# P221/S2002/Problem 3A - ~ ) ```P221/S2002/Problem 3A
Assume that the magnitude of the attractive gravitational force between two particles (1
and 2) of masses 1 and 2 , a distance apart, is given by - ~ ) 12.5 2 , where ) is
the new gravitational constant. Note that this force is inversely proportional to 2.5 , not
2 !!!
(a) [4 points] Calculate the potential energy &lt; &sup2;&sup3; in this case, choosing it to be 0 at
infinity.
We need the function whose negative derivative with respect to is the expression for the force
given above. That means &lt; &sup2;&sup3;  &deg;1.5 since the derivative of &deg;1.5 will be proportional to
/2.5 (remember that &deg; ~ c1 ). With the correct factors,
&lt; &sup2;&sup3; ~ c )1 2 &deg;1.51.5 &Agrave;
No constant need to be added to this if we want to have &lt; &sup2;B&sup3; ~ 0.
Assume for the rest of this problem that particle 1 moves in a circular orbit of radius 9
about particle 2. (2  1 )
(b) [4 points] Calculate the kinetic energy of particle 1 in terms of ) , 1 , 2 , and 9 .
Applying Newton's second law of motion to this system, as we did in deriving Kepler's third law,
)
1 2
9 2.5
~ 1 #2 &deg;9 so 1 #2 ~ )
1 2
9 1.5
and
2
1 #
~)
1 2
29 1.5
.
(c) [3 points] Show that the total mechanical energy of particle 1 is given by
1 2
.
, ~ c )
69 1.5
We have the kinetic energy from part (b) and the potential energy from part (a), so
, ~ 2 b&lt; ~)
1 2
29 1.5
1 2
4
c ) 1.5
9 1.5 ~ )
1 2
9 1.5
54
2
c
1.5
5~ c)
1 2
69 1.5
.
(d) [3 points] Calculate the period ; of particle 1 in an orbit of radius around particle 2.
From part (b) we have 1 #2 ~ )
; ~ 29&deg;# ~ 29&deg;&sup2;m
1 2
9 1.5
)2
9 1.5
so # ~ m
)2
9 1.5
&Agrave; Then
&sup3; ~ 2l9 3.5 &deg;)2
(e) [3 bonus points] Calculate the escape speed of particle 1 (the speed it would need to
escape, starting at rest from ~ 9 ).
The energy that must be added to the particle to enable it to escape (to have , ~ 0) is just
1 2
2
c &lt; ~ ) 1.5
9 1.5 . Setting this equal to 1 #esc , we have
2
#esc
~)
22
1.59 1.5
so #esc ~ m
4)2
39 1.5
.
P221/S2002/Problem 3B
A 2.0-kg block which moves on a frictionless horizontal surface is attached to a horizontal
spring for which the force constant ~ 200 N/m. It is held at an extension of 5.0 cm and
then released at ! ~ 0.
(a) [2 points] What is the maximum amplitude of the subsequent oscillations?
The maximum amplitude ( will just be 5.0 cm, the initial extension from which it is released at
rest.
(b) [2 points] What is the angular frequency in rad/s?
~ l&deg; ~ l&sup2;200 N/m&sup3;&deg;&sup2;2.0 kg&sup3; ~ l100 rad/s ~ 10 rad/s
(c) [4 points] Find the displacement %&sup2;!&sup3; as a function of time, using the boundary
conditions of the values of % and #% at ! ~ 0.
At ! ~ 0, % ~ 5.0 cm and #% ~ 0. The form of %&sup2;!&sup3; that meets these conditions, with an
amplitude and angular frequency as calculated in parts (a) and (b), is
%&sup2;!&sup3; ~ ( cos ! ~ (5.0 cm) cos ((10 rad/s) !)
(d) [3 points] Find the velocity component #% when % ~ (&deg;2.
#% &sup2;!&sup3; ~ %&sup2;!&sup3;&deg;! ~ c (5.0 cm)(10 rad/s) sin ((10 rad/s)!)
~ c (50 cm/s) sin ((10 rad/s)!)
When % ~ (&deg;2, cos ((10 rad/s) !) ~ 1/2 so sin ((10 rad/s)!) ~ l 3 &deg;2  0.9 so
#%  c (50 cm/s)(0.9)  c 45 cm/s.
(e) [3 points] Find the acceleration component % when % ~ (&deg;2.
It is possible to differentiate #% &sup2;!&sup3;, but it is simpler to note that for simple harmonic
motion, % &sup2;!&sup3; ~ c 2 %&sup2;!&sup3; so the answer is
% ~ c (10 rad/s)2 (5.0 cm&deg;2) ~ c 250 cm/s2 ~ c 2.5 m/s2
(f) [2 bonus points] Find the total mechanical energy , in joules for this block-spring
system.
, ~ (2 ~ 2
(2 ~ &sup2;0.5&sup3;&sup2;2.0 kg&sup3;&sup2;10 rad/s&sup3;2 &sup2;0.05 m&sup3;2 ~ 0.25 J
P221/S2002/Problem 3C
A small sphere of mass ~ 0.50 d 10c3 g
carries a charge ~ b 8.85 d 10c9 C. It
hangs in the Earth's gravitational field from
a silk thread of negligible mass that makes an
angle ~ 45&deg; with a large vertical nonconducting sheet, as shown in the figure.
Notice that is infinitesimally small.
(a) [5 points] Use Gauss' law to show that the electric field , due to a nonconducting
infinite plane with uniform charge per unit area is equal to &deg;20.
By symmetry, the electric field must be
directed perpendicular to the plane.
Furthermore, by symmetry, the electric field
magnitude can depend only on the distance 9
from the plane, and must be the same at the
same distance 9 on each side.
For a Gaussian surface, use a small &quot;pillbox&quot; of cross-sectional area ( with ends located
a distance 9 from the nonconducting plane, one end on each side.
The electric flux through the sides of the pillbox is zero, since the electric field will be
parallel to the sides at all points. The electric flus through each end is ,( and the total electric
flux through the Gaussian surface is 2,(.
The electric charge on the portion of the plane within the Gaussian surface is (.
Gauss' law then leads to 2,( ~ (&deg;0 so , ~ &deg;0 . Note that it does not depend on
9 at all.
(b) [3 points] Draw a free-body diagram for the small sphere, identifying the forces on the
diagram.
There are three forces acting on the small
sphere: the gravitational force W directed
down, the tension force ;W directed up the string,
and the electric force , ~ &deg;0 directed to
the right.
(c) [3 points] Calculate the uniform charge density (in C/m2 ) for the sheet.
The force conditions for static equilibrium tell us that
vertical direction:
; cos 45&deg; ~ horizontal direction: ; sin 45&deg; ~ &deg;0
Since sin 45&deg; ~ cos 45&deg;, the left-hand sides of these two equations are equal, so the right-hand
sides must also be equal. Then ~ &deg;20 so
~ 20 &deg;
~ 2&sup2;0.50 d 10c6 kg &sup3;&sup2;10 N/kg&sup3;&sup2;8.85 d 10c C /N h m &sup3;&deg;&sup2;8.85 d 10c9 C&sup3;
~ 1 d 10c8 C/m2 &Agrave;
(d) [3 points] Calculate the tension ; (in N) of the silk thread.
The tension is most easily calculated using the vertical direction condition from part (c):
; cos 45&deg; ~ so ; ~ (0.50 d 10c6 kg)(10 m/s2 )/0.7 ~ 0&Agrave;7 d 10c5 N
MULTIPLE-CHOICE QUESTIONS
Answer these questions (worth 3 points each) on the machine-gradeable form. Note
that the multiple-choice questions on this exam are numbered 21 through 30.
Remember: no use of calculators on any part of this exam.
An object of mass ~ 0.20 kg is
constrained to move in a horizontal circle of
radius 9 ~ 0.10 m by a string of negligible
mass. At a particular instant the object has
speed # ~ 3.0 m/s and is experiencing a
W which
torque of 4.0 N h m due to a force acts on the object in a direction tangential to
the circle.
W?
21. What is the magnitude (in N) of the force (A) 10
(B) 20
(C) 30
&Egrave;(D) 40
(E) 50
The torque ~ - 9 sin 90&deg; ~ - 9 so - ~ &deg;9 ~ &sup2;4 N h m&sup3;&deg;&sup2;0.10 m&sup3; ~ 40 N
22. At what rate (in W) is the force doing work?
(A) 40
(B) 60
(C) 80
(D) 100
&Egrave;(E) 120
The rate at which the force is doing work is
7 ~ &gt; &deg;! ~ - &deg;! ~ - # ~ #&deg;9 ~ (4.0 N h m)(3.0 m/s)/(0.10 m) ~ 120 W.
23. What is the angular acceleration (in rad/s2 ) of the object?
&Egrave;(A) 2000
(B) 3000
(C) 4000
(D) 5000
(E) 6000
~ 0 so ~ &deg;0 ~ &deg;4 9 2 ~ &sup2;4.0 N h m&sup3;&deg;&sup2;0.20 kg&sup3;&sup2;0.10 m&sup3;2 ~ 2000 rad/s2 .
A playground merry-go-round is a a uniform
disk (0 ~ 4 9 2 ) of radius 9 and mass
4 ~ 10. It is free to rotate about a
frictionless vertical axle through its center.
When the merry-go-round has angular speed
a child of mass running with speed #
tangential to the disk leaps onto the edge.
Suppose # ~ 39 .
24. What is the angular momentum of the child and merry-go-round system about the
center?
(A) 39 2
(B) 59 2
&Egrave;(C) 89 2
(D) 119 2
(E) 139 2
The angular momentum of the child is #9 ~ &sup2;39 &sup3;9 ~ 39 2 and that of the merry-goround is 0 ~ &sup2; 4 9 2 &sup3; ~ &sup2;10&sup3;9 2 ~ 592 , so the total angular momentum is the sum
of these, 89 2 .
25. What is the total moment of inertia (rotational inertia) of the system about the axle?
(A) 9 2
B! 49 2
C! 59 2
&Egrave;(D) 69 2
(E) 89 2
The total moment of inertia is the moment of inertia of the child, 9 2, plus that of the merry-goround, 4 9 2 ~ &sup2;10&sup3;92 ~ 59 2 , for a total of 69 2 .
26. By what factor would the angular speed of the system increase if the child were to
crawl to the center of the merry-go-round?
(A) 1.1
&Egrave;(B) 1.2
(C) 1.3
(D) 1.4
(E) 1.5
If the child were to crawl to the center, the child's moment of inertia would become 0, and the
total moment of inertia of the system would just be the moment of inertia of the merry-go-round,
59 2 . Angular momentum would be conserved, so 0i i ~ 0f f and
2
2
f &deg; i ~ 0i &deg;0f ~ 69 &deg;59 ~ 6&deg;5 ~ 1.2. This is the factor asked for.
A torsional pendulum has a torsional torque constant ~ 12 N h m/radian. The moment
of inertia of the rotating object is 0 ~ 0.040 kg h m2 . Oscillations are set up with an
27. What is the frequency of the oscillations?
&Egrave;(A) 2&Agrave;8 Hz
(B) 3.2 Hz
(C) 3.6 Hz
(D) 4.0 Hz
(E) 4.4 Hz
~ &deg;2 ~ l&deg;0&deg;2 ~ l12/0.040&deg;6.28 ~ l300&deg;6.28 ~ 17/6.28 ~ a little less than 3.
28. What is the total mechanical energy of the system, assuming that there is zero potential
energy at the equilibrium position?
(A) 0.5 J
(B) 1.0 J
&Egrave;(C) 1.5 J
(D) 2.0 J
(E) 2.5 J
The mechanical energy of an elastic system is (2 . In the case of a torsional pendulum, 2
places the role of , and max plays the role of (, so , ~ max
~ (0.5)(12)(0.5)2 J ~ 1.5 J.
29. A meter stick of mass is suspended by two ropes, one at the 20-cm mark and one at
the 60-cm mark. Determine the tension ;1 in the rope at the 20-cm mark.
(A) 6 ;1
&sect;
;2
&sect;
20
50 60
&Egrave;(B) 4 (C) 3 &uml;
(D) 2 (E) The net torque is zero. Take the net torque about the 60-cm mark: (10 cm) c (40 cm);1 ~ 0
so ;1 ~
4
.
30. Which transition shown below represents the absorption of a photon with the least
energy?
E is correct. The transitions representing absorption of photons are those pointed up, A and E.
Of these, A is longer and thus has the greater energy, while E is shorter and thus has the lesser
energy.
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