Circular Motion: Angular Variables
• Arc Length:
The arc length s is related to the angle θ (in radians = rad)
as follows:
s
(360o = 2π rad)
θ=
s = rθ
r
• Angular Displacement and Angular Velocity:
Δθ = θ f − θ i
Δθ dθ
ω = lim
=
Δt →0 Δt
dt
(radians/second)
• Tangential Velocity and Angular Velocity:
r ds
vt = vt =
= rω
dt
vt = rω
Tangential
Velocity
• Period and Frequency (constant ω):
2πr 2π
1
C = 2πr = vtT T =
=
f =
vt
ω
T (revolutions/second)
R. Field 2/4/2014
University of Florida
PHY 2053
Page 1
Circular Motion: Radial Acceleration
• Radial (centripetal) Acceleration:
Centripetal acceleration = “toward the center”
r
Δvt = −(vt Δθ )rˆ
Radial
Acceleration
r
aradial
r
Δvt
Δθ
)rˆ = −(vtω )rˆ = −(rω 2 )rˆ
= lim
= −(vt
Δt →0 Δt
Δt
aradial
2
v
r
t
= aradial = rω 2 =
r
R. Field 2/4/2014
University of Florida
r̂
Magnitude = (vt)2/r
Direction = toward the center of the circle
PHY 2053
Page 2
Example Problem
• A puck of mass m slides in a circle of
radius r = 0.5 m on a frictionless table
while attached to a hanging cylinder of
mass M = 2m by a cord through a hole in
the table. What speed of the mass m keeps
the cylinder at rest?
Mgr
Answer:
v=
≈ 3.13m / s
m
Solution:
FT − Mg = 0
FT = maradial
v2
=m
r
v2
m = Mg
r
R. Field 2/4/2014
University of Florida
FT
Mgr
2mgr
=
= 2 gr
m
m
FT
= 2(9.8m / s 2 )(0.5m) ≈ 3.13m / s
Mg
v=
PHY 2053
Page 3
Example: Conical Pendulum
• A stone of mass m is connected
to a cord with length L and
negligible mass. The stone is
undergoing uniform circular
motion in the horizontal plane. If
the cord makes an angle φ with
the vertical direction, what is the
period of the circular motion?
x-component:
T sin φ = ma x = maradial = mrω 2
y-component:
(1)
T cos φ − mg = ma y = 0
T cos φ = mg
(2)
Divide (1) by (2)
rω 2 ( L sin φ )ω 2
=
tan φ =
g
g
R. Field 2/4/2014
University of Florida
ω=
g
L cos φ
PHY 2053
T period =
2π
ω
= 2π
L cos φ
g
Page 4
Exam 1 Spring 2012: Problem 20
• A conical pendulum is constructed from a stone
of mass M connected to a cord with length L and L
negligible mass. The stone is undergoing
uniform circular motion in the horizontal plane
M
as shown in the figure. If the cord makes an
angle θ = 30o with the vertical direction and the
period of the circular motion is 4 s, what is the y-axis
2
L
length L of the cord (in meters)? tan φ = 4π R
2πR
T=
v
2 2
FT cos φ − Mg = 0 (1)
R
4
π
2
2
v
=
v
T2
FT sin φ = Ma x = M
(2)
R
R
Divide (2) by (1)
2
sin
φ
=
v
L
tan φ =
gT 2
gT 2 tan φ
R=
4π 2
R
L=
sin φ
Answer: 4.59
% Right: 34%
Rg
R. Field 2/4/2014
University of Florida
φ
φ
φ
FT
x-axis
R
Mg
gT 2
(9.8m / s 2 )(4s ) 2
L=
=
≈ 4.59m
2
2
o
4π cos φ
4π cos(30 )
PHY 2053
Page 5
Example Problem: Unbanked Curves
• A car of mass M is traveling in a circle
with radius R on a flat highway with
speed v. If the static coefficient of
friction between the tires and the road is
μs, what is the maximum speed of the car
such that it will not slide?
x-component:
f s = Ma x = Maradial
y-component:
N − Mg = Ma y = 0
v
2
max
R
y-axis
v2
=M
R
fs ≤ μ s N
vtangential
aradial
N
R
x-axis
fs
R
R
R
=
( f s ) max =
(μs N ) =
( μ s Mg ) = μ s gR
M
M
M
vmax = μ s gR
R. Field 2/4/2014
University of Florida
PHY 2053
Page 6
Exam 2 Spring 2011: Problem 4
• Near the surface of the Earth, a car is traveling at a
constant speed v around a flat circular race track with a
radius of 50 m. If the coefficients of kinetic and static
friction between the car’s tires and the road are μk = 0.1,
μs = 0.4, respectively, what is the maximum speed the car
can travel without slipping?
2
y-axis
Answer: 14 m/s
% Right: 74%
v
2
max
f s = Ma x = Maradial
v
=M
R
FN − Mg = Ma y = 0
R
=
( f s ) max
M
fs ≤ μ s FN
x-axis
R
R
=
( μ s FN ) =
( μ s Mg ) = μ s gR
M
M
FN
R
fs
vmax = μ s gR = (0.4)(9.8m / s 2 )(50m) = 14m / s
R. Field 2/4/2014
University of Florida
PHY 2053
Page 7
Example Problem: Banked Curves
• If the car in the previous problem is
traveling on a banked road (angle θ),
what is the maximum speed of the car
such that it will not slide?
x-component:
f s cos θ + N sin θ = Maradial
v2
=M
R
N cos θ − f s sin θ − Mg = Ma y = 0
Mg
N=
( f s ) max = μ s N
cos θ − μ s sin θ
y-component:
= − Mg
R
R
[( f s ) max cos θ + N sin θ ] = ( μ s cosθ + sin θ ) N
v =
M
M
( μ cos θ + sin θ )
( μ + tan θ )
Rg ( μ s + tan θ )
= Rg s
= Rg s
vmax =
(cos θ − μ s sin θ )
(1 − μ s tan θ )
(1 − μ s tan θ )
2
max
R. Field 2/4/2014
University of Florida
PHY 2053
Page 8
Rolling Without Slipping:
Rotation & Translation
• If a cylinder of radius R rolls without
slipping along the x-axis then:
v
θ
x = s = rθ
R
s
x-axis
x
dx
dθ
v=
=R
= Rω
dt
dt
Translational
Speed
R. Field 2/4/2014
University of Florida
Rotational
Speed
PHY 2053
Page 9