CIRCULAR MOTION - GRAVITATION - Physics-matters.net

CHAPTER 5
CIRCULAR MOTION - GRAVITATION
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Units
• Kinematics of Uniform Circular Motion
• Dynamics of Uniform Circular Motion
• Highway Curves, Banked and Unbanked
• Nonuniform Circular Motion
• Centrifugation
• Newton’s Law of Universal Gravitation
• Gravity Near the Earth’s Surface; Geophysical Applications
• Satellites and “Weightlessness”
• Kepler’s Laws and Newton’s Synthesis
Circular Motion
An object moving in a circle must have a force acting on it;
otherwise it would move in a straight line.
The direction of the force is towards the center of the circle.
Kinematics of Uniform Circular Motion
Looking at the change in velocity in the limit that the time interval becomes infinitesimally small, we
see that 2
v
aR
=
r
1

This acceleration is called the centripetal, or radial, acceleration, and it
points towards the center of the circle.
Forces in Accelerating Reference Frames
• Distinguish real forces from fictitious forces
• “Centrifugal” force is a fictitious force
• Real forces always represent interactions between objects
Example 1: A 150-g ball at the end of a string is revolving
uniformly in a horizontal circle of radius 0.600 m. The ball makes
2.00 revolutions in a second. What is its centripetal acceleration
2π
r 2(3.14)(0.600 m)
v= = = 7.54 m/
s
T (0.500 s)
a
R
2 2
= v (7.54 m/ s)
94.7 m/
r
= (0.600 m)
= s
2
Example 2: The Moon’s orbit about the Earth has a radius of 384,000 km and a period T of 27.3
days. Determine the acceleration of the Moon toward the Earth.
2 2 2 2 8
v (2 r) 4 r 4 (3.84x10 m)
aR
= = π = π = π = 2.72x10 − m/
s
2 2 6 2
r T r T (2.36x10 s)
2
In terms of g ( 9.80 m/
s ):
−3 2⎛
g ⎞
−4
a= 2.72x10 m/ s ⎜
2.78x10
2 ⎟=
g
⎝9.80 m/
s ⎠
3 2
Dynamics of Uniform Circular Motion
There is no centrifugal force pointing
outward; what happens is that the
natural tendency of the object to
move in a straight line must be
overcome.
If the centripetal force vanishes, the
object flies off tangent to the circle.
2

This force may be provided by the tension in a string, the normal
force, or friction, among others.
Example 3: Estimate the force a person exerts on a string attached to a 0.150-kg ball to make the
ball revolve in a horizontal circle of radius 0.600 m. The ball makes 2.00 revolutions per second (T
= 0.50s)
F = ma a = v 2 / r v=
2 π r/
T
R
v= 2 π (0.600 m) /(0.500 s) = 7.54 m/
s
2 2
v
(7.54 m/ s)
FT
= m = (0.150 kg) ≈14N
r
(0.600 m)
Vertical Circle
• Look at the forces at the top of the circle
• The minimum speed at the top of the circle can be found
vtop
= gR
Example 4: A 0.150-kg ball on the end of a 1.1-m-long cord is swung
in a vertical circle. (a) Determine the minimum speed the ball must
have at the top of its arc so that the ball continues moving in a circle.
v
= gr
v m s m
2
= (9.80 / )(1.10 ) = 3.28 m/
s
3

(b) Calculate the tension in the cord at the bottom of the arc, assuming the ball is moving at twice
the speed of part (a).
2
v
FT
= m + mg
r
2
(6.56 m/ s)
2
FT
= (0.150 kg) + (0.150 kg)(9.80 m/ s ) = 7.34N
(1.10 m)
Highway Curves, Banked and Unbanked
When a car goes around a curve, there must be a net force towards the center of the circle of
which the curve is an arc. If the road is flat, that force is supplied by friction.
If the frictional force is insufficient, the car will tend to move
more nearly in a straight line, as the skid marks show.
As long as the tires do not slip, the friction is static. If the
tires do start to slip, the friction is kinetic, which is bad in
two ways:
1. The kinetic frictional force is smaller than the
static.
2. The static frictional force can point towards the
center of the circle, but the kinetic frictional force
opposes the direction of motion, making it very
difficult to regain control of the car and continue
around the curve.
Banking the curve can help keep cars from skidding. In fact, for
every banked curve, there is one speed where the entire
centripetal force is supplied by the horizontal component of the
normal force, and no friction is required. This occurs when:
4

Example 5: A 1000-kg car rounds a curve on a flat road of radius 50
m at a speed of 14 m/s. Will the car follow the curve, or will it skid
(a) Assume the pavement is dry and the coefficient of static friction is
μ = 0.60 .
s
FN
on the car is equal to the weight mg since the road is flat.
2
FN
= mg = (1000 kg)(9.8 m/ s ) = 9800N
The net horizontal force required to keep the car moving in a circle
around the curve is
2 2
v
(14 m/ s)
FR
= maR
= m = (1000 kg) = 3900N
r
(50 m)
The maximum friction force attainable is
F = μ F = (0.60)(9800 N) = 5900N
fr S N
Since the force of only 3900 N is needed the car can follow the
curve.
(B) If the pavement was icy and μ
S
= 0.25 then the maximum static friction force possible is
F = μ F = (0.25)(9800 N) = 2500N
fr S N
The car will skid because the ground cannot exert sufficient force (3900N) to keep it moving in a
curve.
Example 6: What is the banking angle needed for an expressway
off-ramp with a curve radius of 50 m at a design speed of 50 km/h
2
v
tanθ =
rg
2
(14 / )
tan θ = m s
= 0.40 = 21.8
2
(50 m)(9.80 m/ s )
o
Nonuniform Circular Motion
If an object is moving in a circular path but at varying speeds, it
must have a tangential component to its acceleration as well as
the radial one. This concept can be used for an object moving
along any curved path, as a small segment of the path will be
approximately circular.
5

Centrifugation
A centrifuge works by spinning very fast. This means there must
be a very large centripetal force. The object at A would go in a
straight line but for this force; as it is, it winds up at B.
Newton’s Law of Universal Gravitation
If the force of gravity is being exerted on objects on Earth, what
is the origin of that force
Newton’s realization was that the force must come from the
Earth.
He further realized that this force must be what keeps the Moon
in its orbit.
• Every particle in the Universe attracts every other
particle with a force that is directly proportional to the
product of the masses and inversely proportional to the
square of the distance between them.
mm
1 2
F = G r
2
The gravitational force on you is one-half of a Third Law pair: the Earth exerts a downward force on
you, and you exert an upward force on the Earth.
When there is such a disparity in masses, the reaction force is
undetectable, but for bodies more equal in mass it can be
significant.
Therefore, the gravitational force must be proportional to both
masses.
By observing planetary orbits, Newton also concluded that the
gravitational force must decrease as the inverse of the square
of the distance between the masses.
6

Gravitation Constant
• Determined experimentally
• Henry Cavendish
– 1798
• The light beam and mirror serve to amplify the motion
Example 6: A 50-kg person and a 75-kg person are sitting next to each other (0.5m) in the
classroom. Estimate the magnitude of the gravitational force each exerts on the other.
−11 2
F G mm
1 2
(6.67x10 N⋅
m )(50 kg)(75 kg)
−6
= = r 2 2
= 110 x N
(0.5 m )
Example 7: Three 0.300-kg pool balls are placed on a table at the
corners of a right triangle. Find the net gravitational force on the
cue ball.
Find the net gravitational force on the cue ball.
21
F on m 2 on m 1
mm
2 1
F21 = G r
2
21
−11 0.3 kg)(0.3 kg)
−11
= (6.67x10 ) = 3.75x10
N
2
(0.4 m)
F 31
by m 3 on m 1
mm
3 1
−11 (0.3 kg)(0.3 kg)
−11
F31 = G = (6.67x10 ) = 6.67x10
N
2 2
r
(0.3 m)
31
F = F + F = (6.67) + (3.75) X10 N = 7.65x10
N
2 2 2 2 −11 −11
X Y
7

Example 8: An astronaut standing on the surface of Ceres, the largest asteroid drops a rock from
a height of 10.0 m. It takes 8.06 s to hit the ground.
(a) Calculate the acceleration of gravity on Ceres:
1
Δ x = vt
o
+ at
2
2
1
− 10.0m= 0 + [ − g (8.06 s) ] = 0.308 m/
s
2 C
2 2
(b) Find the mass of Ceres, given that the radius of Ceres is 510 km.
=
M m
C
mgC
G R
2
C
g R
= = g
G
2
C C
21
M
C
1.20x10
k
Gravity near the Earth’s Surface; Geophysical Applications
Now we can relate the gravitational constant to the local acceleration of gravity. We know that, on
the surface of the Earth:
Solving for g gives:
Now, knowing g and the radius of the Earth, the mass of the Earth can be calculated:
Kg
Gravity Near the Earth’s Surface; Geophysical Applications
The acceleration due to gravity varies over the Earth’s surface
due to altitude, local geology, and the shape of the Earth, which
is not quite spherical.
Example 9: Estimate the effective value of g on top of Mt.
Everest, 8850 m above sea level.
−11 2 2 24
mE
(6.67x10 N⋅
m / kg )(5.98x10 kg)
g = G = = 9.77 m/
s
2 6 2
r
(6.389x10 )
r = 6380 km + 8.9 km = 6389 km
2
8

Satellites and “Weightlessness”
Satellites are routinely put into orbit around the Earth. The
tangential speed must be high enough so that the satellite
does not return to Earth, but not so high that it escapes
Earth’s gravity altogether.
The satellite is kept in orbit by its speed – it is continually falling, but
the Earth curves from underneath it.
Objects in orbit are said to experience weightlessness. They do have
a gravitational force acting on them, though!
The satellite and all its contents are in free fall, so there is no normal
force. This is what leads to the experience of weightlessness.
r
Example 10: A geosynchronous satellite is one that stays above
the same point on the Earth, which is possible only if it is above
a point on the equator. Determine the height above the Earth’s
surface such a satellite must orbit.
2
3 GmET
r =
2
4π
T = 1 day = (24h)(3600s/h)=86,400s
Gm T (6.67x10 )(5.98x10 )(86,400)
4π
4π
2 −11 24 2
3 E
22 3
= = = 7.54x10
m
2 2
7
r = 4.23x10
m or 42,300km − 6380km = 36,000km
The satellites speed has to be
−11 24
GmE
(6.67x10 )(5.98x10 )
v= = = 3070 m/
s
7
r
(4.23x10 )
Escape Speed
• The escape speed is the speed needed for an object to soar off into space and not return
v
esc
=
2GM
E
R
E
9

• For the earth, v esc is about 11.2 km/s
• Note, v is independent of the mass of the object
Various Escape Speeds
• The escape speeds for various members of the solar
system
• Escape speed is one factor that determines a
planet’s atmosphere
Johannes Kepler (1571 – 1630)
was a German mathematician, astronomer and astrologer,
and a key figure in the 17 th century astronomical revolution.
He is best known for his laws of planetary motion.
Kepler’s Laws and Newton's Synthesis
Kepler’s laws describe planetary motion.
1. The orbit of each planet is an ellipse, with
the Sun at one focus.
10

2. An imaginary line drawn from each planet to the Sun
sweeps out equal areas in equal times.
3. The square of the orbital period of any planet is
proportional to cube of the average distance from the Sun
to the planet.
2 3
T = Kr
– For orbit around the Sun, K = K S = 2.97x10 -19 s 2 /m 3
– K is independent of the mass of the planet
Example 11: Mar’s period was noted by Kepler to be about 687 days (Earth days), which is
(687d/365d) = 1.88yr. Determine the distance of Mars from the Sun using the Earth as a reference.
11
rES
= 1.50x10
m
2 2
r ⎛ 3 3
MS
T ⎞
M
⎛1.88yr⎞
= ⎜ ⎟ = ⎜ ⎟ = 1.52
rES
⎝TE
⎠ ⎝ 1yr
⎠
11
So Mars is 1.52 times the Earth’s distance from the sun or 2.28x10
m .
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CHAPTER 5
CIRCULAR MOTION & GRAVITATION
CONCEPTS
1. Two masses, A and B, move in circular paths as shown to the right. The
centripetal acceleration of mass A, compared to that of mass B, is one-half
as great.
2. The diagram to the right represents a 4.0 x 10 2 kg satellite, S, in a circular orbit
at an altitude of 5.6 x 10 6 m. The orbital speed of the satellite is 5.7 x 10 3 m/s
and the radius of the Earth, R, is 6.4 x 10 6 m. If the altitude decreased, its
centripetal acceleration would increase.
3. The diagram to the right represents a ball undergoing uniform circular motion as it
travels clockwise on a string. At the moment shown in the diagram, the correct
directions of both the velocity and centripetal acceleration of the ball is B.
4. A convertible car with its top down is traveling at constant speed around a
circular track. When the car is at point A, a passenger in the car throws a ball
straight up, the ball could land at point C.
5. A car going around a curve is acted upon by a centripetal force, F, if the speed of the car
were twice as great, the centripetal force necessary to keep it moving in the same path would
be 4F.
Concepts 6 – 8 refer to the following: The diagram on the right shows an
object with a mass of 1.0 kg attached to a string 0.50 m long. The object is
moving at a constant speed of 5.0 m/s in a horizontal circular path with
center at point O.
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6. While the object is undergoing uniform circular motion, its acceleration is directed toward
the center of the circle.
7. If the string is lengthened while the speed of the object remains constant, the centripetal
acceleration of the object will decrease.
8. If the string is cut when the object is at position shown, the path the object will travel from
this position will be a straight line tangent to the circle.
Concepts 9 – 13 refer to the following: The diagram to the right represents a ball of
mass M attached to a string. The ball moves at a constant speed around a flat
horizontal circle of radius R.
9. The centripetal acceleration of the ball is constant in magnitude, but changing in
direction.
10. If the velocity of the ball is doubled, the centripetal accelration is quadrupled.
11. If the string is shortened while the speed of the ball remains the same, the centripetal
acceleration will increase.
12. When the ball is in the position shown, the directioin of the centripetal force is toward point D.
13. If the mass of the ball is decreased, the centripetal force required to keep it moving in the
same circle at the same speed decreases.
Concepts 14 & 15 refers to the following: A 60-kg adult and a 30-kg child are passengers
on a rotor ride. When the rotating hollow cylinder reaches a certain constant speed, v, the
floor moves downward. Both passenger stay “pinned” against the wall of the rotor, as
shown to the right.
14. The magnitude of the frictional force between the adult and the wall
of the spinning rotor is F. The magnitude of the frictional force
between the child and the mass of the spinning rotor is 1/2 F.
15. Compared to the magnitude of the acceleration of the adult, the
magnitude of the acceleration of the child is the same.
16. The best description of the force keeping the ball in the circular path is that it is
perpendicular to the circle and directed toward the center of the circle.
17. The shape of the path of the Earth about the Sun is an ellipse
with the Sun at one focus.
18. The shapes of the paths of the planets about the Sun are all
ellipses with the Sun at one focus.
13

19. The increase in a planet’s speed as it approaches the Sun is described by
Kepler’s Second Law. The best explanation for this empirical law is that
the gravitational potential energy lost as the planet nears the Sun becomes
kinetic energy.
20. As the distance of a satellite from Earth’s surface increases, the time
the satellite takes to make one revolution around the Earth increases.
21. The condition required for a satellite to be in a geosynchronous orbit
around the Earth’s is that the period of revolution of the satellite must
be the same as the rotational period of the Earth.
22. The banking angle in a turn on the Olympic bobsled track is not constant,
But increases upward from the horizontal. Coming around a turn, the
bobsled team will intentially “climb the wall”, then go lower coming out of
the turn. They do this to take the turn at a faster speed.
23. A roller coaster car is on a track that forms a circular loop in the vertical
plane. If the car is to just maintain contact with the track at the top of the
loop, the minimum value for its speed at that point is rg
24. A pilot executes a vertical dive, and then follows a semi-circular arc
until it is going straight up. Just as the plane is at its lowest point, the
force on him is more than g, and pointing up.
25. A spaceship is traveling to the moon. It is never beyond the pull of Earth’s
gravity.
14

26. An astronaut is inside a space capsule in orbit around the Earth. He is able to
float inside the capsule because he and his capsule move with the same
constant acceleration.
27. A car goes around a curve of radius r at a constant speed v. The direction of the net
force on the car is toward the curve’s center.
28. Two objects attract each other gravitationally. If the distance between their centers
doubles, the gravitational force is cut a fourth.
29. Two objects gravitationally attract each other. If the distance between their centers is
cut in half, the gravitational force between them quadruples.
30. The acceleration of gravity on the moon is one-sixth what it is on Earth. An object of mass
72 kg is taken to the moon. Its mass there is 72 kg.
15