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Lecture Lecture IVIV
• Centripetal Force
•Law of Gravitation UNIFORM CIRCULAR UNIFORM CIRCULAR MOTIONMOTION
2D KINEMATICS
PERIODPERIOD
Period ( T )
-time for one revolution
-time for a particle to go around a closed path
-unit: Seconds (s)
r : radius of the
circular arc
v : speed of the
particle
2 rT
v
π=
FREQUENCYFREQUENCY
Frequency
-number of revolutions per unit time
--unit: Hertz (Hz) unit: Hertz (Hz) –– 1/second1/second
1f
T=
motion of an object following a circular path while moving at a constant speed
UNIFORM CIRCULAR MOTION UNIFORM CIRCULAR MOTONUNIFORM CIRCULAR MOTON
� Constant speed but velocity changes direction
� Experiences centripetal acceleration.
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Tangential speed is directly proportional to rotational
speed/angular velocity and the distance from the axis (radial distance)
v= ωr
The speed of something moving along a circular path can be called tangential speed because the direction of motion is always tangent to the circle.
UNIFORM CIRCULAR MOTON
Velocity and acceleration
in uniform circular
motion at angular rate ω;
the speed is constant, but
the velocity is always
tangent to the orbit; the
acceleration has constant
magnitude, but always
points toward the center of
rotation.
Period of revolution (Period)
Frequency:
Centripetal acceleration:
Using v from (1):
CENTRIPETAL ACCELERATIONCENTRIPETAL ACCELERATION
2 rT
v
π=
1f
T=
2
c
va
r=
( )2 22 4
c
r ra
T r T
π π= =
Centripetal Acceleration and Angular Frequency
(Angular velocity/ Angular frequency)Unit: rad/s (Radians/second)
1 rev/s = 2 rad/sec
Note: 1 rad=360o/2π=57.3o
2 f
v
r
ω π
ω
=
= π
22
c
va r
rω= =
EXAMPLE 1EXAMPLE 1
In a carnival ride, the passengers travel at In a carnival ride, the passengers travel at constant speed in a circle of radius 5.0m. constant speed in a circle of radius 5.0m. They make one complete circle in 4.0s. What They make one complete circle in 4.0s. What is their velocity and acceleration?is their velocity and acceleration?
SOLUTION 1SOLUTION 1
T
Rv
π2=
( )s
mv
0.4
0.52π=
smv /9.7=
R
vac
2
=
( )m
smac
0.5
/9.72
=
2/12 smac =
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Example 2
What is the centripetal acceleration of an automobile driving at 40km/h on a circular track of radius 20m?
Given:
v = 40km/h,
r = 20m
Required: centripetal acceleration, ac
Equation to use: ac = v2/r
ac=(40000m/3600s)2
20m
= 6.17 m/s2
EXAMPLE 3EXAMPLE 3
The period of a stone swung in a horizontal circle on a 2.00-m radius is 1.00s.
(a) What is its angular velocity in rad/s?
Given: r = 2.00 m, T = 1.00s
Required: ω
Equations to use:
A.ω = 2πf
B.T = 1/f
C.ω=2π/T= 2π rad/s
The period of a stone swung in a horizontal circle on a 2.00-m radius is 1.00s.
(b) What is its linear speed in m/s?
Given: ω (from the previous problem)
Required: v
Equation to use:
ω = v/r
therefore
v = ωr
v = (2π rad/s)2.00m= 4π m/s
The period of a stone swung in a horizontal circle on a 2.00-m radius is 1.00s.
(c) What is its radial acceleration in m/s2?
Given: v (from the previous problem), r
Required: ac
Equation to use: ac = v2/r
ac = v2/r=(4π m/s) 2/2.00m
=78.5m/s2
Period and Frequency
Equations to remember!Equations to remember!
v= ωr (Tangential speed or linear speed)
Centripetal Acceleration
Angular Frequency/velocity
Unit: rad/s (Radians/second)
22
c
va r
rω= =
( )2 2
2 2
2 4c
r ra
T r T
π π= =
v
rω = 2 fω π=
1f
T=
2 rT
v
π=
CENTRIPETAL FORCE
�Centripetal Force is a force that tends to keep
object in moving around a circular arc or path
�The magnitude of the centripetal force is the
product of an object’s mass and its centripetal acceleration as it moves around the circular
path
�The direction of the centripetal force is
always directed towards the center of the circle.
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Centripetal acceleration
� An object traveling in a circle, even though it moves with a constant speed, will have an acceleration (since velocity changes direction)
� This acceleration is called centripetal (“center-seeking”).
� The acceleration is directed toward the center of the circle of motion
Centripetal acceleration and
the angular velocity
� The angular velocity and the
linear velocity are related
(v = ωr)
� The centripetal acceleration can also be related to the
angular velocitySimilar
triangles!
, but
a
v s vv s
v r r
v v sa
t r t
∆ ∆= ⇒ ∆ = ∆
∆ ∆= ⇒ =
∆ ∆
uurr
22orC C
va a r
rω= =
Thus:
Similar
triangles!
Total acceleration
� What happens if linear velocity also changes?
� Two-component acceleration:
� the centripetal component of the acceleration is due to changing direction
� the tangential component of the acceleration is due to changing speed
� Total acceleration can be found from these components:
slowing-down car
2 2
t Ca a a= +
Forces Causing Centripetal
Acceleration
� Newton’s Second Law says that the centripetal acceleration is accompanied by a force
� F stands for any force that keeps an object following a circular path
� Force of friction (level and banked curves)
� Tension in a string
� Gravity
2
C
vF ma m
r= =∑
F
F
F
F
Centripetal Force
Net inward force to provide centripetal acceleration
Due to contact and/or gravitational forces
Direction:
towards the center
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Centripetal Force
This is not a new kind of force, but merely a name for the force needed for circular motion.
The centripetal force maybe due to:string, spring, or other contact force such as normal force and friction, action-at-a-distance forces such as gravity;
SOME EXAMPLES
For a rock whirled on the end of a string, the
centripetal force is the force of tension in the string.
For an object sitting on a rotating turntable, the
centripetal force is friction.
For the motion of the Earth around the Sun, the
centripetal force is gravity.
centripetal force could be a combination of
two or more forces. For example, as a
Ferris-wheel rider passes through the
lowest point, the centripetal force on her is
the difference between the normal force
exerted by the seat and her weight.
Example 1: level curves
Consider a car driving at 20 m/s (~45 mph) on a level circular turn of radius 40.0 m. Assume the car’s mass is 1000 kg.
1. What is the magnitude of frictional force experienced by car’s tires?
2. What is the minimum coefficient of friction in order for the car to safely negotiate the turn?
Example a:
Given:
masses: m=1000 kg
velocity: v=20 m/s
radius: r = 40.0m
Find:
1. f=?
2. µ=?
1. Draw a free body diagram,
introduce coordinate frame and
consider vertical and horizontal
projections
�
0y
F N mg
N mg
= = −
=
∑
( )
2
2
420
1000 1.0 1040
xF ma f
vf ma m
r
m skg N
m
= = −
= − = −
= − = − ×
∑
2. Use definition of friction
force:
�
Lesson: µ for rubber on dry concrete is 1.00!
rubber on wet concrete is 0.2!
driving too
fast…
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4
2
10 , thus
1.0 101.02
1000 9.8
vf mg m N
r
N
kg m s
µ
µ
= = =
×= ≈
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Assignment (1/2): banked curves
Consider a car driving at 20 m/s (~45 mph) on a 30°banked circular curve of radius 40.0 m. Assume the car’s mass is 1000 kg.
1. What is the magnitude of frictional force experienced by car’s tires?
2. What is the minimum coefficient of friction in order for the car to safely negotiate the turn?
NEWTON’S LAW OF GRAVITATION
Newton’s Law of Universal
Gravitation� 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.
2
21
r
mmGF =
�� G is the universal gravitational constantG is the universal gravitational constant
�� G = 6.673 x 10G = 6.673 x 10--1111 N m² /kg²N m² /kg²
�� This is an example of an This is an example of an inverse square lawinverse square law
Gravitation Constant
� Determined experimentally
� Henry Cavendish
� 1798
� The light beam and mirror serve to
amplify the motion
Example
Question: Calculate gravitational attraction between two students 1 meter apart. Assume the student 1 has a mass of 70 kg while the other one has a mass of 90 kg.
( )
21 11 2
22 2
7 0 9 06 .6 7 1 0
1
m m N m kg kgF G
r kg m
−= = ×
74.2 10F N−≈ × Extremely small
compared to the
weight (F = mg).
Applications of Universal
Gravitation 1: Mass of the Earth� Use an example of an object
close to the surface of the earth
� r ~ RE
G
gRM E
E
2
=
11 2
E
E
GM mm g
R=
7
Applications of Universal
Gravitation 2: Acceleration Due
to Gravity� g will vary with altitude
2r
MGg E=
mgr
MGm
r
mMGF EE =
==
22
Escape Speed
� The escape speed is the speed needed for an
object to soar off into space and not return
� For the earth, vesc is about 11.2 km/s
� Note, v is independent of the mass of the object
E
Eesc
R
GMv
2=
PHYSICS FAIR
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