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Chapter 3 Homework
Due: 10:00pm on Friday, February 14, 2014
You will receive no credit for items you complete after the
assignment is due. Grading Policy
Exercise 3.7
The coordinates of a bird flying in the xy-plane are given by
and , where
and .
Part A
Calculate the velocity vector of the bird as a function of
time.
Give your answer as a pair of components separated by a comma.
For example, if you think the xcomponent is 3t and the y component
is 4t, then you should enter 3t,4t. Express your answer using
twosignificant figures for all coefficients.
ANSWER:
Correct
Part B
Calculate the acceleration vector of the bird as a function of
time.
Give your answer as a pair of components separated by a comma.
For example, if you think the xcomponent is 3t and the y component
is 4t, then you should enter 3t,4t. Express your answer using
twosignificant figures for all coefficients.
ANSWER:
Correct
Part C
Calculate the magnitude of the bird's velocity at .
Express your answer using two significant figures.
ANSWER:
Correct
40 C0 50 ND0
C NT
D NT
= 02 0 NT
= 0,-2.4 0 NT
0 T
= 5.4 2 NT
Typesetting math: 43%
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Part D
Let the direction be the angle, that the vector makes with the
+x-axis measured counterclockwise. Calculate thedirection of the
bird's velocity at .
Express your answer in degrees using two significant
figures.
ANSWER:
Correct
Part E
Calculate the magnitude of the bird's acceleration at .
Express your answer using two significant figures.
ANSWER:
Correct
Part F
Calculate the direction of the bird's acceleration at .
ANSWER:
Correct
Part G
At , is the bird speeding up, slowing down or moving at constant
speed?
ANSWER:
0 T
= -63 J
0 T
2.4 NT
0 T
= -90 J
0 T
speeding up
slowing down
moving at constant speed
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Correct
Conceptual Problem about Projectile Motion
Learning Goal:
To understand projectile motion by considering horizontal
constant velocity motion and vertical constant accelerationmotion
independently.
Projectile motion refers to the motion of unpowered objects
(called projectiles) such as balls or stones moving near thesurface
of the earth under the influence of the earth's gravity alone. In
this analysis we assume that air resistance canbe neglected.
An object undergoing projectile motion near the surface of the
earth obeys the following rules:
1. An object undergoing projectile motion travels horizontally
at a constant rate. That is, the x component ofits velocity, , is
constant.
2. An object undergoing projectile motion moves vertically with
a constant downward acceleration whosemagnitude, denoted by , is
equal to 9.80 near the surface of the earth. Hence, the y component
of
its velocity, , changes continuously.
3. An object undergoing projectile motion will undergo the
horizontal and vertical motions described abovefrom the instant it
is launched until the instant it strikes the ground again. Even
though the horizontal andvertical motions can be treated
independently, they are related by the fact that they occur for
exactly thesame amount of time, namely the time the projectile is
in the air.
The figure shows the trajectory (i.e., the path) of a
ballundergoing projectile motion over level ground. The time
corresponds to the moment just after the ball islaunched from
position and . Its launch
velocity, also called the initial velocity, is .
Two other points along the trajectory are indicated in
thefigure.
One is the moment the ball reaches the peak of itstrajectory, at
time with velocity . Its position at
this moment is denoted by or
since it is at its maximum height.
The other point, at time with velocity ,
corresponds to the moment just before the ball strikesthe ground
on the way back down. At this time itsposition is , also known as (
since it is at its maximum horizontal range.
Projectile motion is symmetric about the peak, provided the
object lands at the same vertical height from which is waslaunched,
as is the case here. Hence .
Part A
How do the speeds , , and (at times , , and ) compare?
ANSWER:
2
4
# NT
2
5
0
T0
N4
N5
2
0
2
4
5
4
5
NBY
0
2
4
5
4
NBY
5
N5
5
2
2
2
0
0
0
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Correct
Here equals by symmetry and both exceed . This is because and
include vertical speed as
well as the constant horizontal speed.
Consider a diagram of the ball at time . Recall that refersto
the instant just after the ball has been launched, so it isstill at
ground level ( ). However, it is already
moving with initial velocity , whose magnitude is and direction
is
counterclockwise from the positive x direction.
Part B
What are the values of the intial velocity vector components and
(both in ) as well as the acceleration
vector components and (both in )? Here the subscript 0 means "at
time ."
Hint 1. Determining components of a vector that is aligned with
an axis
If a vector points along a single axis direction, such as in the
positive x direction, its x component will be itsfull magnitude,
whereas its y component will be zero since the vector is
perpendicular to the y direction. Ifthe vector points in the
negative x direction, its x component will be the negative of its
full magnitude.
Hint 2. Calculating the components of the initial velocity
Notice that the vector points up and to the right. Since "up" is
the positive y axis direction and "to the
right" is the positive x axis direction, and will both be
positive.
As shown in the figure, , , and are three sides of a right
triangle, one angle of which is . Thus
and can be found using the definition of the sine and cosine
functions given below. Recall that
= = > 0
= > = 0
= > > 0
> > > 0
> > = 0
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
0
0
N4
5
2
NT2
J EFHSFFT
2
4
2
5
NT
4
5
NT
0
2
2
4
2
5
2
4
2
5
2
J
2
4
2
5
NT
J EFHSFFT
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and and note that
,
What are the values of and ?
Enter your answers numerically in meters per second separated by
a comma.
ANSWER:
ANSWER:
Correct
Also notice that at time , just before the ball lands, its
velocity components are (the same
as always) and (the same size but opposite sign from by
symmetry). The
acceleration at time will have components (0, -9.80 ), exactly
the same as at , as required by Rule
2.
The peak of the trajectory occurs at time . This is the point
where the ball reaches its maximum height . At the
peak the ball switches from moving up to moving down, even as it
continues to travel horizontally at a constant rate.
Part C
What are the values of the velocity vector components and (both
in ) as well as the acceleration
vector components and (both in )? Here the subscript 1 means
that these are all at time .
ANSWER:
NT2
J EFHSFFT
TJOJ
MFOHUI PG PQQPTJUF TJEF
MFOHUI PG IZQPUFOVTF
2
5
2
DPTJ
MFOHUI PG BEKBDFOU TJEF
MFOHUI PG IZQPUFOVTF
2
4
2
2
4
2
5
15.0,26.0 NT
30.0, 0, 0, 0
0, 30.0, 0, 0
15.0, 26.0, 0, 0
30.0, 0, 0, -9.80
0, 30.0, 0, -9.80
15.0, 26.0, 0, -9.80
15.0, 26.0, 0, +9.80
0
NT2
4
NT2
5
2
5
0
NT
0
0
5
NBY
2
4
2
5
NT
4
5
NT
0
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Correct
At the peak of its trajectory the ball continues traveling
horizontally at a constant rate. However, at thismoment it stops
moving up and is about to move back down. This constitutes a
downward-directed change invelocity, so the ball is accelerating
downward even at the peak.
The flight time refers to the total amount of time the ball is
in the air, from just after it is launched ( ) until just before
itlands ( ). Hence the flight time can be calculated as , or just
in this particular situation since .Because the ball lands at the
same height from which it was launched, by symmetry it spends half
its flight timetraveling up to the peak and the other half
traveling back down. The flight time is determined by the initial
verticalcomponent of the velocity and by the acceleration. The
flight time does not depend on whether the object is
movinghorizontally while it is in the air.
Part D
If a second ball were dropped from rest from height , how long
would it take to reach the ground? Ignore air
resistance.
Check all that apply.
Hint 1. Kicking a ball of cliff; a related problem
Consider two balls, one of which is dropped from rest off the
edge of a cliff at the same moment that theother is kicked
horizontally off the edge of the cliff. Which ball reaches the
level ground at the base of the clifffirst? Ignore air
resistance.
Hint 1. Comparing position, velocity, and acceleration of the
two balls
Both balls start at the same height and have the same initial y
velocity ( ) as well as the
same acceleration ( downward). They differ only in their x
velocity (one is zero, the other
nonzero). This difference will affect their x motion but not
their y motion.
ANSWER:
0, 0, 0, 0
0, 0, 0, -9.80
15.0, 0, 0, 0
15.0, 0, 0, -9.80
0, 26.0, 0, 0
0, 26.0, 0, -9.80
15.0, 26.0, 0, 0
15.0, 26.0, 0, -9.80
0
0
0
0
0
0
5
NBY
2
5
#
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ANSWER:
Correct
In projectile motion over level ground, it takes an object just
as long to rise from the ground to the peak as ittakes for it to
fall from the peak back to the ground.
The range \texttip{R}{R} of the ball refers to how far it moves
horizontally, from just after it is launched until just before
itlands. Range is defined as x_2 - x_0, or just \texttip{x_{\rm
2}}{x_2} in this particular situation since x_0 = 0.
Range can be calculated as the product of the flight time
\texttip{t_{\rm 2}}{t_2} and the x component of the
velocity\texttip{v_{\mit x}}{v_x} (which is the same at all times,
so v_x = v_{0,x}). The value of \texttip{v_{\mit x}}{v_x} can
befound from the launch speed \texttip{v_{\rm 0}}{v_0} and the
launch angle \texttip{\theta }{theta} using trigonometricfunctions,
as was done in Part B. The flight time is related to the initial y
component of the velocity, which may also befound from
\texttip{v_{\rm 0}}{v_0} and \texttip{\theta }{theta} using trig
functions.The following equations may be useful in solving
projectile motion problems, but these equations apply only to
aprojectile launched over level ground from position (x_0 = y_0 =
0) at time t_0 = 0 with initial speed \texttip{v_{\rm 0}}{v_0} and
launch angle \texttip{\theta }{theta} measured from the horizontal.
As was the case above, \texttip{t_{\rm 2}}{t_2} refers to the
flight time and \texttip{R}{R} refers to the range of the
projectile.
flight time: \large{t_2 = \frac{2 v_{0, y}}{g} = \frac{2 v_0
\sin(\theta)}{g}}
range: \large{R = v_x t_2 = \frac{v_0 2^ \sin(2\theta)}{g}}
In general, a high launch angle yields a long flight time but a
small horizontal speed and hence little range. A low launchangle
gives a larger horizontal speed, but less flight time in which to
accumulate range. The launch angle that achievesthe maximum range
for projectile motion over level ground is 45 degrees.
Part E
Which of the following changes would increase the range of the
ball shown in the original figure?
Check all that apply.
ANSWER:
The ball that falls straight down strikes the ground first.
The ball that was kicked so it moves horizontally as it falls
strikes the ground first.
Both balls strike the ground at the same time.
0
0
0
0
0
0
0
0
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Correct
A solid understanding of the concepts of projectile motion will
take you far, including giving you additionalinsight into the
solution of projectile motion problems numerically. Even when the
object does not land at thesame height from which is was launched,
the rules given in the introduction will still be useful.Recall
that air resistance is assumed to be negligible here, so this
projectile motion analysis may not be thebest choice for describing
things like frisbees or feathers, whose motion is strongly
influenced by air. The valueof the gravitational free-fall
acceleration \texttip{g}{g} is also assumed to be constant, which
may not be
appropriate for objects that move vertically through distances
of hundreds of kilometers, like rockets ormissiles. However, for
problems that involve relatively dense projectiles moving close to
the surface of theearth, these assumptions are reasonable.
Exercise 3.10
A daring 510-{\rm N} swimmer dives off a cliff with a running
horizontal leap, as shown in the figure .
Part A
What must her minimum speed be just as she leaves the top of the
cliff so that she will miss the ledge at thebottom, which is 1.75
{\rm m} wide and 9.00 {\rm m} below the top of the cliff?
ANSWER:
Increase \texttip{v_{\rm 0}}{v_0} above 30 \rm{m/s}.
Reduce \texttip{v_{\rm 0}}{v_0} below 30 \rm{m/s}.
Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to 45
\rm{degrees}.
Reduce \texttip{\theta }{theta} from 60 \rm{degrees} to less
than 30 \rm{degrees}.
Increase \texttip{\theta }{theta} from 60 \rm{degrees} up toward
90 \rm{degrees}.
v_0 = 1.29 {\rm m/s}
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Correct
Exercise 3.12
A rookie quarterback throws a football with an initial upward
velocity component of 15.9{\rm m/s} and a horizontalvelocity
component of 19.9{\rm m/s} . Ignore air resistance.
Part A
How much time is required for the football to reach the highest
point of the trajectory?
Express your answer using three significant figures.
ANSWER:
Correct
Part B
How high is this point?
Express your answer using three significant figures.
ANSWER:
Correct
Part C
How much time (after it is thrown) is required for the football
to return to its original level?
Express your answer using three significant figures.
ANSWER:
Correct
Part D
How does this compare with the time calculated in part (a).
t_1 = 1.62 {\rm s}
h = 12.9 {\rm m}
t_2 = 3.24 {\rm s}
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Express your answer using three significant figures.
ANSWER:
Correct
Part E
How far has it traveled horizontally during this time?
Express your answer using three significant figures.
ANSWER:
Correct
Exercise 3.13
A car comes to a bridge during a storm and finds the bridge
washed out. The driver must get to the other side, so hedecides to
try leaping it with his car. The side the car is on is 22.4{\rm m}
above the river, while the opposite side is amere 1.6{\rm m} above
the river. The river itself is a raging torrent 57.0{\rm m}
wide.
Part A
How fast should the car be traveling just as it leaves the cliff
in order just to clear the river and land safely on theopposite
side?
ANSWER:
Correct
Part B
What is the speed of the car just before it lands safely on the
other side?
ANSWER:
Correct
\large{\frac{t_2}{t_1}} = 2.00
x = 64.6 {\rm m}
v_0 = 27.7 {\rm m/s}
v = 34.3 {\rm m/s}
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Exercise 3.15
Inside a starship at rest on the earth, a ball rolls off the top
of a horizontal table and lands a distance D from the foot ofthe
table. This starship now lands on the unexplored Planet X. The
commander, Captain Curious, rolls the same ball offthe same table
with the same initial speed as on earth and finds that it lands a
distance 2.43 D from the foot of thetable.
Part A
What is the acceleration due to gravity on Planet X?
ANSWER:
Correct
Exercise 3.19
In a carnival booth, you win a stuffed giraffe if you toss a
quarter into a small dish. The dish is on a shelf above the
pointwhere the quarter leaves your hand and is a horizontal
distance of 2.1 {\rm m} from this point (the figure ). If you toss
thecoin with a velocity of 6.4 {\rm{ m/s}} at an angle of 60
\^circabove the horizontal, the coin lands in the dish. You
canignore air resistance.
Part A
What is the height of the shelf above the point where the
quarter leaves your hand?
Express your answer using two significant figures.
ANSWER:
g_{\rm X} = 1.66 {\rm m/s 2^}
H = 1.5 {\rm m}
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Correct
Part B
What is the vertical component of the velocity of the quarter
just before it lands in the dish?
Express your answer using two significant figures.
ANSWER:
Correct
Problem 3.51
A jungle veterinarian with a blow-gun loaded with a tranquilizer
dart and a sly 1.5-{\rm kg} monkey are each a height25{\rm m} above
the ground in trees a distance 70{\rm m} apart. Just as the hunter
shoots horizontally at the monkey,the monkey drops from the tree in
a vain attempt to escape being hit.
Part A
What must the minimum muzzle velocity of the dart have been for
the hunter to hit the monkey before it reached theground?
Express your answer using two significant figures.
ANSWER:
Correct
Circular Launch
A ball is launched up a semicircular chute in such a way that at
the top of the chute, just before it goes into free fall, theball
has a centripetal acceleration of magnitude 2 \texttip{g}{g}.
v_y = -0.89 {\rm m/s}
v = 31 {\rm m/s}
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Part A
How far from the bottom of the chute does the ball land?
Your answer for the distance the ball travels from the end of
the chute should contain \texttip{R}{R}.
Hint 1. Speed of ball upon leaving chute
How fast is the ball moving at the top of the chute?
Hint 1. Equation of motion
The centripetal acceleration for a particle moving in a circle
is a_c = v^ 2/r, where \texttip{v}{v} is its
speed and \texttip{r}{r} is its instantaneous radius of
rotation.
ANSWER:
Incorrect; Try Again; 5 attempts remaining
Hint 2. Time of free fall
How long is the ball in free fall before it hits the ground?
Express the free-fall time in terms of \texttip{R}{R} and
\texttip{g}{g}.
Hint 1. Equation of motion
There is constant acceleration due to gravity, so you can use
the general expression\large{y(t) = y_0 + v_{y0}t + \frac{a_y t
2^}{2}}.
\texttip{v}{v} = \sqrt{\left(2 g R\right)}
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Write the values of \texttip{y_{\rm 0}}{y_0}, \texttip{v_{\rm
y0}}{v_y0}, and \texttip{a_{\mit y}}{a_y}
(separated by commas) that are appropriate for this situation.
Use the standard convention that \texttip{g}{g} is the magnitude of
the acceleration due to gravity. Take y = 0 at the ground, and
take
the positive y direction to be upward.
ANSWER:
Hint 2. Equation for the height of the ball
To find the time in free fall before the ball hits the ground,
\texttip{t_{\rm f \hspace{1 pt}}}{t_f}, set the
general equation for the height equal to the height of the
ground.
Answer in terms of \texttip{t_{\rm f \hspace{1 pt}}}{t_f},
\texttip{R}{R}, and \texttip{g}{g}.
ANSWER:
ANSWER:
Hint 3. Finding the horizontal distance
The horizontal distance follows from D = x(t_{\rm f \hspace{1
pt}}) = x_0 + v_{x0}t_{\rm f \hspace{1 pt}},
where x_0=0. \texttip{v_{\rm x0}}{v_x0} and \texttip{t_{\rm f
\hspace{1 pt}}}{t_f} were found in Parts i and ii
respectively.
ANSWER:
Correct
Direction of Acceleration of Pendulum
Learning Goal:
To understand that the direction of acceleration is in the
direction of the change of the velocity, which is unrelated to
thedirection of the velocity.
The pendulum shown makes a full swing from -\pi/4 to + \pi/4.
Ignore friction and assume that the string is massless.The eight
labeled arrows represent directions to be referred to when
answering the following questions.
\texttip{y_{\rm 0}}{y_0}, \texttip{v_{\rm y0}}{v_y0},
\texttip{a_{\mit y}}{a_y} = 2 R,0,-g
y(t_{\rm f \hspace{1 pt}})=0 = \large{2 R-\frac{g}{2} {t_{f}}
{^2}}
\texttip{t_{\rm f \hspace{1 pt}}}{t_f} = \large{2
\sqrt{\frac{R}{g}}}
\texttip{D}{D} = \large{\sqrt{19.6 R} \sqrt{\frac{4R}{9.8}}}
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Part A
Which of the following is a true statement about the
acceleration of the pendulum bob, \texttip{\vec{a}}{a_vec}.
ANSWER:
Correct
Part B
What is the direction of \texttip{\vec{a}}{a_vec} when the
pendulum is at position 1?
Enter the letter of the arrow parallel to
\texttip{\vec{a}}{a_vec}.
Hint 1. Velocity at position 1
What is the velocity of the bob when it is exactly at position
1?
ANSWER:
Correct
Hint 2. Velocity of bob after it has descended
\texttip{\vec{a}}{a_vec} is equal to the acceleration due to
gravity.
\texttip{\vec{a}}{a_vec} is equal to the instantaneous rate of
change in velocity.
\texttip{\vec{a}}{a_vec} is perpendicular to the bob's
trajectory.
\texttip{\vec{a}}{a_vec} is tangent to the bob's trajectory.
\texttip{v_{\rm 1}}{v_1} = 0 {\rm m/s}
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What is the velocity of the bob just after it has descended from
position 1?
ANSWER:
Correct
ANSWER:
Correct
Part C
What is the direction of \texttip{\vec{a}}{a_vec} at the moment
the pendulum passes position 2?
Enter the letter of the arrow that best approximates the
direction of \texttip{\vec{a}}{a_vec}.
Hint 1. Instantaneous motion
At position 2, the instantaneous motion of the pendulum can be
approximated as uniform circular motion.What is the direction of
acceleration for an object executing uniform circular motion?
ANSWER:
Correct
We know that for the object to be traveling in a circle, some
component of its acceleration must be pointingradially inward.
Part D
What is the direction of \texttip{\vec{a}}{a_vec} when the
pendulum reaches position 3?
Give the letter of the arrow that best approximates the
direction of \texttip{\vec{a}}{a_vec}.
very small and having a direction best approximated by arrow
D
very small and having a direction best approximated by arrow
A
very small and having a direction best approximated by arrow
H
The velocity cannot be determined without more information.
H
C
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Hint 1. Velocity just before position 3
What is the velocity of the bob just before it reaches position
3?
ANSWER:
Hint 2. Velocity of bob at position 3
What is the velocity of the bob when it reaches position 3?
ANSWER:
ANSWER:
Correct
Part E
As the pendulum approaches or recedes from which position(s) is
the acceleration vector \texttip{\vec{a}}{a_vec}
almost parallel to the velocity vector
\texttip{\vec{v}}{v_vec}.
ANSWER:
Correct
Exercise 3.25
The earth has a radius of 6380 {\rm km} and turns around once on
its axis in 24 {\rm h}.
very small and having a direction best approximated by arrow
B
very small and having a direction best approximated by arrow
C
very small and having a direction best approximated by arrow
H
The velocity cannot be determined without more information.
\texttip{v_{\rm 3}}{v_3} = 0 {\rm m/s}
F
position 2 only
positions 1 and 2
positions 2 and 3
positions 1 and 3
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Part A
What is the radial acceleration of an object at the earth's
equator? Give your answer in {\rm{m/s}} 2^.
ANSWER:
Correct
Part B
What is the radial acceleration of an object at the earth's
equator? Give your answer as a fraction of {\it g}.
ANSWER:
Correct
Part C
If a_{{\rm{rad}}} at the equator is greater than {\it g} ,
objects would fly off the earth's surface and into space. What
would the period of the earth's rotation have to be for this to
occur?
ANSWER:
Correct
Exercise 3.29
A Ferris wheel with radius 14.0 {\rm m} is turning about a
horizontal axis through its center (the figure ). The linear
speedof a passenger on the rim is constant and equal to 6.40{\rm
{\rm m/s}} .
a_{\rm rad} = 3.40102 {\rm{m/s}} 2^
a_{\rm rad} = 3.40103 {\it g}
T = 5070 {\rm s}
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Part A
What is the magnitude of the passenger's acceleration as she
passes through the lowest point in her circularmotion?
ANSWER:
Correct
Part B
What is the direction of the passenger's acceleration as she
passes through the lowest point in her circular motion?
ANSWER:
Correct
Part C
What is the magnitude of the passenger's acceleration as she
passes through the highest point in her circularmotion?
ANSWER:
a = 2.93 {\rm m/s 2^}
towards the center
outwards the center
a = 2.93 {\rm m/s 2^}
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3/3/2014 Chapter 3 Homework
http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2766014
20/21
Correct
Part D
What is the direction of the passenger's acceleration as she
passes through the highest point in her circularmotion?
ANSWER:
Correct
Part E
How much time does it take the Ferris wheel to make one
revolution?
ANSWER:
Correct
Exercise 3.30
At its Ames Research Center, NASA uses its large 20-G centrifuge
to test the effects of very large accelerations(hypergravity) on
test pilots and astronauts. In this device, an arm 8.84 {\rm m}
long rotates about one end in ahorizontal plane, and the astronaut
is strapped in at the other end. Suppose that he is aligned along
the arm with hishead at the outermost end. The maximum sustained
acceleration to which humans are subjected in this machine
istypically 12.5 {\it g}.
Part A
How fast must the astronaut's head be moving to experience this
maximum acceleration?
ANSWER:
Correct
Part B
towards the center
outwards the center
T = 13.7 {\rm s}
v = 32.9 {\rm m/s}
-
3/3/2014 Chapter 3 Homework
http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2766014
21/21
What is the difference between the acceleration of his head and
feet if the astronaut is 2.00 {\rm m} tall?
ANSWER:
Correct
Part C
How fast in rpm \left( {\rm rev/min} \right) is the arm turning
to produce the maximum sustained acceleration?
ANSWER:
Correct
Problem 3.88
A projectile is thrown from a point P. It moves in such a way
that its distance from P is always increasing.
Part A
Find the maximum angle above the horizontal with which the
projectile could have been thrown. You can ignore
airresistance.
ANSWER:
Correct
Score Summary:
Your score on this assignment is 98.3%.You received 13.77 out of
a possible total of 14 points.
\Delta a = 27.7 {\rm m/s 2^}
\large{\frac{1}{T}} = 35.5 {\rm rpm}
\phi = 70.5 \^circ