Transcript Dynamics of circular motion

Physics 7C lecture 04
Dynamics: Circular Motion
Thursday October 10, 8:00 AM – 9:20 AM
Engineering Hall 1200
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Dynamics of circular motion
• If a particle is in
uniform circular
motion, both its
acceleration and the
net force on it are
directed toward the
center of the circle.
• The net force on the
particle is Fnet = mv2/R.
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What if the string breaks?
• If the string breaks, no net force acts on the ball, so it
obeys Newton’s first law and moves in a straight
line.
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Avoid using “centrifugal force”
• Figure (a) shows the
correct free-body
diagram for a body in
uniform circular
motion.
• Figure (b) shows a
common error.
• In an inertial frame of
reference, there is no
such thing as
“centrifugal force.”
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Force in uniform circular motion
• A sled on frictionless ice is kept in uniform circular
motion by a rope.
• Follow Example 5.19.
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Q5.11
A sled moves on essentially frictionless
ice. It is attached by a rope to a vertical
post set in the ice. Once given a push,
the sled moves around the post at
constant speed in a circle of radius R.
If the rope breaks,
A. the sled will keep moving in a circle.
B. the sled will move on a curved path, but not a circle.
C. the sled will follow a curved path for a while, then move
in a straight line.
D. the sled will move in a straight line.
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A5.11
A sled moves on essentially frictionless
ice. It is attached by a rope to a vertical
post set in the ice. Once given a push,
the sled moves around the post at
constant speed in a circle of radius R.
If the rope breaks,
A. the sled will keep moving in a circle.
B. the sled will move on a curved path, but not a circle.
C. the sled will follow a curved path for a while, then move
in a straight line.
D. the sled will move in a straight line.
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A conical pendulum
• A bob at the end of a wire moves in a horizontal
circle with constant speed.
• Follow Example 5.20.
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Q5.12
A pendulum bob of mass m is attached
to the ceiling by a thin wire of length L.
The bob moves at constant speed in a
horizontal circle of radius R, with the
wire making a constant angle  with
the vertical. The tension in the wire
A. is greater than mg.
B. is equal to mg.
C. is less than mg.
D. is any of the above, depending on the bob’s speed v.
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A5.12
A pendulum bob of mass m is attached
to the ceiling by a thin wire of length L.
The bob moves at constant speed in a
horizontal circle of radius R, with the
wire making a constant angle  with
the vertical. The tension in the wire
A. is greater than mg.
B. is equal to mg.
C. is less than mg.
D. is any of the above, depending on the bob’s speed v.
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A car rounds a flat curve
• A car rounds a flat unbanked curve. What is its
maximum speed?
• Follow Example 5.21.
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A car rounds a flat curve
• If the car is moving at speed v, what is the
coefficient of friction?
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A car rounds a flat curve
• f = m v2/R
• f = μ n = μ mg
• we have: μ = v2/ (R g)
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A car rounds a banked curve
• At what angle should a curve be banked so a car can
make the turn even with no friction?
• Follow Example 5.22.
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A car rounds a banked curve
• If the car is moving at speed v, what is the tilt of the
track?
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A car rounds a banked curve
• n cos β = mg
• n sin β = m v2/R
• we have: tan β = v2/(R g)
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Uniform motion in a vertical circle
• A person on a Ferris wheel moves in a vertical circle.
• Follow Example 5.23.
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Q5.13
A pendulum of length L with a bob of mass m swings back and
forth. At the low point of its motion (point Q ), the tension in the
string is (3/2)mg. What is the speed of the bob at this point?
A. 2 gL
B. 2gL
C. gL
gL
D.
2
gL
E.
2
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P
R
Q
A5.13
A pendulum of length L with a bob of mass m swings back and
forth. At the low point of its motion (point Q ), the tension in the
string is (3/2)mg. What is the speed of the bob at this point?
A. 2 gL
B. 2gL
C. gL
gL
D.
2
gL
E.
2
© 2012 Pearson Education, Inc.
P
R
Q
The fundamental forces of nature
• According to current understanding, all forces are
expressions of four distinct fundamental forces:
• gravitational interactions
• electromagnetic interactions
• the strong interaction
• the weak interaction
• Physicists have taken steps to unify all interactions
into a theory of everything.
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Dynamics of circular motion
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Dynamics of circular motion
difference between time 0 and t
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Dynamics of circular motion
difference between time 0 and t
Δv
when t is small,
Δv = v * ω t = v2 t / R
(since ω = v / R)
let t = dt :
dv = v2/R dt
a = dv / dt = v2/R
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Six artificial satellites
Six artificial satellites complete one
circular orbit around a space station in
the same amount of time. Each satellite
has mass m and radius of orbit L. The
satellites fire rockets that provide the
force needed to maintain a circular orbit
around the space station. The
gravitational force is negligible.
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Six artificial satellites
Six artificial satellites complete one
circular orbit around a space station in
the same amount of time. Each satellite
has mass m and radius of orbit L. The
satellites fire rockets that provide the
force needed to maintain a circular orbit
around the space station. The
gravitational force is negligible.
ω is the same for all satellites.
T = ω2 m L, scaling as the product of m and L
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