Download lecture 14 circular motion

LECTURE 14
CIRCULAR MOTION
Instructor: Kazumi Tolich
Lecture 14
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Reading chapter 6-5
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Velocity
Uniform circular motion
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Acceleration and speed
Centripetal and tangential acceleration of non-uniform circular motion
Dynamics of circular motion
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Centripetal force
Unbanked and banked curves
Loop-the-Loop
Circular motion
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Motion along a circular
path or a segment of a
circular path is called
circular motion.
¨  An object in a circular
motion is always
accelerating.
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Velocity of circular motion
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Consider the motion of a pendulum.
The direction of the velocity is always tangential to the path,
perpendicular to the circle’s radius.
Uniform circular motion
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Motion in a circle at constant speed (not constant
velocity!).
Constantly accelerating since direction of the
velocity is changing.
The direction of acceleration in uniform circular
motion is always toward the center of the circle,
“centripetal” (center-seeking) acceleration.
The magnitude of the centripetal acceleration is
given by
v2
acp =
r
Centripetal acceleration
v
ac
Why don’t satellites fall into Earth?
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Q.
A.
Why don’t satellites fall into Earth
because of Earth’s gravity?
They do.
The tangential velocity of the satellite
is fast enough so that the distance the
satellite falls towards Earth, and the
distance it travels in the tangential
direction in a given time follow the
satellite’s orbit.
Example: 1
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A spinner runs at
v = 9.2 m/s around a
circular track with a
centripetal acceleration
of acp = 3.8 m/s2.
What is the track
radius, r?
Acceleration of non-uniform circular motion
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The total acceleration of a particle moving in a
circular path with tangential acceleration is given by
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atotal = acp + at
http://www.youtube.com/watch?v=vT1mLpD6H_c
Clicker question: 1
Centripetal force
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To keep an object in a circular motion, force must be
applied toward the center of the circle.
Any form of such force (normal force, tension, gravity, etc) is
called “centripetal force.”
Centripetal force is just a name, so you should NEVER label
a “centripetal force” in a FBD.
v2
∑ Fcp = m r
Demo 1
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Circle with Gap
Direction of travel
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Imagine a rock is whirled around in a uniform circular motion by a
string.
The tension in the string supplies the centripetal force necessary for
the rock to follow the circular path.
The velocity of the rock is always tangent to the path.
If the string breaks, the rock will move along the straight line,
tangent to the circular path. Recall Newton’s first law.
no
Clicker question: 2
Example: 2
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Igor is a cosmonaut-engineer
in a spacecraft orbiting
Earth at an altitude
h = 520 km with a speed
v = 7.6 km/s. Igor’s mass is
m = 79 kg.
a) 
b) 
What is his acceleration?
What (centripetal)
gravitational force does
Earth exert on Igor?
Unbanked curves
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On a unbanked road, static friction by the road on the
car provides the centripetal force necessary for the
car to follow a curve.
¨  It is static friction because no slipping occurs at the
point of contact between road and tires.
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!
N
Unbanked curves: 2
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The maximum speed the car can go is limited by the
maximum static friction the road can exert on the tire.
x direction : − fs,max = − µs N = −macp,max
2
vmax
= −m
r
y direction : N − mg = 0 → N = mg
2
vmax
µs N = m
r
2
vmax
µs ( mg ) = m
r
vmax = µs gr
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N
!
N
+x
Example: 3
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A bicyclist travels at a
constant speed of
v = 9.00 m/s in a circle of
radius r = 25.0 m on a
flat ground. The combined
mass of the bicycle and
rider is m = 85.0 kg.
Calculate the magnitude
of the force of friction
exerted by the road on
the bicycle.
Banked curves
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On a banked road, the normal force of the road will
have a component in the centripetal direction.
¨  Banking is useful for an icy road.
¨  The banking angle is usually chosen so that no friction
is needed for a car to complete the curve at the
specified speed.
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N
Banked curves: 2
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Banking angle:
v2
x direction : N sin θ = macp = m
r
y direction : N cosθ − mg = 0
→ N cosθ = mg
v2
N sin θ m r
=
N cosθ
mg
v2
tan θ =
rg
Banking angle
N
!
N
Loop-the-Loop
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If you are going fast enough, the
normal force on the bike is nonzero, i.e, the bike is touching the
track.
If you are barely making it, the
normal force at the top is zero.
At the top : N = 0
2
vmin
∑ F = N + mg = 0 + mg = m r
vmin = gr
At the top
mg
N
Loop-the-loop failure
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Loop-the-loop fails when you do not have
enough speed at the top.
¤  http://www.youtube.com/watch?v=tzQJNeqiGG4
Example: 4
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The radius of curvature of
the track at the top of a
loop-the-loop on a rollercoaster ride is r = 12.0 m.
At the top of the loop, the
force that the seat exerts on
a passenger of mass m is
0.40mg. How fast is the
roller-coaster car moving as
it moves through the highest
point of the loop?