Answer, Key – Homework 6 – David McIntyre – 45123 – Mar 25, 2004
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Chapters 5 and 6 problems.
001 (part 1 of 2) 5 points
An amusement park ride consists of a rotating
circular platform 11.4 m in diameter from
which 10 kg seats are suspended at the end
of 1.18 m massless chains. When the system
rotates, the chains make an angle of 43.7◦ with
the vertical.
The acceleration of gravity is 9.8 m/s2 .
002 (part 2 of 2) 5 points
If a child of mass 62.5 kg sits in a seat, what is
the tension in the chain (for the same angle)?
Correct answer: 982.756 N.
Explanation:
M = 62.5 kg
From the first part we have
T cos θ = (m + M ) g
(m + M ) g
T =
cos θ
(10 kg + 62.5 kg) (9.8 m/s2 )
=
cos 43.7◦
= 982.756 N .
d
l
1
θ
What is the speed of each seat?
Correct answer: 7.81126 m/s.
Explanation:
In the vertical direction we have
003 (part 1 of 1) 0 points
A figure of a dancer on a music box moves
counterclockwise at constant speed around
the path shown below. The path is such
that the lengths of its segments, P Q, QR,
RS, and SP , are equal. Arcs QR and SP are
semicircles.
S
R
P
T cos θ = m g ,
where T is the tension in the chain. In the
horizontal direction we have
T sin θ =
m v2
.
r
Since
r = ` sin θ +
d
2
= (1.18 m) sin 43.7◦ +
Q
Which of the following best represents the
magnitude of the dancer’s acceleration as a
function of time t during one trip around the
path, beginning at point P ?
a
1.
O
tP tQ tR tS tP
11.4 m
2
2.
t
a
= 6.51524 m ,
we have
p
v = g r tan θ
q
= (9.8 m/s2 ) (6.51524 m) tan 43.7◦
= 7.81126 m/s .
O
tP tQ tR tS tP
3.
t
a
O
tP tQ tR tS tP
t
Answer, Key – Homework 6 – David McIntyre – 45123 – Mar 25, 2004
2
p
a
3. v = 2 µ g R
4.
s
gR
t
correct
4. v =
O
tP tQ tR tS tP
µ
1p
a
5.
5. v =
gR
µ
p
6.
v
=
2
gR
t
O
tP tQ tR tS tP
p
7. v = µ g R
correct
p
Explanation:
8. v = µ 2 π g R
During RS and P Q, the acceleration is zero
p
because the velocity is a constant both in mag9. v = 2 µ g R
nitude and in direction. During QR and SP ,
p
even though the magnitude of the velocity is
10. v = g R
a constant, it changes its direction, which will
Explanation:
v2
result in an acceleration of , where v is the
Basic Concepts:
r
m v2
velocity and r is the radius of the arc. So the
Centripetal force: F =
acceleration is fixed at a non-zero constant
r
Frictional force: fs ≤ µ N = fsmax
during SP and QR.
Solution: The maximum frictional force
due
to friction is fmax = µ N , where N is the
004 (part 1 of 2) 0 points
inward directed normal force of the wall of
An amusement park ride consists of a large
the cylinder on the person. To support the
vertical cylinder that spins about its axis
person vertically, this maximal friction force
fast enough that any person inside is held
fsmax must be larger than the force of gravity
up against the wall when the floor drops away
m g so that the actual force, which is less
(see figure). The coefficient of static friction
than
µN , can take on the value m g in the
between the person and the wall is µ and the
positive
vertical direction. Now, the normal
radius of the cylinder is R.
v2
force supplies the centripetal acceleration
R
R
on the person, so from Newton’s second law,
N =
m v2
.
R
Since
ω
What is the minimum tangential velocity
needed to keep the person from slipping downward?
1p
gR
2
p
2. v = µ 2 g R
1. v =
fsmax
µ m v2
= µN =
≥ mg,
R
the minimum speed required to keep the person supported is at the limit of this inequality,
which is
2
µ m vmin
= m g, or
R
µ
¶1
gR 2
vmin =
.
µ
Answer, Key – Homework 6 – David McIntyre – 45123 – Mar 25, 2004
005 (part 2 of 2) 0 points
Suppose a person whose mass is m is being
held up against the wall with a constant tangential velocity v greater than the minimum
necessary.
Find the magnitude of the frictional force
between the person and the wall.
1. F =
m v2
µR
3
Noting that the maximum centripetal force is
required when the race car is moving at the
maximum allowed speed without slipping, we
have
2
m vmax
= µs m g
R
p
=⇒ vmax = µs g R
q
= (0.293)(9.8 m/s2 )(59.1 m)
= 13.0269 m/s .
2. F = m g correct
m v2
3. F =
R
µ m v2
4. F =
R
µ m v2
R
m v2
= µmg +
R
2
µmv
− mg
=
R
mg
=
µ
m v2
=
− µmg
R
5. F = m g +
6. F
7. F
8. F
9. F
10. F = µ m g
Explanation:
The vertical friction force must equal m g,
in order to balance the force of gravity and not
have any acceleration in the vertical direction.
006 (part 1 of 2) 5 points
A car of mass 629 kg travels around a flat,
circular race track of radius 59.1 m. The coefficient of static friction between the wheels
and the track is 0.293.
The acceleration of gravity is 9.8 m/s2 .
What is the maximum speed v that the car
can go without flying off the track?
Correct answer: 13.0269 m/s.
Explanation:
When the race car is going around a curved
track, the maximum centripetal force that can
be applied by the frictional force is fs = µs mg.
007 (part 2 of 2) 5 points
The same car now travels on a straight track
and goes over a hill with radius 194 m at the
top.
What is the maximum speed that the car
can go over the hill without leaving the road?
Correct answer: 43.6028 m/s.
Explanation:
m v2
= mg − N
r
where N is the normal force acting on the
car from the ground. The car will fly off the
ground just when N = 0 so the maximum
speed allowed will be
√
vmax = g r
q
= (9.8 m/s2 )(194 m)
= 43.6028 m/s .
008 (part 1 of 4) 3 points
The following figure shows a Ferris wheel that
rotates 5 times each minute and has a diameter of 20 m.
The acceleration of gravity is 9.8 m/s2 .
What is the centripetal acceleration of a
rider?
Correct answer: 2.74156 m/s2 .
Answer, Key – Homework 6 – David McIntyre – 45123 – Mar 25, 2004
Explanation:
The period of the Ferris wheel is
T = 60 s/5 = 12 s .
The speed of the wheel is
2πr
T
2 π (10 m)
=
12 s
= 5.23599 m/s ,
v=
so the centripetal acceleration is
a=
4
What force (magnitude) does the seat exert
on a rider when the rider is halfway between
top and bottom?
Correct answer: 620.751 N.
Explanation:
In this case, the force exerted by the seat
has two components: the vertical one balancing the gravity and the horizontal one providing the centripetal force. Thus we have
p
Fm = m g 2 + a2
q
= (61 kg) (9.8 m/s2 )2 + (2.74156 m/s2 )2
= 620.751 N .
v2
r
(5.23599 m/s)2
=
10 m
= 2.74156 m/s2 .
009 (part 2 of 4) 3 points
What force does the seat exert on a 61 kg
rider at the lowest point of the ride?
Correct answer: 765.035 N.
Explanation:
The force exerted by the seat balances the
gravity and provides the centripetal force, so
Fl = m [g + a]
= (61 kg) (9.8 m/s2 + 2.74156 m/s2 )
= 765.035 N
010 (part 3 of 4) 2 points
What force does the seat exert on a 61 kg
rider at the highest point of the ride?
Correct answer: 430.565 N.
Explanation:
The gravity is partly balanced by the force
exerted by the seat and this resultant provides
the centripetal force, so
Fl = m [g − a]
= (61 kg) (9.8 m/s2 − 2.74156 m/s2 )
= 430.565 N .
011 (part 4 of 4) 2 points
012 (part 1 of 1) 0 points
A car traveling on a straight road at 7.9 m/s
goes over a hump in the road. The hump
may be regarded as an arc of a circle of radius
10 m.
The acceleration of gravity is 9.8 m/s2 .
What is the apparent weight of a(n) 381 N
woman in the car as she rides over the hump?
Correct answer: 138.365 N.
Explanation:
There are two forces acting on the woman
of mass m. One is the gravity, W; the other
is the supporting force from the chair, which
is equal to the apparent weight Wa in magnitude. Their resultant force provides the
centripetal force so
W − W a = Fc =
m v2
r
then
W v2
gr
(381 N) (7.9 m/s)2
= 381 N −
(9.8 m/s2 ) (10 m)
= 138.365 N .
Wa = W − F c = W −
013 (part 1 of 2) 0 points
A planet similar to the Earth has a radius
7.6 × 106 m and has an acceleration of gravity of 10 m/s2 on the planet’s surface. The
planet rotates about its axis with a period of
Answer, Key – Homework 6 – David McIntyre – 45123 – Mar 25, 2004
25 h. Imagine that the rotational speed can
be increased.
If an object at the equator is to have zero
apparent weight, what is the new period?
Correct answer: 1.52154 h.
Explanation:
When the apparent weight is zero, the gravity is equal to the centripetal force
µ ¶2
2π
2
mg = mω R = m
R
Tf
so this faster period, Tf , is
s
R
Tf = 2 π
g
s
7.6 × 106 m
= 2π
10 m/s2
= 5477.55 s
= 1.52154 h .
Two identical stars, a fixed distance D
apart, revolve in a circle about their mutual
center of mass, as shown below. Each star has
mass M and speed v.
v
M
v
Which of the following is a correct relationship among these quantities?
1. v 2 =
2. v 2 =
3. v 2 =
014 (part 2 of 2) 0 points
By what factor would the speed of the object
be increased when the planet is rotating at
the higher speed?
Correct answer: 16.4307 .
Explanation:
The velocity is
4. v 2 =
The period is T = 25 h. We know the ratio of
vf
the corresponding velocities, R =
, is the
v
inverse of two different periods, which is, for
this case,
R=
vf
T
=
v
Tf
25 h
1.52154 h
= 16.4307 .
=
015 (part 1 of 1) 0 points
Given: G is the universal gravitational constant.
M
D
where R is the mean radius of the planet.
distance
2πR
=
.
v=
time
T
5
5. v 2 =
6. v 2 =
7. v 2 =
GM
D
4GM
D
2GM
D
2 G M2
D
GM
D2
4 G M2
D
GM
correct
2D
8. v 2 = M G D
Explanation:
From the basic formulae for the circular orbital movement, the centripetal acceleration
v2
is a =
.
D
2
Using the Newton’s second law of motion,
F
we know the acceleration is a =
, where F
M
is the force between two stars and is totally
supplied by the universal force.
So we obtain
F
GM
2 v2
=a=
=
D
M
D2
Answer, Key – Homework 6 – David McIntyre – 45123 – Mar 25, 2004
=⇒
v2 =
GM
.
2D
016 (part 1 of 2) 5 points
On the way to the moon the Apollo astronauts
reach a point where the Moon’s gravitational
pull is stronger than that of Earth’s.
Determine the distance of this point from
the center of the Earth. The masses of the
Earth and the Moon are respectively 5.37 ×
1024 kg and 7.36 × 1022 kg. The distance from
the Earth to the Moon is 3.66 × 108 m.
Correct answer: 3.27642 × 108 m.
Explanation:
If re is the distance from this point to the
center of the Earth and rm is the distance
from this point to the center of the Moon,
then from the formula
G m Mm
G m Me
=
2
2
re
rm
we obtain
rm
q=
=
re
r
Mm
Me
s
7.36 × 1022 kg
5.37 × 1024 kg
= 0.117072 .
=
On the other hand,
re + r m = R .
Eliminating rm from the last two equalities,
we obtain
R
q+1
3.66 × 108 m
=
0.117072 + 1
= 3.27642 × 108 m .
re =
017 (part 2 of 2) 5 points
What is the acceleration due to the Earth’s
gravity at this point? The universal gravitational constant G is 6.672 × 10−11 N m2 /kg2 .
Correct answer: 0.00333757 m/s2 .
Explanation:
6
From the relation
a=
F
G Me
=
m
re2
we obtain that the acceleration a due to the
Earth’s gravity at this point is
a = 6.672 × 10−11 N m2 /kg2
5.37 × 1024 kg
×
(3.27642 × 108 m)2
= 0.00333757 m/s2 .
018 (part 1 of 3) 0 points
Given:
A solar system similar to our Sun and
Earth, where
Mearth
Rearth
Msun
sun
Rearth
= 4.09 × 1024 kg
= 6.38 × 106 m
= 1.84 × 1030 kg
= 1.56 × 1011 m .
Determine the magnitude of the change ∆F
in gravitational force that the Sun exerts on
a 73.1 kg woman standing on the equator at
noon and midnight.
Hint: Since ∆r is so small, use differentials.
Assume: The Sun and the Earth are the
only masses acting on the woman.
Correct answer: 6.03305 × 10−5 N.
Explanation:
Basic Concepts The gravitational force
exerted by m2 on m1 is
~ 21 = −G m1 m2 r̂12
F
(1)
2
r12
and the magnitude of the force is
m1 m2
F =G
.
(2)
r2
If we want to find a change ∆F due to a
change ∆r, we write
∆F
dF
∆r ≈
∆r
(3)
∆r
dr
since using “differentials” essentially amounts
to approximating
∆F =
dF
∆F
≈
∆r
dr
(4)
Answer, Key – Homework 6 – David McIntyre – 45123 – Mar 25, 2004
for small changes.
Solution: Differentiating F with respect to
r, we find
dF
m Ms
= −2 G
(5)
dr
r3
so, using differential approximations,
m Ms
∆F ≈ −2 G
∆r .
(6)
r3
The change in distance ∆r is equal to the
twice the radius of the Earth (i.e., the diameter of the Earth), since the woman moves
from the closest point (noon) to the furthest
point (midnight).
∆r = 2 RE
= 2 (6.38 × 106 m)
= 1.276 × 107 m .
Therefore
m Ms
∆r .
(6)
∆F ≈ −2 G
r3
≈ −2 (6.67259 × 10−11 N m2 /kg2 )
(73.1 kg) (1.84 × 1030 kg)
×
(1.56 × 1011 m)3
× (1.276 × 107 m)
= −6.03305 × 10−5 N
|∆F | = 6.03305 × 10−5 N .
019 (part 2 of 3) 0 points
Determine the fractional percent change in
∆F
the Sun’s gravitational force
% on the
F
woman due to the rotation of the Earth in the
12 hours between noon and midnight.
Assume: The Sun and the Earth are the
only masses acting on the woman.
Correct answer: 0.016359 %.
Explanation:
Using Eqs. (2) and (6), the fractional
change is
m Ms
−2 G
∆r
∆F
r3
=
(7)
m Ms
F
G
r2
2 ∆r
=−
r
2 (1.276 × 107 m)
=−
1.56 × 1011 m
= −0.00016359 ,
7
so the percentage decrease is
¯
¯
¯
¯ ∆F ¯ ¯¯
¯
¯ = ¯−0.00016359¯¯ 100
¯ F ¯
= 0.016359 % .
020 (part 3 of 3) 0 points
This part is the same as the previous part except we are concerned with the Moon’s gravitation force instead of the Sun’s gravitational
force on the woman.
Given: A system similar to our Earth and
Moon, where
Mearth
Rearth
Mmoon
moon
Rearth
= 4.09 × 1024 kg
= 6.38 × 106 m
= 7.22 × 1022 kg
= 3.89 × 108 m .
Determine the fractional percent change in
∆F
% on the
the Moon’s gravitational force
F
woman due to the rotation of the Earth in the
12 hours between noon and midnight.
Assume: The Earth and the Moon are the
only masses acting on the woman.
Correct answer: 6.56041 %.
Explanation:
The fractional change is
m Mmoon
∆r
3
∆F
r
=
m Mmoon
F
G
r2
2 ∆r
=−
r
2 (1.276 × 107 m)
=−
3.89 × 108 m
= −0.0656041 ,
−2 G
so the percentage decrease is
¯
¯
¯
¯ ∆F ¯ ¯¯
¯
¯ = ¯−0.0656041¯¯ 100
¯ F ¯
= 6.56041 % .