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Dynamics of Uniform Circular Motion
Chapter 5
Learning Objectives- Circular motion and rotation Uniform circular motion Students should understand the uniform circular motion of a particle, so they
can: Relate the radius of the circle and the speed or rate of revolution of the
particle to the magnitude of the centripetal acceleration. Describe the direction of the particle’s velocity and acceleration at any
instant during the motion. Determine the components of the velocity and acceleration vectors at
any instant, and sketch or identify graphs of these quantities. Analyze situations in which an object moves with specified
acceleration under the influence of one or more forces so they can determine the magnitude and direction of the net force, or of one of the forces that makes up the net force, in situations such as the following:• Motion in a horizontal circle (e.g., mass on a rotating merry-go-
round, or car rounding a banked curve).• Motion in a vertical circle (e.g., mass swinging on the end of a
string, cart rolling down a curved track, rider on a Ferris wheel).
Table Of Contents
5.1 Uniform Circular Motion
5.2 Centripetal Acceleration
5.3 Centripetal Force
5.4 Banked Curves
5.5 Satellites in Circular Orbits
5.6 Apparent Weightlessness and Artificial Gravity
5.7 Vertical Circular Motion
Chapter 5:
Dynamics of Uniform Circular Motion
Section 1:
Uniform Circular Motion
Other Effects of Forces
Up until now, we’ve focused on forces that speed up
or slow down an object.
We will now look at the third effect of a force
Turning
We need some other equations as the object will be
accelerating without necessarily changing speed.
Uniform circular motion is the motion of an object traveling at a constant speed on a circular path.
DEFINITION OF UNIFORM CIRCULAR MOTION
Let T be the time it takes for the object totravel once around the circle.
vr
T
2
r
Example 1: A Tire-Balancing Machine
The wheel of a car has a radius of 0.29m and it being rotatedat 830 revolutions per minute on a tire-balancing machine. Determine the speed at which the outer edge of the wheel ismoving.
revolutionmin102.1minsrevolution830
1 3
s 072.0min 102.1 3 T
sm25
s 072.0
m 0.2922
T
rv
Newton’s Laws
1st
When objects move along a straight line the sideways/perpendicular forces must be balanced.
2nd When the forces directed perpendicular to velocity become
unbalanced the object will curve. 3rd
The force that pulls inward on the object, causing it to curve off line provides the action force that is centripetal in nature. The object will in return create a reaction force that is centrifugal in nature.
Chapter 5:
Dynamics of Uniform Circular Motion
Section 2:
Centripetal Acceleration
In uniform circular motion, the speed is constant, but the direction of the velocity vector is not constant.
9090
r
tv
v
v
r
v
t
v 2
r
vac
2
The direction of the centripetal acceleration is towards the center of the circle; in the same direction as the change in velocity.
r
vac
2
Conceptual Example 2: Which Way Will the Object Go?
An object is in uniform circular motion. At point O it is released from its circular path. Does the object move along the straightpath between O and A or along the circular arc between pointsO and P ?
Straight path
Example 3: The Effect of Radius on Centripetal Acceleration
The bobsled track contains turns with radii of 33 m and 24 m. Find the centripetal acceleration at each turn for a speed of 34 m/s. Express answers as multiples of .sm8.9 2g
rvac2
m 33
sm34 2
ca
m 24
sm34 2
ca
2sm35 g6.3
2sm48 g9.4
Chapter 5:
Dynamics of Uniform Circular Motion
Section 3:
Centripetal Force
Recall Newton’s Second Law
aF
mm
F
a
When a net external force acts on an objectof mass m, the acceleration that results is directly proportional to the net force and hasa magnitude that is inversely proportional tothe mass. The direction of the acceleration isthe same as the direction of the net force.
r
vmmaF cc
2
Recall Newton’s Second Law
Thus, in uniform circular motion there must be a net force to produce the centripetal acceleration.
The centripetal force is the name given to the net force required to keep an object moving on a circular path.
The direction of the centripetal force always points toward the center of the circle and continually changes direction as the object moves.
Problem Solving Strategy – Horizontal Circles
1. Draw a free-body diagram of the curving object(s).
2. Choose a coordinate system with the following two axes.
a) One axis will point inward along the radius (inward is positive direction).
b) One axis will point perpendicular to the circular path (up is positive direction).
3. Sum the forces along each axis to get two equations for two unknowns.
a) FRADIUS: +FIN FOUT = m(v2)/ r
b) F : FUP FDOWN = 0
4. Do the math of two equations with two unknowns.
R
Just in case…
The third dimension in these problems would be a direction tangent to the circle and in the plane of the circle.
We choose to ignore this direction for objects moving at constant speed.
If an object moves along the circle with changing speed then the forces tangent to the circle have become unbalanced.
You can sum the tangential forces to find the rate at which speed changes with time, aTAN.
The linear kinematics equations can then be used to describe motion along or tangent to the circle. FTAN: FFORWARD FBACKWARD = m aTAN
R
tan
Example 5: The Effect of Speed on Centripetal Force
The model airplane has a mass of 0.90 kg and moves at constant speed on a circle that is parallel to the ground. The path of the airplane and the guideline lie in the samehorizontal plane because the weight of the plane is balancedby the lift generated by its wings. Find the tension in the 17 mguideline for a speed of 19 m/s.
r
vmTFc
2
m 17
sm19kg 90.0
2
T N 19
5.3.1. A boy is whirling a stone at the end of a string around his head. The string makes one complete revolution every second, and the tension in the string is FT. The boy increases the speed of the stone, keeping the radius of the circle unchanged, so that the string makes two complete revolutions per second. What happens to the tension in the sting?
a) The tension increases to four times its original value.
b) The tension increases to twice its original value.
c) The tension is unchanged.
d) The tension is reduced to one half of its original value.
e) The tension is reduced to one fourth of its original value.
Chapter 5:
Dynamics of Uniform Circular Motion
Section 4:
Banked Curves
On an unbanked curve, the static frictional forceprovides the centripetal force.
Unbanked curve
On a frictionless banked curve, the centripetal force is the horizontal component of the normal force. The vertical component of the normal force balances the car’s weight.
Banked Curve
r
vmFF Nc
2
sin
mgFN cosrg
v2
tan
Example 8: The Daytona 500
The turns at the Daytona International Speedway have a maximum radius of 316 m and are steely banked at 31degrees. Suppose these turns were frictionless. As what speed would the cars have to travel around them?
rg
v2
tan tanrgv
31tansm8.9m 316 2v
mph 96 sm43v
Chapter 5:
Dynamics of Uniform Circular Motion
Section 5:
Satellites in Circular Orbits
There is only one speed that a satellite can have if the satellite is to remain in an orbit with a fixed radius.
Don’t worry, it’s only rocket science
2r
mMG E
r
GMv E
r
vmFc
2
Example 9: Orbital Speed of the Hubble Space Telescope
Determine the speed of the Hubble Space Telescope orbitingat a height of 598 km above the earth’s surface.
m10598m1038.6
kg1098.5kgmN1067.636
242211
v
mph 6,9001 sm1056.7 3 v
22MACHv
r
GMv E
T
r
r
GMv E 2
EGM
rT
232
Period to orbit the Earth
hours 24T
EGM
rT
232
Geosynchronous Orbit
3
2
243
112
2
109742.51067.6400,86
kgxskg
mxs
r
3
2
2
2EGMT
r 3
2
2
2EGMT
r
mxr 71022.4 mxrE61038.6
milesmxh 000,221058.3 7
Chapter 5:
Dynamics of Uniform Circular Motion
Section 6:
Apparent Weightlessness and Artificial Gravity
Conceptual Example 12: Apparent Weightlessness and Free Fall
In each case, what is the weight recorded by the scale?
Example 13: Artificial Gravity
At what speed must the surface of the space station moveso that the astronaut experiences a push on his feet equal to his weight on earth? The radius is 1700 m.
mgr
vmFc
2
2sm80.9m 1700
rgv
smv 130
Chapter 5:
Dynamics of Uniform Circular Motion
Section 7:
Vertical Circular Motion
Circular Motion
In the previous lesson the radial and the perpendicular forces were emphasized while the tangential forces were ignored. Each class of forces serves a different function for objects moving along a circle.
Class of Force Purpose of the Force
Radial Forces Curves the object off a straight-line path.
Perpendicular Forces Holds the object in the plane of the circle.
Tangential Forces Changes the speed of the object along the circle.
Circular Motion
Most of the horizontal, circular problems occurred at constant speed so that we could ignore the tangential forces. The vertical, circular problems have objects moving with and against gravity so that speed changes. Tangential forces become significant. The good news is that perpendicular forces can now be ignored unless hurricanes are present.
Problem Solving Strategy for Vertical Circles
1. Draw a free-body diagram for the curving objects.
2. Choose a coordinate system with the following two axes.
a) One axis will point inward along the radius.
b) One axis points tangent to the circle in the circular plane, along the direction
of motion.
3. Sum the forces along each axis to get two equations for two unknowns.
a) FRADIUS: +FIN FOUT = m(v2)/ r b) FTAN : FFORWARD FBACKWARDS = ma
4. You can generally expect the weight of the object to have components in both
equations unless the object is exactly at the top, bottom or sides of the circle.
5. If the object changes height along the circle you may need to write a
conservation of energy statement. This goes well with centripetal forces since
there is an {mv2} in both kinetic energy terms and in centripetal force terms.
6. Do the math with 3(a) and 4 or perhaps 3(a) and 3(b).
Minimum/Maximum Speed Problems
Sometimes the problem addresses “the minimum speed” that an object can move through the top of the circle or “maximum speed” that an object can move along the top of the circle.
If the bucket of water turns too slowly you get wet. If a car tops a hill too quickly it leaves the ground. Allowing v2/r to equal g can solve many of these questions. By solving for v you will find a critical speed.
WC FF mgr
mv
2
rgv
Conceptual Example: A Trapeze Act
In a circus, a man hangs upside down from a trapeze, legsbent over and arms downward, holding his partner. Is it harderfor the man to hold his partner when the partner hangs straight down and is stationary of when the partner is swingingthrough the straight-down position?
r
vmmgFN
21
1
r
vmmgFN
23
3
r
vmFN
22
2
r
vmFN
24
4
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