Circular Motion
Tangential & Angular Acceleration
vt
• Tangential Acceleration:
The arc length s is related to the angle θ (in radians = rad)
as follows: s = r
θ
The tangential velocity vt is related to the angular velocity
ω as follows:
vt = rω
Tangential
Velocity
The tangential acceleration at is related to the angular acceleration α as follows:
Δω dω
dvt
dω v = r ω
=
at =
=r
= rα α = lim
Δt →0 Δt
dt
dt
dt
• Overall Acceleration:
t
(radians/s2)
r
r
r
atot = aradial + at = −aradial rˆ + atθˆ
r
2
atot = atot = aradial
+ at2
Rick Field 2/6/2014
University of Florida
PHY 2053
at
Radial Axis
ar
Tangential
Acceleration
r
Radial
Acceleration
Page 1
Angular Equations of Motion
• Angular Equations of Motion (constant α):
If the angular acceleration α is constant then
α (t ) = α
(radians/s2)
ω (t ) = ω0 + αt (radians/s)
θ (t ) = θ 0 + ω0t + 12 αt 2 (radians)
ω 2 (t ) − ω02 = 2α (θ (t ) − θ 0 )
a
2
aradial (t ) = rω (t ) (m/s )
at (t ) = rα
(m/s2)
vt (t ) = vt 0 + at t
(m/s)
s (t ) = s0 + vt 0t + at t
1
2
2
(m)
v t2 (t ) − vt20 = 2at (s (t ) − s0 )
t
aradial (t ) = v t (t ) / r
2
2
Radial Axis
(m/s2)
ar
Tangential
Acceleration
r
Radial
Acceleration
Rick Field 2/6/2014
University of Florida
PHY 2053
Page 2
Angular Equations of Motion
• Angular Equations of Motion (constant α):
θ (t )
N (t ) =
2π
ω (t )
f (t ) =
2π
Let N = Number of revolutions (rev)
Let f = Number of revolutions per second
α (t ) = α
α
2π
(rad/s2)
ω (t ) = ω0 + αt (rad/s)
θ (t ) = θ 0 + ω0t + 12 αt 2 (rad)
ω 2 (t ) − ω02 = 2α (θ (t ) − θ 0 )
Rick Field 2/6/2014
University of Florida
(frequency)
(rev/s2)
f (t ) = f 0 + ( 2απ )t
(rev/s)
N (t ) = N 0 + f 0t + ( )t
1
2
α
2π
2
(rev)
f 2 (t ) − f 02 = 2( 2απ )( N (t ) − N 0 )
PHY 2053
Page 3
Angular Equations: Examples
• A disk rotates about its central axis starting from rest at t = 0 and
accelerates with constant angular acceleration. At one time it is
rotating at 4 rev/s; 60 revolutions later, its angular speed is 16 rev/s.
Starting at t = 0, what is the time required to complete 64 revolutions?
f 2 (t ) − f 02
(16rev / s ) 2 − (4rev / s ) 2
α
=
= 2rev / s 2
=
2π 2( N (t ) − N 0 )
2(60rev )
N (t ) − N 0 = ( )t
1
2
α
2π
2
t=
Answer: t = 8 seconds
2( N (t ) − N 0 )
2(64rev)
=
= 8s
( 2απ )
(2rev / s 2 )
• An astronaut is being tested in a centrifuge. The centrifuge has a
radius R and, in starting from rest at t = 0, rotates with a constant
angular acceleration α = 0.25 rad/s2 . At what time t > 0 is the
magnitude of the tangential acceleration equal to the magnitude of
the radial acceleration (i.e. centripetal acceleration)?
aradial (t ) = Rω 2 (t ) = Rα 2t 2 = at = Rα
t=
Rick Field 2/6/2014
University of Florida
1
α
=
Answer: t = 2 seconds
1
= 2s
0.25rad / s 2
PHY 2053
Page 4
Exam 2 Spring 2011: Problem 2
• A race car accelerates uniformly from a speed of 40 m/s to
a speed of 58 m/s in 6 seconds while traveling around a
circular track of radius 625 m. When the car reaches a
speed of 50 m/s what is the magnitude of its total
acceleration (in m/s2)?
Answer: 5
% Right: 49%
Rick Field 2/6/2014
University of Florida
v2 − v1 (58m / s ) − (40m / s )
= 3m / s
=
at =
t 2 − t1
6s
v 2 (50m / s ) 2
=
= 4m / s
ar =
R
625m
atot = at2 + ar2 = 5m / s
PHY 2053
Page 5
Gravitation: Circular Orbits (M >> m)
For circular orbits the gravitational force is perpendicular to the
velocity and hence the speed of the mass m is constant. The force
Fg is equal to the mass times the radial (i.e. centripetal)
acceleration as follows:
GM
GmM
v2
2
r
=
(radius of the orbit, constant)
Fg =
=
ma
=
m
=
mr
ω
radial
r2
r
v2
GM
r3
(angular velocity, constant)
r
r3
2πr
T=
= 2πr
= 2π
v
GM
GM
r
Fg
Assume M >> m so that the
position of M is fixed!
GM
v=
(speed, constant)
r
ω=
M
For circular orbits r, v, and
ω are also constant.
(period of rotation)
In general both masses rotate
about the center-of-mass and the
formulas are more complicated!
• Kepler’s Third Law:
T2 =
4π r
GM
2 3
The period squared is
proportional to the radius cubed.
Rick Field 2/6/2014
University of Florida
vm
M
CM
rM
×
Fg
m
rm
VM
PHY 2053
Page 6
v
m
Circular Orbits: Example
• Two satellites are in circular orbit around the Earth. The first satellite
has mass M1 and is travelling in a circular orbit of radius R1. The
second satellite has mass M2 = M1 is travelling in a circular orbit of
radius R2 = 4R1. If the first satellite completes one revolution of the
Earth in time T, how long does it take the second satellite to make one
revolution of the Earth?
Answer: 8T
2 3
4
π
R1
T12 =
GM 1
2 3
2
3
2 3
π
π
π
4
R
4
(
4
R
)
4
R1
2
2
1
T2 =
=
= 64
= 64T12
GM 2
GM 1
GM 1
T2 = 8T1 = 8T
Rick Field 2/6/2014
University of Florida
PHY 2053
Page 7
Circular Orbits: Example
• Two diametrically opposed masses m revolve around a
circle of radius R. A third mass M = 2m is located at the
center of the circle. What is the period T of rotation for
this system of three masses?
3
Answer: T =
Fgrav
4π
3
M = 2m
m
m
R
R
Gm
2
GmM Gm 2 Gm
v
=
+
= 2 (M + 14 m ) = maradial = m
2
2
R
(2 R)
R
R
m
F
M = 2m
m
R
G ( M + 14 m)
v=
R
2πR
R
R
4R
4π
= 2πR
=
=
=
π
π
T=
2
R
2
R
v
G ( M + 14 m)
G (2m + 14 m)
9Gm
3
Rick Field 2/6/2014
University of Florida
PHY 2053
R3
Gm
Page 8