General Physics – ph 212, sec2
Midterm I (Ch 12 – 14)
April 28, 2006
Name:____________
Type A
Exam is closed book and closed notes. Use only your note card.
Write all work and answers in the color papers provided.
Show all your work and explain your reasoning (No credit will be given for an answer
that does not include the necessary solution or explanation, except for true/false or
multiple choice questions)
Partial credit may be awarded for a correct method of solution, even if the answer is wrong.
Part I – True or False (3 points each): For questions 1 – 7, state whether each statement is true or false.
1. The "shimmy" or wobble of a car's wheel can be corrected by static balancing. False (wheel must be
balanced while rotating)
2. If you are standing on a weighing scale and all of a sudden the atmosphere vanished, then the
reading on the scale would decrease. False
3. An object in motion with constant speed along a straight line can never have nonzero angular
momentum. False (L = mvr)
4. A x B is the same as B x A False (A x B = – B x A)
5. An object is in equilibrium only when there are no forces acting on it. False
6. The energy of a damped, undriven oscillator decreases exponentially with time. True
Part II – Multiple Choice (2 points each): Choose the one correct answer for each of the following
questions that best answers or completes the question.
9.
Suppose you are at the center of a large freely-rotating horizontally platform in an amusement park. As
you crawl to the edge, the angular momentum of you and the platform
A) decreases
B) increases
C) remains the same as well as the angular speed.
D) decreases in the direction proportional to your decrease in angular speed.
E) remains the same but the angular speed decreases.
10. Which of the following is an example of an object in neutral equilibrium?
A) A chair resting on the floor.
C) A sphere resting atop another sphere
B) A pencil standing on its point.
D) A sphere resting on a flat horizontal surface
neutral equilibrium The state of equilibrium possessed by a body that will stay at rest if moved into a new position; it will
neither move back to its original position nor move on any further. A sphere placed on a horizontal surface is an example
11. What is the magnitude of the vector cross product of two vectors each of magnitude 2 that add up to
zero?
A) 2
B) 4
C) square root of 2
D) square root of 8
E) 0
Under this condition A = –B  A + B = 0 , cross product of parallel vectors is zero
12. Under what condition does the equation
= dL/dt apply for a system of particles?
A) Under all conditions.
B) Only if the system consists of a single particle.
C) Only if the center of mass of the system is at rest.
D) If the quantities are calculated in a rotating reference frame or about the center of mass.
E) If the quantities are calculated in an inertial reference frame or about the center of mass.
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1. Pendulum A has a bob of mass MA and length LA while pendulum B has a bob mass of MB and
length LB. If TA = 2TB, then
A) LA = 4LB and MA = 4MB
C) LA = 4LB for any mass ratio MA/MB.
B) LA = 2LB and MA = 2MB
D) LA = √2 LB for any mass ratio MA/MB.
13. The acceleration functions for five different objects are given below. Which object will move
with SHM?
A) a = –7t
D) a = 7cos(–2x)
B) a = –7
E) a = 7x
C) a = –7x
14. If A is the amplitude of a vibrating mass on a spring, then the mass in one period travels a
distance:
A) 0
B) A
C) 2A
D) 4A
Part III – Problems: Show your work clearly and completely for each of the following problems.
15. (1 point) What happens to you if the problems on this test scared half to death twice?
16. (16 points)The particle of mass m starts from a height h and slides down the frictionless ramp
and collides with a uniform vertical rod, sticking to it. The rod has a length l and mass M and
is pivoted at its end as shown. After collision, the rod swings through an angel of before
momentarily coming to rest. In terms of h, m, M, and l :
A) find the speed of the particle just before colliding
with the rod.
Total mechanical energy is conserved between points (1)
and (2) (just before collision)
E1 = E2  ½ mv12 + mgy1 = ½ mv22 + mgy2
0 + mgh = ½ mv22 + 0
v1
2 gh
B) find .
Here the total angular momentum (f the particle-rod system) is conserved. Between
points (2) and (3) (right after collision):
L2 = L3  (Lparticle + Lrod)2 = (L particle + L rod)2
mv2r + Irod 2 = Iparticle 2 + Irod 2
m 2 gh l
2
0 ml 2
2
1 2
Ml
3
2
3m( 2 gh )
(3m M )l
Note: acceleration is NOT constant for rotation of rod and particle (i.e you can not use
kinematics eqs).
To find use conservation of energy between points 2 and 3, or Newton’s 2nd law for
rotation. I’ll use the latter, but conservation of energy is easier.
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I
I
d
dt
I
d d
d dt
I
d
d
Which chain rule was used and the fact that d /dt =
l
( Mg sin
mgl sin )
2
Mgl
(
sin
mgl sin )d
0
2
d
d
I
e
0
I d
2
2 0
(M
2m) gl cos
2I
0
2
2
2
(M
2m) gl cos
(M
2m) gl
2m 2 h
(3m M )
( M 2m)l
1 cos
I
cos
1
( Ml 2
3
2
2
1
1
(M
3m( 2 gh )
ml )
(3m M )l
2
6m 2 h
2m)(M
3m)l
17. (15 points)Consider a ladder with a painter climbing up it as
shown. If the mass of the ladder is 12.0 kg, the mass of the
painter is 55.0 kg, and the ladder begins to slip at its base when
her feet are 70% of the way up the length of the ladder, what is the
coefficient of static friction between the ladder and the floor?
Assume the wall is frictionless.
All forces and their components are nicely drawn for you.
f smax
s
f smax
n
F Gx
and
n
F Gy
F Gx
s
F Gy
Fx
0
F Gx
FW
0
Fy
0
F Gy
mg
Mg
F Gx
0
FW
F Gy
(m
M )g
To calculate Fw, calculate total torques about point A (the foot of the ladder):
0 with
rF sin
mg (1.5m) Mg (2.1m)
s
s
or
FW (4.0m) 0
mg (1.5m) Mg (2.1m)
4(m M ) g
0.50
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rF
s
(12.0kg )(1.5m) 55.0kg (2.1m)
4(12.0 55.0)kg
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18. (20 points) (Recall mother of all labs) A 250-gram mass is attached to a spring and oscillates
in SHM along an x-axis with its displacement (in meters) given by:
x(t) = 0.25 + 0.50 cos(2.0t + /6)
a. Make a sketch of the displacement of the object as a
Position vs time
function of time for several periods.
0.90
Xeq = (xmax + |xmin|)/2 = (0.75m +0.25)/2
Xeq = 0.25m
0.75
Position (m)
b. What is the equilibrium position of the
object? Equilibrium is at the horizontal
line where the graph is symmetric about
(the mass oscillates about this point)
0.60
0.45
0.30
0.15
0.00
-0.150.00
3.00
6.00
9.00
12.00
15.00
-0.30
time (sec)
c. What is the amplitude, period, maximum and minimum
displacement for this motion?
By comparing (inspection) the given equation with the general equation:
x(t) = Xeq + A cos( t + )
x(t) = 0.25 + 0.50 cos(2.0t + /6)
A = 0.50m
= 2.0 rad/s  T = 2 /
 T = sec
xmax = 0.75m
xmin = – 0.25m
d.
Determine the velocity and acceleration of the object as a function of time.
v = dx/dt

v(t) = –1.0 sin(2.0t + /6)
a = dv/dt

a(t) = –2.0 cos(2.0t + /6)
e. What is the force acting on the object both as a function of time and also as a function
of displacement x?
F(t) = ma(t) 
F(t) = (0.25kg)(–2.0 cos(2.0t + /6))
F(t) = –.50 cos(2.0t + /6)
Since x(t) = 0.25 + 0.50 cos(2.0t + /6)  x(t) – 0.25 = 0.50 cos(2.0t + /6)
Therefore
F(t) = –.50 cos(2.0t + /6)  F(x) = –( x(t) – 0.25 )
f. What is the total mechanical energy of the object?
E = ½ kA2
where
k
m
k 1.0
k
0.25kg
2.0 sec
1
kg
sec 2
E = ½ (1.0kg/sec2)(0.50m)2  E = 0.125 J (costant)
g. Determine the position, velocity, acceleration, force, and total mechanical energy of the
object at t = 1.0 s.
Insert t = 1.0 in x(t), v(t), a(t), and F(t) equations (show your work)
Ans: x(1.0) = – 0.16 m
v(1.0) = – 0.58 m/s a(1.0) = 1.6 m/sec2
F(1.0) = 0.41 N
E = 0.125 J
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Do either problem 19 or 20; please note the points on each problem.
Everyone has a photographic memory. Some just don't have film.
19. (18 points) A piston of mass m is placed at the midpoint of a closed cylinder of cross sectional
area of A and length 2L as shown. Assume the pressure P on either side of the piston obeys
Boyle’s law (Boyle’s Law: At constant
temperature the product of pressure and
volume of a gas is always constant; i.e.
P0V0 = P1V1). Let at x = 0, the pressure of
the gas on both sides of the piston be Po.
A) Draw the freebody diagram showing all
forces acting on the piston when it is
displaced a small amount x.
F2 – F1 = Fnet
As the piston moves an amount x to the right, pressure on its right side increases (air is
compressed) while on its left side decreases.
(this is in Ch 15): P = F/A
B)
Find the equation of motion of the piston when it is disturbed from its equilibrium position
x = 0. Is the equation of motion linear or non-linear? Explain.
F2 – F1 = Fnet
Fnet ma
F2 F1 ma
F
but P
P2 A P1 A ma
A
Boyle' s Law :
P0V0 P1V1 P2V2
and V1 A( L x) and V2 A( L x) with V0 AL
Now
P2 A P1 A
P0V0
V2
ma
P0V0
A
V1
m
d 2x
dt 2
2 P0 ALx
d 2x
m
Not SHM
L2 x 2
dt 2
Since acceleration is not linearly proportional to –x , the piston will not execute SHM.
x
Expanding the non-linear term ( 2
) in a Taylor series about equilibrium position (x =
L x2
0). Linearize the equation of motion by keeping only the linear term of the expansion. What
type of motion does your new equation of motion execute? Explain.
D) What is the period of oscillation? Make sure your units work out correctly.
x
Expanding 2
using Taylor’s expansion about x = a = 0 will result:
L x2
Finally insert for V ' s :
C)
( x a) f (a) ( x a) 2 f (a)
....
1!
2!
x
1( L2 x 2 ) 2 x( x)
f ( x)
and
f
(
x
)
L2 x 2
( L2 x 2 ) 2
Since x is small, we will ignore all higher derivatives after first:
Taylor’s Expansion: f ( x)
f ( x)
f (a)
f (0)
( x 0) f (0)
1!
f ( x) 0
x
L2
x
L2
Inserting the above in the equation of motion will result:
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2 P0 ALx
L2 x 2
m
d 2x
dt 2
2 P0 ALx
L2
m
This in now a simple harmonic motion with:
So the period of oscillation is:
d 2x
dt 2
d 2x
dt 2
d 2x
dt 2
T=2 / 
2
T
2
x
2
2 P0 AL
x
m L2
2 P0 A
mL
mL
P0 A
20. This problem is deleted from this exam. But don’t worry; you’ll see it on your test on Monday
(it is similar to above).
This problem will not need any approximation, but uses triangle trigonometry and theory of
gravitation.
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