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Discussion Question 1A
P212, Week 1
P211 Review: 2-D Motion with Uniform Force
The mathematics and physics of the problem below are similar to problems you will encounter in
P212, where the force is due to the action of an electric field on a charged particle.
A point particle of mass m travels freely in the x-direction with uniform velocity v0. At x = 0, it
enters a region between two plates oriented perpendicular to the y-axis; the plate spacing is w,
and then plate length in the x-direction is L. The particle enters on the mid-plane y = 0. While
between the plates, it experiences a constant, spatially uniform force F in the +y-direction. After
exiting the plates the particle again moves freely.
y
L
+w/2
v0
m
x
F
–w/2
(a) Our first task will be to obtain an expression for the y-coordinate of the point at which the
particle exits the plates. We will assume that the plate spacing is wide enough that the particle
never strikes either plate. But before we start, consider these possible solutions:
(1) y
F
m(v0
L)
(2) y
FL
mv0
(3) y
Fw 2
mv02
Could any of them be correct? Why or why not? Remember, units and limiting behavior ! In
fact, from only those two considerations, you can write down the correct answer to within a
factor of 2 without using any formulas at all. Want to give it a try? (This procedure is called
dimensional analysis and physicists use it all the time, especially when developing new
theories.)
None of the three answers is correct. As you’ll see in the next problem, this dimensional
analysis has almost given us the complete answer, and has provided a valuable check on our
upcoming work. But to get the complete solution we have to perform a calculation.
(b) Now go ahead and calculate the correct expression for y.
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y = 1/2 a t
2
where a = F / m and t = L / v0 … therefore, y
1F L
2 m v0
2
(c) Next, find an expression for the maximum value of the force F for which the particle passes
through the force region without striking either plate.
ymax = w / 2 and y = F L2 / (2 m v02)
therefore Fmax = m v02 w / L2
(d) For the conditions of part (c), find an expression for tan where is the angle of deflection
at which the particle exits the force region. (Did you draw a sketch? Did you check your units?)
tan
vy / vx.
vy = ay t
tan
= w / L.
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Discussion Question 1B
P212, Week 1
P211 Review: Gravitational Forces and Superposition
mm
G 12 2 rˆ12 , 1
r12
F1 2
r12
rˆ12
2
This problem develops skills you will need in P112 in finding the electric fields created by
sets of point charges.
Four particles of equal mass M are fixed at the corners of a square with sides of length a. A fifth
y
M
M
a
M
x
m
a
M
M
a
M
a
M
m
M
a
M
M
m
X
a
M
M
particle has mass m and moves under the gravitational forces of the other four.
(a) Find the x- and y-components of the net gravitational force on m due to the other four
masses when m is located at the center of the square (left-hand figure).
Hint: Draw a sketch! Use superposition and draw a vector diagram consisting of four
vectors, each representing the force exerted by one of the corner particles on m. For ease
of reference, label the four (equal) corner masses M1, M2, M3, and M4. Label the
corresponding force vectors F1, F2 , etc. With the vector diagram in hand, it is vastly easier
to calculate the requested components of the total force.
y
M
M
A sketch immediately reveals that the forces on the test mass m
cancel, giving a net force of zero.
a
M
x
m
M
a
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(b) Find the x- and y-components of the net gravitational force on m when it is located at the
center point of the right-hand side of the square (middle figure). Use the same solution
procedure that was recommended above for part (a).
Fx = F3x + F4x.
|F3| = |F4| = G M m / r2, with r2 = (a/2)2 + a2 = 5a2/4
M3
M1
a
m
M4
M2
F3x = F3cos , where cos
The final answer is:
Fy = 0 and Fx = -16 G M m / (5 5a2)
(c) Check your answer to part (b) by testing at least 3 examples of limiting behavior. Do you
get the results you expect?
Now, an important point: suppose you were given numerical values in this problem:
M = 3 kg, m = 1 kg, and a = 5 cm. If you had plugged those numbers into your equations
right from the start, you’d get the final result Fx = 1.15 x 10-7 N. Would you be able to
check the limiting behavior of this answer? No! We’ve learned an important lesson:
Never plug in numbers until the end of your calculation.
Fx goes to infinity as m or M go to infinity (infinite masses exert infinite force)
Fx goes to zero as m or M go to zero (without mass, there is no gravitational force)
Fx goes to infinity as a goes to zero (infinite force exists when the masses are right on top
of each other)
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Fx goes to zero as a goes to infinity (no force if the masses are infinitely far apart)
(d) When mass m is located on the x-axis a distance X large compared to a
(right-hand figure), one can use a simple physical argument to see that the net force on m due to
the other four particles is approximately Fx CGMmX 2 and Fy 0 where C is a numerical
constant. (This is a long-distance approximation: it gets better and better as X increases.) What
is the applicable physical argument? Use it to find the value of the dimensionless constant C.
The square of masses will look like a single object with mass 4M. The constant C in the
approximate formula is therefore –4 (- sign needed since the force is attractive).
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Discussion Question 1C
P212, Week 1
Review: Potential Energy
Four particles of equal mass M are fixed at the corners of a square with sides of length a. A fifth
particle has mass m and moves under the gravitational forces of the other four.
Consider the gravitational potential energy U of the mass m as a function of its position (x,y). This
potential energy map U(x,y) is an extremely useful way of representing the effect of the
gravitational force on the mass m:
If m is held at rest at some point (x1,y1) and you let it go, it will always move toward a point
of lower potential energy. You can therefore think of the function U(x,y) as a topographical
map: the particle m will always “roll downhill”.
If the mass m moves from point (x1,y1) to point (x2,y2), the potential energy difference
U(x1,y1) – U(x2,y2) tells you exactly how much work was done by gravity, and exactly how
much kinetic energy the particle gained as a result.
Let’s visualize all this. The figures below show the potential energy map U(x,y) for the square
configuration of four masses M illustrated above. The height of the surfaces indicates the magnitude
of U at each point. Which one is correct?
The correct map is the bottom middle one. The top three can be eliminated since they indicate
higher potential energy near the masses M. That can’t be: the gravitational force attracts masses
toward each other, so the potential energy must be lower closer to the masses. The bottom left
1
figure corresponds to the potential around a single point mass M, not four of them. Finally, we can
see that the bottom middle figure is the correct one by considering superposition: the potential
energy map for the four masses is just the superposition (sum) of four maps for a single mass, like
the bottom-left figure. The bottom right figure shows additional features (ridges) in the map that
would not be caused by our simple collection of 4 masses.
Now let’s perform some calculations using potential energy. Here’s the master formula for the
gravitational potential energy between two masses m1 and m2 separated by a distance r12:
U12
mm
G 1 2
r12
(a) Show that the net gravitational potential energy of particle m when at the origin is
U(0,0)
4 2GMm / a.
Hint: The equation above gives the gravitational potential energy of mass m1 in the presence of
mass m2, or vice versa. Use it and superposition to find the net potential energy of mass m in
the presence of the four fixed masses M. As long as the masses M remained fixed, it is not
necessary to consider their mutual potential energies.
y
M
a
M
M
m
a
x
M
By superposition, the potential energy of m in the presence of the four masses (which we’ll call M1
through M4) is just the sum of the potential energy due to each individual mass. In symbols, U = U1
+ U2 + U3 + U4. We used superposition of forces in problem 1 … notice how much easier it is with
potential energy since that quantity is a scalar (just a number), not a vector, and so does not involve
any components. From the left-hand diagram, we see that the distance r between m and each of the
M’s is the same: r2 = (a/2)2 + (a/2)2 = a2/2, and so r = a / 2. From the potential energy formula, we
get U = 4.U1 = -4 G M m 2 / a.
(b) If the particle m is released at rest from infinity and passes through the origin, calculate the
magnitude of its velocity there.
If the particle “falls” from infinity to the origin, it loses potential energy in the amount U that you
calculated above. This energy has to go somewhere it is converted to kinetic energy = m v2 / 2.
Rearranging the terms in -U = m v2 / 2 we can find the particle’s final velocity: v
2
8 2GM
a
(c) Calculate the net potential energy of m when it is at the mid-point of the right-hand side of the
square.
y
M
M
m
a
M
x
M
a
Again by superposition, U = U1 + U2 + U3 + U4. Since all the masses are the same, the only
difference between the Ui’s is the distance from the mass Mi to the test mass m. For the two righthand masses, r1 = a/2. For the two left-hand masses, r22 = (a/2)2 + a2 = 5a2/4, and so r2 = 5a/2.
U = –2 G M m / r1 – 2 G M m / r2 = -4 G M m / a (1 + 1/ 5)
(d) If the particle m is released at rest from the mid-point of the right-hand side of the square, does
it reach the origin? Support your answers with physical arguments.
From part (a), U at origin = -4 2 G M m / a = -5.657 G M m / a.
From part (b), U on right-hand side = -4 ( 1 + 1/ 5) G M m / a = -5.789 G M m / a.
The potential energy is thus lower on the right-hand side than at the origin. That means that the
mass m prefersto be on the right-hand side than at the origin. Further, it cannot reach the origin
unless it starts with enough kinetic energy to overcome the potential difference between the two
positions. Since our particle is starting from rest, no it will not reach the origin.
(e) In Discussion Question B, you calculated the force on the mass m when it is at the mid-point of
the right-hand side of the square. If all went well, you found that this force points in the –x
direction, i.e. towards the origin. Given this fact, plus your answer to part (d), can you make a
rough sketch of the potential energy function U(x) for points along the x-axis? Give it a try!
U
y
M
M
a
M
x
–a/2
3
a/2
x
x
a
M
Discussion Question 1D
P212, Week 1
P211 Review: Uniform Circular Motion
F1 2
r12
mm
G 12 2 rˆ12 , 1
r12
2
rˆ12
In P112 you will encounter problems where charged particles move in uniform
circular motion. The forces involved may be electric or magnetic in nature. The
answer to part(e) contains the secret of the cyclotron.
Kepler’s Third Law (K-III) for planetary motion about the sun for circular orbits is
T 2 CR 3 where T is the period, R is the radius of the planet’s orbit and C is a constant.
v
R
m
M
(a) Derive K-III for a circular orbit and in the process find an algebraic expression for C
in terms the mass of the sun MS, the universal gravitational constant G, and numerical
factors.
F = -G m MS / R2
F = ma = -m v2 / R
Set equal to each other and substitute period for velocity.
T2 = 4 2 R3/ (G MS)
(b) Using your answer from part (b), re-express K-III as a relationship between the
angular frequency of the motion and the radius R of the form:
2
= f(R, G, MS).
= G MS / R3
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(c) Consider uniform circular motion of a body of mass m about a central force which
a b
depends on velocity: F Dv R where D is a constant and a and b are known
exponents. Derive K-III for this force, again expressing your answer as a relationship
between and R, (The constant D will necessarily appear in your final expression, but v
must not.)
2-a
= D Ra + b – 1/m
(d) For the case a = 0 and b = –2, verify that your answer to part(c) collapses to that for
part (b) (this is an excellent limiting behavior check).
(e) Evaluate your answer to part (c) for the case a = 1 and b = 0. How does the angular
frequency depend on radius for a force of this nature?
= D/m, angular frequency independent of radius.
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