Uniform Circular Motion
Horizontal Circle
Vertical Circle
Banked Curve
Unbanked Curve
Newton’s Law of Universal Gravitation
Kepler’s Laws
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Uniform Circular Motion

Motion of an object traveling at a
constant speed in a circular path.
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Uniform Circular Motion

r
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Horizontal Circle
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Velocity
V


Velocity is constant
in magnitude only.
The direction is
changing with time.
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Centripetal Acceleration

ac = V2
R

ac

Acceleration is
occurring because
the direction of the
velocity is changing.
Acceleration is
always toward the
center of the circular
path.
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Centripetal Force

Fc = mac

Fc

Net force (Fc) and
acceleration always
act in the same
direction.
The centripetal force
keeps the object
moving in a circle
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Centrifugal Force

Fictitious force. Doesn’t exist

Force and acceleration acting outward.
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Derived Equations for
ac and Fc
a = V2
R
F = ma
F = m V2
R
a = (2πR/T)²
R
F = m 4π²R
T²
a = 4π²R
T²
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Example 1. Period
How long does it take a plane traveling at a speed of 110 m/s, to
fly once around a circle whose radius is 2850 m?
Given:
V = 110 m/s
V = 2πr
R = 2850 m
T
T=?
T = 2πr
V
= 2π(2850 m)
110 m/s
T = 160 s
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Example 2. Frequency
It takes 2 min to twirl a ball in a circle 50 times.
What is the frequency?
f=
50
2 min. X 60 s
min
f = 0.4/s
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Example 3. Velocity
A stopper is twirled in a horizontal circle whose radius is
0.75 m. The stopper travels once around the circle in
0.30 s.
What is its velocity?
Given:
R = 0.75 m
V = 2πr
T = 0.30 s
T
= 2π(0.75 m)
0.30 s
V = 16 m/s
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Example 4. Acceleration


A runner moving at a speed of 7.5 m/s rounds a
bend with a radius of 25 m. Determine the
centripetal acceleration of the runner.
Givens:
a = V2
v = 7.5 m/s
R
r = 25 m
= (7.5 m/s)²
a=?
25 m
a = 2.3 m/s²
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Example 5. Acceleration
A race car travels at a constant speed around a circular track
whose radius is 2.6 km. If the car goes around the track in
360 s, what is the magnitude of the centripetal acceleration of
the car?
Given:
ac = 4π² r
R = 2.6 x 10³ m
T²
T = 360 s
= 4 π² (2.6 x 10³ m )
(360s)²
ac = 0.79 m/s²
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Example 6. Centripetal Force
A 0.015 kg ball is shot from the plunger of a
pinball machine. Because of a centripetal
force, the ball follows a circular arc whose
radius is 0.25 m at a speed of 0.68 m/s.
Calculate the centripetal force.
Given:
M = 0.015 kg
Fc = m V²
R = 0.25 m
V = 0.68 m/s
Fc = ?
r
= (0.015 kg)(0.68 m/s)²
0.25 m
Fc = 0.028 N
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Example 7. Centripetal Force
A child is twirling a 0.0120 kg ball on a string in a horizontal circle
whose radius is 0.l00 m. The ball travels once around the circle
in 0.500 s. Determine the centripetal force on the ball.
Given:
M = 0.0120 kg
R = 0.100 m
T = 0.500 s
Fc = m 4π² r
T²
= (0.0120 kg)4 π² (0.100 m)
(0.500 s)²
Fc = 0.189 N
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Example 8. Centripetal Force
A 0.350 kg ball is twirled in a horizontal circle.
The centripetal acceleration is 0.75 m/s².
Calculate the centripetal force.
Given:
m = 0.350 kg
a = 0.75 m/s²
Fc = mac
= (0.350 kg)(0.75 m/s²)
Fc = 0.26 N
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Vertical Circle
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Direction of forces when object is at
top and bottom of vertical circle
Fc = Fnet
Fnet
Top
=T+W
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Bottom
Fnet = T - W
F
c
equations for vertical circle
Fc = Fnet
Object at Top
T + W = mV2
R
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Object at Bottom
T – W = mV2
R
Example 9.
Top of Vertical Circle
A 2.0 kg object is attached to a 1.5 m long string and swung in a
vertical circle at a constant speed of 12 m/s. What is the tension
in the string when the object is at the top of its path?
Given:
m = 2.0 kg
r = 1.5 m
V = 12 m/s
T=?
T + W = mV²
R
T = mV² - W
R
= (2.0 kg)(12 m/s)² - (2.0 kg)(9.80 m/s²)
1.5 m
T = 1700 N
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Example 10. Vertical Circle
What is the tension in the string when the object is at
the bottom of its path?
Given:
m= 2.0 kg
V = 12 m/s
r = 1.5 m
T – W = m V²/r
T = m V² + W
R
= (2.0 kg)(12m/s)²+ 19.6 N
1.5 m
T = 2100 N
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Unbanked Curve
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F c is the Static Friction, (fs)
The static friction acts toward the center of circular motion.
fs = Fc
µsFN = m V²/R
µsmg = m V²/R
µsg = V²/R
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Unbanked Curve
Racing on a flat track, a car going 32 m/s rounds a
curve 56 m in radius. What would be the minimum
coefficient of static friction between tires and road
that would be needed for the car to round the curve
without skidding?
Given:
V = 32 m/s
r = 56 m
µs g = V²/R
µs =(32 m/s)² /(56 m)(9.80 m/s²)
µs = 1.87
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Banked Curve
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Fc = Wx
The x-component of the weight is acting
towards the center of the circular path
Wx = Fc
mg sin Ө = Fc
mg sin Ө = m V²/R
g sin Ө = V²/R
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Example 11. Banked Curve
A 60.0 kg speed skater comes into a curve of 20.0 m
radius and banked at 12º. Calculate the speed
needed to negotiate the curve?
Given:
m = 60.0 kg
R = 20.0m
V=?
g sin Ө = V²/R
V² = Rg sin Ө
=(20.0 m)(9.80 m/s²)(sin 12º)
V = 6.38 m/s
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Motion and Forces on a
Banked Curve


These problems deal with an object (a car, for example)
navigating a circular curve at constant speed. The radius of the
curve is fixed at 10 m, but the speed and mass of the object,
the angle at which the curve is banked, the coefficient of static
friction, and the gravitational field strength can all be varied.
Two views of the object are shown: i) overhead showing the
circular path with acceleration and velocity vectors, ii) ground
level showing the forces acting on the object. As the values of
input parameters are changed, the vectors change in
response. A "fric-o-meter" gauges the ratio of the actual static
friction to the maximum available static friction. This allows the
user to determine whether the object will skid off the curve or
slide down to the center
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Motion and Forces on a
Banked Curve
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Conical Circular Path
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Conical Circular Path
Fc = Tx
The x-component of T provides the Fc to keep the object moving
in a circular path.
T sinӨ = Fc
T sinӨ = m V²/R
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Using two equations to find V
T cos Ө = mg
T sin Ө = mV2/r
T = mg
cos Ө
T sin Ө
T cos Ө
T = mV2/r
sin Ө
=
mg
mV2/r
V2 = rg tan Ө
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Example 12. Conical Path
The Swing Ride at an amusement park travels at 15.0 m/s and the
12.0 m swings make an angle of 25.0º with the vertical as they
move in a circle. What is the tension in the rope of a swing
carrying a 50.0 kg person?
Given:
V = 15.0 m/s
Ө = 25.0º
M = 50.0 kg
R = 12.0 m
T sinӨ = m V²/R
T = m V²
r sinӨ
= (50.0 kg)(15.0 m/s)²
(12.0 m)(sin 25.0º)
T = 2218 N
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Newton’s Law of
Universal Gravitation
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Newton’s Law of
Universal Gravitation
If the path of a planet
were an ellipse, then
the net force on the
planet varies
inversely with the
square of the
distance between
the planet and the
sun.
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Symbols represent
Fg = gravitational
force of attraction
G = gravitational
constant,
6.67 x 10-11 Nm²/kg²
(Henry Cavendish)
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Fg is directly proportional to m1m2
indirectly proportional to R²



Mass of one object
doubled, Fg is doubled.
Mass of one of the objects
is halved, Fg is halved.
Distance between the
centers of masses is
doubled, Fg is decreased to
¼.
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Kepler’s Laws
Three Laws of Planetary Motion
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Law 1

Path of the planets are ellipses with the
sun at one focus.
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Law 2



An imaginary line extending from the
sun to a planet will sweep out equal
areas in equal amounts of time.
Planets move fastest when closest to
the sun.
Planets move slowest when farthest
from the sun.
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Shaded segments = 30 days
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Law 3

The ratio of the squares of the periods
of any two planets in orbit around the
Sun is equal to the ratio of the cubes of
their distances from the sun.
Ta² = ra³
Tb² rb³
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Motion of Planets and
Satellites
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Satellite in Orbit



Uniform circular motion if always at the
same height above Earth
Path has both vertical and horizontal
components
Velocity of horizontal component great
enough… path will follow a curve that
matches the curve of Earth. Projectile
is in orbit.
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Weightlessness

Occurs when the downward force of
gravity is unbalanced and there is no
upward force acting on a satellite in
orbit.
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Deriving Equations
F=F
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Example 1. F = F
F = ma
F = Gm1m2
r²
ma = Gm1m2
r²
a = Gm
r²
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Example 2. F = F
W = mg
Fg = Gm1m2
r²
mg = Gm1m2
r²
g = Gm
r²
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Example 3. F = F
Fc= mV²
r
mV²
r
V²
Fg = Gm1m2
r²
= Gm1m2
r²
= Gm
r
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Example 4. F = F
Fc = m4π²r
T²
m4π²r
T²
Fg = Gm1m2
r²
= Gm1m2
r²
m = 4π²r³
GT²
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