rate of change of A with respect to B.
The area beneath the segment is a trapezoid, which average 'altitude' is the average value of A and whose average 'width' is the change in the value of B.
The area therefore represents (average value of A) * (change in value of B).
The units of the area are therefore the product of the units of A and the units of B.
If A represents the rate of change of some quantity Q with respect to B, then the average value of A is the approximate average value of the rate of change of Q with respect to B, which by the definition of rate of change is (average rate) = (change in Q) / (change in B). The area would therefore represent (change in Q) / (change in B) * (change in B) = (change in Q).
You should know the above by the time you take the Major Quiz.
8. What is the definition of work and what are its units?
Answer:
The work done by force F acting through displacement `ds is `dW = F * `ds, provided both the force acts along the same line as the displacement.
The unit of force is the kg * m/s^2, or Newton.
The unit of displacement is the meter.
The unit of F * `ds is
therefore the kg m/s^2 * m = kg m^2 / s^2, or alternatively the Newton * meter.
This unit is called the Joule.
9. What is the definition of kinetic energy and what are its units?
Answer:
The kinetic energy of an object is
kinetic energy = 1/2 m
v^2, where m is the mass of the object and v its velocity.
The units of kinetic energy are kg * (m/s)^2 = kg m^2 / s^2, or Joules.
10. What is the definition of potential energy and what are its units?
Answer:
The definition is:
The change in the potential energy of an object is the work done by the object against the conservative forces acting on it.
In application it is often
useful to note that this is equal and opposite to the work done by the
conservative forces acting on the object.
11. What does the work-kinetic energy theorem say?
Answer:
The theorem says that the work done on an object by the net force acting on it is equal to the change in its kinetic energy.
In symbols this can be expressed as
`dW_net_ON = `dKE.
Note that the net force on an object is the sum of all forces acting on the object.
12. What does the impulse-momentum theorem say?
Answer:
The impulse of a force is the product F * `dt of the force and the time interval during which it is applied.
The momentum of an object is the product m * v of the object's mass and velocity.
The impulse-momentum theorem says that the impulse of the net force acting on an object is equal to the change in its momentum.
In symbols this is
F_net `dt = `d (m v)
Provided mass is constant, `d(m v) = m * `dv and F_net `dt = m `dv.
13. How are the work-energy theorem and the impulse-momentum theorem related to the equations of uniformly accelerated motion?
Answer:
Both theorems are derived, at least for the case of uniform acceleration, by substituting a = F / m (from Newton's Second Law) into equations of uniformly accelerated motion:
If we substitute a = F / m for acceleration in the fourth equation of motion and rearrange we get the work-kinetic energy theorem.
If we substitute a = F / m for acceleration in the second equation of motion and rearrange we get the impulse-momentum theorem.
14. What is the work-energy theorem in terms of work done by nonconservative forces, change in KE and change in PE, and how does this theorem make sense?
Answer:
The total work done by the nonconservative forces acting on an object is equal to the change in its KE and PE:
`dW_noncons_ON = `dKE + `dPE.
This makes sense because `dW_net_ON = `dKE, and the net force acting on an object is the sum of the conservative forces and the nonconservative forces acting on it. So `dW_net_ON = `dW_noncons_ON + `dW_cons_ON. Since `dPE = `dW_cons_BY = -`dW_cons_ON, `dW_net_ON = `dW_noncons_ON - `dPE. This is equal to `dKE so `dW_noncons_ON = `dKE + `dPE.
15. How do you find the components of a vector given its direction and magnitude? How do you find the direction and magnitude of a vector given its components?
Answer:
If you know magnitude and direction, find the angle theta of the vector with the positive x axis, measured counterclockwise from that axis. Then
x component = magnitude * cos(theta) and
y component = magnitude * sin(theta).
If you know the components then
magnitude = sqrt( (x component)^2 + (y component)^2) and
angle as measured counterclockwise from the positive x axis = arcTan( (y component) / (x component) ), plus 180 deg of the x component is negative.
Alternatively you can use right-angle trigonometry if you know it well enough and are careful with the signs of your results.
You should know the above by the time you take Test 1.
16. How do you find the magnitude and direction of the gravitational force between two masses, given the positions of the masses?
Answer:
17. How do you find the velocity of a given circular orbit of a satellite about a planet of given mass?
Answer:
18. How do you find the gravitational potential energy of a given mass at a given distance from the center of a given planet?
Answer:
19. For a planet in an elliptical orbit with the planet at one focus of the ellipse, how do you find the change in kinetic energy between the point of closest approach and the point at which the satellite is at its greatest distance from the planet?
Answer:
20. What is the definition of a radian and how do we convert from radians to degrees, and vice versa?
Answer:
21. What is the definition of torque and what are its units?
Answer:
22. What is the definition of moment of inertia and what are its units?
Answer:
23. If you apply a given torque to an object with a given moment of inertia, how does the object respond?
Answer:
24. What are the details of the analogy between linear dynamics and rotational dynamics?
Answer:
You should know the above by the time you take Test 2.
25. When the radial line to a point moving about a circle of radius r is at angle theta, what are the coordinates of the point?
Answer:
26. When a point moving with angular velocity omega about a circle of radius r is at angle theta, what are the direction and magnitude of its velocity vector and what are the coordinates of the this vector?
Answer:
27. When a point moving with angular velocity omega about a circle of radius r is at angle theta, what are the direction and magnitude of its velocity vector and what are the x and y components of the this vector?
Answer:
28. When a point moving with angular velocity omega about a circle of radius r is at angle theta, what are the direction and magnitude of its centripetal acceleration vector and what are the components of the this vector?
Answer:
29. What is the angular position of a point moving with angular velocity omega around a circle of radius r at clock time t, assuming the point started at the angular position theta = 0? What therefore are the coordinates of its position, the components of its velocity vector and the components of its centripetal acceleration vector at this instant?
Answer:
You should know the above
by the time you take the final exam.