Physics 2514
Lecture 19
P. Gutierrez
Department of Physics & Astronomy
University of Oklahoma
Physics 2514 – p. 1/14
Goals
Start the discussion of circular motion.
We will define the kinematic variables needed to describe
circular motion.
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Circular Motion
We will start with the case of uniform circular motion. (This
refers to circular motion with a constant speed.)
Some definitions
Period (T ) time interval to complete one orbit, therefore
speed is given by
v=
circumference
2πr
=
period
T
r is the radius of the circle.
Physics 2514 – p. 3/14
Circular Motion
Some definitions—Use angular variables to measure
position
θ(radians) =
s
r
θfull circle = 2π
θ increases in counter-clockwise
direction decreases in clockwise
direction.
θ measured from x axis
Physics 2514 – p. 4/14
Circular Motion
Some definitions—Use angular variables to measure speed
Angular Velocity
ωaverage =
∆θ
∆t
dθ
∆θ
=
∆t→0 ∆t
dt
ω = lim
ω > 0 motion counter-clockwise
ω < 0 motion clockwise
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Angular Position vs. Time
Slope of angular position vs. time plot gives angular velocity vs.
time plot.
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Clicker
A particle moves clockwise around a circle at constant speed for
2.0 s. It reverses direction and moves counter clockwise at half
the original speed until it has traveled through the same angle.
Which is the particles angle-versus-time graph?
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Kinematic Equations
For uniform circular motion ω is constant.
From definition of angular velocity
dθ
ω=
dt
⇒
Z
θf
dθ =
θ0
Z
tf
ωdt = θf = θ0 + ω(tf − t0 )
t0
Take t0 = 0, tf = t, then
θ(t) = ωt + θ0
Tangential acceleration is zero (acceleration in direction
objects moves).
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Coordinate System
Tangential velocity and acceleration in terms of angular variables
Recall: s = rθ ⇒ vt =
ds
dt
= r dθ
dt = rω ω is angular velocity.
Therefore: vt = rω ⇒ at =
acceleration.
dvt
dt
= r dω
dt = rα α is angular
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Kinematic Equations
The kinematic equations for uniform circular motion were
derived earlier, here we consider nonuniform motion
Motion along arc is 1-D with tangential
acceleration and velocity determining motion
s = s0 + vot t +
1
at t 2
2
vt = v0t + at t
Now divide by r
1
1
(s = s0 + v0t t + at t2 )
r
2
1
(vt = v0t + at t)
r
θ = θ 0 + ω0 t +
ω = ω0 + αt
1 2
αt
2
α = at /r
Physics 2514 – p. 10/14
Centripetal Acceleration
Assume constant tangential speed (|~v|)
Acceleration points to center of circle (~ar ⊥ ~v)
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Centripetal Acceleration
Calculate average acceleration
CB = ∆~r2 −∆~r1 = ~
v2 ∆t−~
v1 ∆t = ∆~
v∆t
Angles
ABO: θ + α + α = 180
DAC: φ + α + α = 180
)
⇒
θ=φ
Similar triangles
AB
CB
=
AB
AO
⇒
|∆~
v|∆t
v∆t
=
v∆t
r
Average radial acceleration
average
ar
|∆~
v|
v2
ar = lim
=
∆t→0 ∆t
r
|∆~
v|
v2
=
=
∆t
r
Physics 2514 – p. 12/14
Clicker
A ball is lodged in a hole in the floor of a merry-go-round that is
turning at constant speed. Which kinematic variable or variables
change with time, assuming that the position is measured
relative to a fixed coordinate system with its origin at the center
of the merry-go-round?
1. the position of the ball only;
2. the velocity of the ball only;
3. the acceleration of the ball only;
4. both the position and velocity of the ball;
5. the position and velocity and acceleration of the ball.
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Assignment
Continue reading Chapter 7
Will discuss dynamics in next lecture
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