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Circular Motion and Gravitation
v
Ball rolling in a
straight line (inertia)
Centripetal Acceleration – acceleration towards
the center of a circle.
v
– a.k.a. Radial Acceleration (aR)
aR
v
aR = v2
r
If you twirl a yo-yo and let go of the string,
what way will it fly?
If you are on a carousel
at constant speed, are
you experiencing
acceleration?
Period and Frequency
Formulas
Period (T)
– Time for one complete
– seconds
aR
Same ball, hooked to
a string
(360o)
revolution
Frequency
– Number of revolutions per second
– rev/s or Hertz (Hz)
aR = v2
r
v = 2pr
T
T=1
f
v = 2prf
T=1
f
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A 150-g ball is twirled at the end of a 0.600 m
string. It makes 2.00 revolutions per second.
Find the period, velocity, and acceleration.
(0.500 s, 7.54 m/s, 94.8 m/s2)
Jupiter is about 778 X 106 km from the sun. It
takes 4332.6 days to orbit the sun.
a) Calculate the circumference of Jupiter’s orbit.
(4.89 X 1012 m)
b) Calculate Jupiter’s period in seconds. (3.74 X
108 s)
c) Calculate Jupiter’s orbital speed. (1.30 X 104
m/s)
d) Calculate Jupiter’s centripetal acceleration
towards the sun. (2.18 X 10-4 m/s2)
The moon has a radius with the earth of about
384,000 km and a period of 27.3 days.
a. Calculate the acceleration of the moon toward
the earth.
b. Convert the answer to g’s.
(Ans: 2.72 X 10-3 m/s2, 2.78 X 10-4 g )
Centripetal Force
Centripetal Force – “center seeking” force that
pulls an object in a circular path.
–
–
–
–
Yo-yo
Planets
Merry-go-round
Car rounding a curve?
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“Centrifugal Force?”
• Doesn’t exist
• “apparent outward force”
• When you let the string go, the ball will
continue in a straight line path. No new
acceleration involved.
• Water in swinging cup example
Direction water
wants to go
Centripetal
Force of
string
SF = maR = mv2
r
A 0.150 kg yo-yo is attached to a 0.600 m string and
twirled at 2 revolutions per second. It is twirled
horizontally.
a) Calculate the circumference of the circle (3.77 m)
b) Calculate the linear speed (7.54 m/s)
c) Calculate the centripetal force in the string (14.2 N)
An electron orbits the nucleus with a radius of 0.5 X
10-10 m. The electron has a mass of 9 X 10-31 kg
and a speed of 2.3 X 106 m/s.
a. Calculate the centripetal force on the electron.
(9.52 X 10-8 N)
b. Calculate the frequency of an electron. (7.32 X
1015 Hz)
c. Convert the velocity to miles/s. (1429 miles/s)
d. What provides the centripetal force?
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A rotating space station is shaped like a wheel. It
has a radius of 30.00 m.
a) Calculate the speed at which it must rotate to provide
1 “g” of gravity. (17.15 m/s)
b) Calculate the centripetal force on a 126 lb man (1 lb is
equivalent to 0.454 kg) (561 N)
c) Calculate the frequency and period of the space
station. (0.0907 Hz, 11.0 s)
d) Where would the “floor” be on such a space station?
Circular Motion: Example 2
Thor’s Hammer
(mjolnir) has a mass
of 10 kg and the
handle and loop have
a length of 50 cm. If
he can swing the
hammer at a speed of
3 m/s, what force is
exerted on Thor’s
hands?
(Ans: 180 N)
What angle will the hammer take with the
horizontal? Also calculate the force of the
handle.
Can Thor swing his hammer so that it is
perfectly parallel to the ground?
FH
mv2 = FHcosq
r
0 = FHsinq – mg
Mass = 10 kg
r = 50 cm
q
v = 3 m/s
mg
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A 0.00500 kg walnut is swung in a horizontal
circle of radius of 50.0 cm. The walnut
makes 2.50 revolutions per second.
a) Calculate the linear speed of the walnut. (7.85
m/s)
b) Draw a free-body diagram of the walnut.
c) Calculate the centripetal force needed to keep it
in a circle. (0.616 N)
d) Calculate the force of tension and the angle the
string makes with the horizontal. (0.618 N,
4.50o)
A 0.25 kg yo-yo is swung in a horizontal radius
of 65.0 cm. The string makes a 10.0o angle
with the horizontal.
a) Draw a free-body diagram of the yo-yo.
b) Calculate the tension in the string. (14.1 N)
c) Calculate the linear speed of the yo-yo. (6.00
m/s)
d) Calculate the period and frequency (0.68 s,
1.47 Hz)
Circular Motion: Example 3
What is the tension in the cord at the bottom
of the arc if the ball moves at the minimum
speed? (v = 3.28 m/s)
A 0.150 kg ball is swung on a 1.10-m string in a
vertical circle. What minimum speed must it
have at the top of the circle to keep moving in
a circle?
mg
T
T
mg
(ANS: 2.94 N)
A rollercoaster vertical loop has a radius of
20.0 m. Assume the coaster train has a
mass of 3,000 kg.
a) Calculate the minimum speed the coaster
needs to have to make the loop. (14.0 m/s)
b) Calculate the normal force the tracks
provide to the train at the bottom of the
curve if the train is travelling at 25.0 m/s.
(123,150 N)
c) Calculate the normal force the tracks
provide at the top of the curve if the train is
travelling at 25.0 m/s. (64, 350 N)
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The ferris wheel at Knoebels has a radius of
16.8 m and travels at a speed of 3.52 m/s.
a) Calculate the frequency and period (0.033 Hz,
30 s)
b) Calculate the normal force that the seat
provides to a 56.0 kg rider at the top. (507 N)
c) Calculate the normal force that the seat
provides to a 56.0 kg rider at the bottom. (590
N)
A car travels over a round hill (radius = 50.0 m).
a) Calculate the maximum speed at which the car
can take the hill. (22.0 m/s)
b) Calculate the normal force on a 1000.0 kg car if
it is travelling over the hill at 10.0 m/s. (7.80 X
103 N)
A 1000-kg car rounds a curve (r=50 m).
Car Rounding a Turn
• Friction provides centripetal force
• Use (ms). Wheels are turning, not sliding, across
the surface
• Wheel lock = kinetic friction takes over. mk is
always less than ms, so the car is much more
likely to skid.
A 15,000-kg truck can safely round a 150 m curve
at a speed of 20 m/s.
a) Calculate the centripetal force needed to keep the truck on the road
(40,000 N )
b) Calculate the coefficient of static friction (0.27)
c) Calculate the maximum speed a 1000 kg Cube car can take the
turn. (20 m/s)
n
fs = Fc
mg
a) Calculate the maximum
speed the car can take if the
road is dry and ms = 0.60 (17
m/s)
b) Calculate the maximum
speed if the road is icy and
ms = 0.25 (11 m/s)
c) If the car is travelling at 14.0
m/s, under which conditions
will they skid? (0.40)
The Rotor
The Rotor at an amusement park has a radius of
7.0 m and makes 30 rev/min.
a) Calculate the speed of the rotor. ( 22.0 m/s)
b) Draw a free body diagram of a person in the
rotor. What causes the normal force?
c) Calculate the coefficient of static friction
between the person and the wall. (0.14)
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The Rotor at an amusement park has a radius of
6.0 m and a ms of 0.20.
a) Draw a free body diagram of a person in the
rotor. What causes the normal force?
b) Calculate the speed of the rotor. (17 m/s)
c) Calculate the period and frequency of the rotor.
(2.22 s, 0.45 rev/s)
The Rotor at an amusement park has a radius of
4.00 m and a ms of 0.250.
a) Draw a free body diagram of a person in the
rotor. What causes the normal force?
b) Calculate the speed of the rotor. (12.5 m/s)
c) Calculate the period and frequency of the rotor.
( 2.00 s, 0.50 rev/s )
Banked Curves
FC = fs + FNsinq
• Banked to reduce the reliance on friction
• Part of the Normal Force now contributes to
the centripetal force
A 1000-kg car rounds a 50 m radius turn at 14
m/s. What angle should the road be banked
so that no friction is required?
(ideally, we bank the
road so that no friction
is required: fs = 0)
Now we will simply work with the Normal
Force to find the component that points to
the center of the circle
First consider the y forces.
SFy = FNcosq - mg
Since the car does not
move up or down:
FNcosq
FN
FN
q
SFy = 0
FNsinq
0 = FNcosq – mg
FNcosq = mg
q mg
q
q mg
q
FN = mg/cosq
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mv2 = FNsinq
r
mv2 = mgsinq
r
cosq
v2 = gtanq
r
v2 = gtanq
r
v2 = tanq
gr
tan q =
(14 m/s)2
(50 m)(9.8m/s2)
=
0.40
q = 22o
A 2,000-kg Nascar car rounds a 300 m radius
turn at 200 miles/hr. (1 mile=1609 m)
a. Convert the speed to m/s. (89.4 m/s)
b. What angle should the road be banked so that
no friction is required? (70o)
c. Suppose a track is only banked at 35.0o,
calculate the maximum speed that a car can
take the turn. (45.3 m/s, 101 mph)
d. Looking at the formula for banking angle, how
could a track designer decrease that angle?
Weightlessness
– True weightlessness exists only very far from
planets
– “Apparent weightlessness” can be achieved on
earth
– Apparent weight is the Normal Force
– Apparent weightlessness (n = 0)
Elevator at Constant Velocity
a= 0
SF = n – mg
0 = n – mg
n = mg
Suppose Chewbacca has a mass
of 102 kg:
n = mg = (102kg)(9.8m/s2)
n = 1000 N
n
mg
a is zero
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Elevator Accelerating Upward
a = 4.9 m/s2
SF = n – mg
ma = n – mg
n = ma + mg
n = m(a + g)
n = (102kg)(4.9m/s2+9.8 m/s2)
n = 1500 N
a = 4.9 m/s2
SF = mg - n
ma = mg - n
n = mg - ma
n = m(g - a)
n = (102kg)(9.8m/s2 – 4.9 m/s2)
n = 500 N
n
mg
a is upward
At what acceleration will he feel weightless?
n=0
SF = mg - n
ma = mg -n
ma = mg - 0
ma = mg
a = 9.8 m/s2
Apparent weightlessness
occurs if a > g
Elevator Accelerating Downward
n
mg
a is down
Calculate the apparent weight of a 56.0 kg man in
an elevator if the elevator is:
a) Accelerating upwards at 2.00 m/s2. (661 N)
b) Accelerating downwards at 2.00 m/s2 (436 N)
c) Accelerating downwards at 9.80 m/s2 (0 N)
d) Accelerating sideways at 9.80 m/s2. (549 N)
n
mg
Mr. Saba is riding an elevator that
accelerates upward at 3.00 m/s2. His
apparent weight is 768 N.
a) Calculate his mass. (60.0 kg)
b) Calculate his apparent weight if the
elevator accelerates downwards at
3.00 m/s2. (408 N)
c) Calculate the acceleration required to
have him weigh 1000 N. Is this up or
down? (6.87 m/s2)
Other examples of apparent
weightlessness
Mr. Fredericks
wishes he were a
God of Math like
me.
Even when you are running, you fell weightless
between strides.
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Why don’t satellites fall back onto the earth?
• Speed
• They are “falling”
due to the pull of
gravity
• Can feel
“weightless” (just
like in the elevator)
A 2.00 kg book is placed in a spinning drum
(rotor) of radius 0.75 m and m of 0.20.
s
Draw a free body diagram
Calculate the speed at which the drum needs to
rotate to suspend the book. (6.06 m/s)
An 1800.0 kg car rounds a turn of 60.0 m radius.
a. If the ms is 0.47, calculate the maximum speed at
which he can take the turn. (16.6 m/s)
b. Suppose you wished to bank the curve so that no
friction is required. Calculate the banking angle.
(25.1o)
Gravitation
Is gravity caused by the earth’s rotation?
Will a man down here fall
off if the earth stops
rotating?
Gravitation
Newton’s Law of Universal Gravitation
1. Every object in the universe is attracted to every
other object. (based on mass)
2. The force drops off with the distance squared.
(As distance increases, the force of gravity drops
very quickly)
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Cavendish proves the law in
1798
Gravitation: Formula
F= GMm
r2
G = 6.67 X 10-11 N m2/kg2
M = mass of one object
m = mass of second object
r = distance from center of objects
Gravitation: The Solar System
Everything in the solar system pulls on everything else.
Sun pulls on Earth
All the other planets
also pull on the
Earth
Some comets/meteors are actually from outside our solar system and
were captured by our sun’s gravity.
What is the force of gravity between two 60.0
kg (132 lbs) people who stand 2.00 m apart?
F = 6.00 X 10-8 N
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What is the force of gravity between a 60 kg person
and the earth? Assume the earth has a mass of
5.98 X 1024 kg and a radius of 6,378,000 m
(~4,000 miles).
A 2000-kg satellite orbits the earth at an altitude of
6380 km (the radius of the earth)above the
earth’s surface. What is the force of gravity on
the satellite?
F= GMm = (6.67 X 10-11 N m2/kg2)(2000kg)(5.98 X 1024 kg)
r2
(6,380,000 m + 6,380,000 m) 2
F = 4900 N
F = 588 N
Calculating the Mass of the Earth
Calculate the mass of the earth knowing that it has a
radius of 6.38 X 106 m. Start using the weight
formula.
Calculating “g”
g = GM
r2
Calculating “g”: Example 1
Calculate the value of g at the top of Mt.
Everest, 8848 m above the earth’s surface.
g = (6.67 X 10-11 N-m2/kg2)(5.98 X 1024 kg)
(6.38 X 106 m)2
g = 9.80 m/s2
g = 9.77 m/s2
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g varies with:
• Altitude
• Location
Objects weigh about 1/6 their weight on Earth
on the Moon. Calculate the mass of the
moon, knowing that the radius of the moon
is 1734 km.
– Earth is not a perfect sphere (roughly 22 km
greater in radius at equator, 13.7 miles)
– Different mineral deposits change density
– “salt domes” - low density salt regions near
petroleum deposits
An object weighs 200 N on earth.
a) Calculate the acceleration of gravity on Mars
(3.71 m/s2)
b) Calculate its weight on Mars (75.5 N)
Rm = 3440 km
Mm = 0.11Me
ME = 5.98 X 1024 kg
Gravitation: Example 3
What is the net force on the moon when it is at a
right angle with the sun and the earth?
Relevant Data:
MM = 7.35 X 1022 kg
ME = 5.98 X 1024 kg
MS = 1.99 X 1030 kg
rMS = 1.50 X 1011 m
rME = 3.84 X 108 m
Pluto has a mass of 1.31 × 1022 kg, and a radius
of 1180 km.
a. Calculate the gravity of Pluto at its surface.
(0.628 m/s2)
b. If a person weighs, 550 N on earth, calculate
his weight on Pluto. (35.2 N)
c. Pluto is 5.9 X 109 km from the sun, and has a
year equal to 248 earth years. Calculate its
orbital speed. (4740 m/s)
Calculate each force separately:
FME = 1.99 X 1020 N
FMS = 4.34 X 1020 N
FR2 = FME2 + FMS2
FME
Earth
q
FMS
FR
FR = 4.77 X 1020 N
tan q = opp = FME
adj
FMS
o
q = 24.6
Sun
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Three 5.00 kg bowling balls are placed at the
corners of an equilateral triangle whose sides are
1.50 m long. Calculate the magnitude and
direction of the gravitational force on the top
ball.
A geosynchronous satellite has a period of one
day. The radius of the Earth is 6380 km and
the mass of the Earth is 5.98 X 1024 kg.
a) Convert the period to seconds (8.64 X 104 s)
b) Calculate the height above the earth that a
geosynchronous satellite must orbit. (Hint: use
mv2/r) (3.59 X 107 m, 4.23 X 107 m total)
c) Calculate the speed of the orbit. (3070 m/s)
Kepler’s Laws (1571-1630)
Four 5.00 kg bowling balls are placed at the
corners of an square whose sides are 1.50 m
long. Calculate the magnitude and direction of
the gravitational force on the lower left ball.
(1.42 X 10-9 N, 45o)
A satellite orbits with a period of 5.00 hours. The
radius of the Earth is 6380 km and the mass of
the Earth is 5.98 X 1024 kg.
a) Calculate the height of the satellite above the
earth. (8.47 X 106 m)
b) Calculate the speed of the orbit. (5.17 X 103 m/s)
1. The orbit of each planet is an
ellipse, with the sun at one focus
1. The orbit of each planet is an ellipse, with the
sun at one focus
2. Each planet sweeps out equal areas in equal
time
3.
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2. Each planet sweeps out equal areas in
equal time
Deriving the Third Law
• Suppose the travel time in both cases is
three days.
• Shaded areas are exactly the same area
m2
Jupiter
Gm1mJ = m1v2
r2
r
GmJ = v2
r
GmJ = 4p2r2
r
T2
2
2
T = 4p
r3 GmJ
Substitute v=2pr
T
A Useful Form
r3
T2
= GM
4p2
ONE SATELLITE
T2 = 4p2
r3 GmJ
T12 = 4p2
r13 GmJ
m1
We can do this for two
different moons
T22 = 4p2
r23 GmJ
The Third Law: Example 1
Mars has a year that is about 1.88 earth years. What
is the distance from Mars to the Sun, using the
Earth as a reference (rES = 1.496 X 108 m)
TWO SATELLITES
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Third Law: Example 2
rM3 = TM2rE3
TE2
rM3
=
(1.88y)2(1.496
(1
X
108
m)3
How long is a year on Jupiter if Jupiter is 5.2
times farther from the Sun than the earth?
y)2
rM3 = 1.18 X 1025 m3
rM = 2.28 X 108 m
Kepler’s Third Law: Example 3
TJ2 = rJ3 TE2 = (5.2)3(1 y)2
rE 3
(1)3
TJ2 = 141 y2
How high should a geosynchronous satellite be
placed above the earth? Assume the satellite’s
period is 1 day, and compare it to the moon,
whose period is 27 days. The average distance
between the earth and the moon is 384,000 km.
TJ = 11.9 y
(3.63 X 107 m, or 3.63 X 104 km)
Third Law: Example 4
What is the mass of the sun, knowing that the
earth is 1.496 X 1011 m from the sun.
T2 = 4p2
r3 GmS
mS= 4p2r3
GT2
mS= 4p2(1.496 X 1011 m)3
(6.67 X 10-11 N-m2/kg2) (3.16 X 107 s)2
mS = 2.0 X 1030 kg
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Calculate the mass of the earth, knowing that
the moon has a period of 27.321 days and
an average distance of 384,400 km.
The dwarf planet Eris is 1.01 X 1010 km from
the sun. The earth is 1.496 X 1011 m from
the sun. Calculate the length of a year on
Eris.
(~555 yrs)
The Gemini 11 spacecraft sent two astronauts to
a height of 1374 km above the earth’s surface.
The radius of the Earth is 6380 km and the
mass of the Earth is 5.98 X 1024 kg.
a) Calculate the period of the satellite (1.89 hour)
b) Calculate the speed of the orbit. (7173 m/s)
The period of Neptune’s moon Galatea is
0.429 days, and the radius is 61,953 km
from the center of Neptune.
a. Calculate the mass of Neptune (1.02 X
1026 kg)
b. Calculate the speed of the orbit. (10,500
m/s)
The Gemini 11 spacecraft had a period of 1.89
hours. The radius of the Earth is 6380 km and
the mass of the Earth is 5.98 X 1024 kg.
a) Calculate the height above the earth that it
orbited. (1374 km)
b) Calculate the speed of the orbit. (7166 m/s)
The mass of Mars is 6.40 X 1023 kg.
Calculate the period (in hours) of its moon
Phobos if Phobos has an orbital radius of
9377 km.
(7.67 h)
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Pluto has a radius of 1150 km and a mass of 1.20
X 1022 kg.
a) Calculate the acceleration of gravity on Pluto
(0.605 m/s2)
b) Calculate the weight of a 70.0 kg person on
Pluto. (42.4 N)
c) Calculate the acceleration of gravity on Pluto in
terms of “g’s” (0.0618 g’s)
The radius of the sun is 695,500 km, and its mass
is 1.99 X 1030 kg.
a. Calculate the acceleration due to gravity at the
surface of the sun. (274 m/s2)
b. Calculate your weight at the surface of the sun.
Choose any planet except earth. Look up its mass,
radius, distance from the sun, and the length of its
day.
A student is given the following data and asked to
calculate the mass of Saturn. The data describes
the orbital periods and radii of several of
Saturn’s moons.
Orbital Period, T
Orbital Radius, R
(seconds)
(m)
4
8.14 X 10
1.85 X 108
5
1.18 X 10
2.38 X 108
1.63 X 105
2.95 X 108
5
2.37 X 10
3.77 X 108
1. Calculate the surface gravity on your planet
2. Calculate the weight of a 56.0 kg person on our
planet.
3. Calculate the height that a satellite must be placed
above the surface of the planet to be in
geosynchronous orbit (T = one day of that planet).
4. The earth is 1.5 X 1011 m from the sun. Calculate
the time it takes (in years) your planet to orbit the
sun. (Use Kepler’s laws)
Let’s use this equation:
T2 = 4p2
r3
GmS
Calculate the following values and graph them.
1
T2
G
4p2 r3
And rearrange it:
GmS= 1
4p2 r3 T2
Once more:
1 = GmS
T2
4p2 r3
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Calculate the slope of the graph
1.60E-10
1.40E-10
1.20E-10
y
= m
x
+b
1.00E-10
8.00E-11
6.00E-11
1 = mS G
T2
4p2 r3
4.00E-11
2.00E-11
0.00E+00
0.00E+00 5.00E-38 1.00E-37 1.50E-37 2.00E-37 2.50E-37 3.00E-37
The Four Fundamental Forces
1.
2.
3.
4.
Gravity
Electromagnetic
Strong Nuclear Force
Weak Nuclear Force
26. 1.62 m/s2
28. 24.5 m/s2
30. 0.91 gsurface
32. 2.0 X 107 m
34. a) 9.8 m/s2
b) 4.3 m/s2
36. 9.6 X 1017 N away from the Sun
38. 2.0 X 1030 kg
40. 3.14 m/s2 upward
42. 5.07 X 103 s (1.41 h), independent of mass
44.a) 58 kg
b) 58 kg
c) 77 kg
d) 39 kg
e) 0
ms = 5.9 X 1026 kg
2. a) 1.52 m/s2, center
b) 38.0 N, center
4. 12 m/s
6. 14 m/s, no effect
8. a) 3.14 N
b) 9.02 N
10. 27.6 m/s, 0.439 rev/s
12. 9.2 m/s
14. 11 rev/min
16.FT1 = 4p2f2(m1r1 + m2r2)
FT2 = 4p2f2m2r2
18.0.20
70. a) 3000 m b) 5500 N
c) 3900 N
44.a) 58 kg (569N) b) 58 kg(569N)
c) 77 kg (755 N) d) 39 kg (382N)
e) 0
52. RIcarus = 1.62 X 1011 m
54. 5.97 X 1024 kg
56. 3.4 X 1041 kg, 1.7 X 1011 “Suns”
58.REuropa = 6.71 X 105 km
RGanymede = 1.07 X 106 km
RCallisto = 1.88 X 106 km
68.1840 rev/day
70. a) 3000 m b) 5500 N
c) 3900 N
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Graphing Centripetal Force A
A 1 kg yo-yo was swung in a circle at a constant speed. The
force on the string was measured as the string was let out
slowly.
Radius (m)
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Force (N)
10.00
5.00
3.33
2.50
2.00
1.67
1.43
1.25
1.11
1.00
Graphing Centripetal Force B
In a second experiment, the speed was
changed while the length of the string (r)
was kept constant
Speed (m/s)
1.0
2.0
3.0
4.0
5.0
6.0
Force (N)
1.0
4.0
9.0
16
25
36
20
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In an experiment, a yo-yo is twirled in a
horizontal circle.
1. Suppose the speed of the yo-yo gradually
increases. Sketch of graph of centripetal
acceleration (aR) versus speed.
2. Suppose the length of the string is
gradually increased. Sketch of graph of
centripetal acceleration (aR) versus radius.
Considering a Sports Management Degree?
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