PHYS101
Uniform Circular Motion
Uniform Circular Motion
Uniform circular motion means, that the particle moves with constant speed on a
circle.
So, the first question that arises is: Is the particle accelerated?
Recall:
d~v(t)
~a(t) =
dt
In order to answer this question, recall that the velocity of a particle is a vector property, i.e. it has a magnitude and a direction. As uniform circular motion means that
the speed is constant, i.e. the magnitude of the velocity, there is no restriction to the
direction. Obviously, the direction is of the velocity is changing.
Let the motion be counter clockwise so, that
the angle ϕ(t) > 0. Then we can write for the
position vector:
~r(t) = R cos ϕ(t) î + R sin ϕ(t) ĵ,
(1)
where R is the constant radius of the circle
and ϕ(t) is the angle that changes over time,
changing also the direction of the position
vector.
Now, let us calculate the velocity, as the second requirement was that the speed of the
particle on the circle is constant.
Figure 1: Particle moving in uniform
circular motion
dϕ(t)
dϕ(t)
r~(t)
= − R sin ϕ(t)
î + R cos ϕ(t)
ĵ =
dt
dt
dt
dϕ(t)
=R
− sin ϕ(t) î + sin ϕ(t) ĵ (2)
dt
~v(t) =
So, we can now calculate the speed, which is the magnitude of the velocity ~v(t) as:
s
dϕ(t) 2
dϕ(t) 2
− R sin ϕ(t)
+ R cos ϕ(t)
|~v(t)| =
dt
dt
s
dϕ(t) 2
dϕ(t) 2
2
2
2
2
=
R sin ϕ(t)
+ R cos ϕ(t)
dt
dt
v
u 2
u
dϕ(t) 2 2
2
u
= tR
sin ϕ(t) + cos2 ϕ(t)
dt
|
{z
}
=1
s
=
R2
dϕ(t)
dt
2
=R
dϕ(t)
= const.
dt
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2015
Department of Physics, Eastern Mediterranean University
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Page 1 of 5
PHYS101
Uniform Circular Motion
As the radius of the circle R is constant, and the speed of the particle must be constant
as well, then the change of the angle with respect to time should be constant as well
dϕ(t)
= const.
dt
(4)
The solution of the so-called differential equation (4) can be obtained very straight
forward as:
ϕ(t) = ωt + ϕ0
(5)
where ω and ϕ0 are constant. We will identify ω later as the angular speed, and ϕ0 as
the initial angle. The angular speed is in the case of uniform circular motion constant.
We can check now, if (5) is the solution of (4), by differentiating ϕ(t) with respect to t
d
d
ϕ(t) =
(ωt + ϕ0 ) = ω = const.
dt
dt
So, we can accept (5) as the solution to (4). So, let us now rewrite the position vector,
the velocity vector, and acceleration vector. So the position vector becomes:
~r(t) = R cos (ωt + ϕ0 ) î + R sin (ωt + ϕ0 ) ĵ.
(6)
The magnitude and the direction are then:
|~r(t)| = R,
θ = ϕ(t) = ωt + ϕ0
The velocity vector is
~v(t) = − Rω sin (ωt + ϕ0 ) î + Rω cos (ωt + ϕ0 ) ĵ,
(7)
where the magnitude and direction of the velocity vector are:
|~r(t)| = Rω,
Using the identity
θ = ϕ(t) + 90◦ = ωt + ϕ0 + 90◦ .
cot θ = tan(90◦ − θ )
we can easily verify the direction as following:
θ = tan
−1
cos(ωt + ϕ0 )
− sin(ωt + ϕ0 )
= tan−1 (− cot(ωt + ϕ0 )) =
−1
◦
= tan
tan 90 − (−(ωt + ϕ0 )) = 90◦ + (ωt + ϕ0 )
Finally, let us consider the acceleration vector:
~a(t) =
d~v(t)
= − Rω 2 cos (ωt + ϕ0 ) î − Rω 2 sin (ωt + ϕ0 ) ĵ =
dt
= −ω 2 R cos (ωt + ϕ0 ) î + R sin (ωt + ϕ0 ) ĵ = −ω 2~r(t) (8)
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2015
Department of Physics, Eastern Mediterranean University
Page 2 of 5
PHYS101
Uniform Circular Motion
Equation (8) obviously shows that the direction of the acceleration vector is in opposite
direction to the position vector, i.e. the acceleration vector points towards the center of the
circle, and is therefore called centripetal acceleration. The word centripetal means towards
the center, or center seeking. Then we can calculate the magnitude and direction of the acceleration vector:
ac = |~a(t)| = ω 2 |~r(t)| = ω 2 R =
v2
ω 2 R2
=
=
R
R
Figure 2: Graphical illustration of the
Summarising, the magnitude of the cenfound results
tripetal acceleration ~ac
ac =
v2
R
(9)
The direction of the acceleration vector can be easily determined as
θ = 180◦ + ωt + ϕ0 .
As the speed is constant, the average speed and the instantaneous speed are equal.
The average speed can be calculated as
d
,
∆t
where d is the distance travelled and ∆t is the time need to travel over the distance
d. So we can easily determine the time T needed for one revolution, i.e. the time for
one complete turn, as the particle travels over the distance d = 2πR with the constant
speed v. So, the time for one revolutions is:
v=
T=
2πR
v
(10)
Example 1. What is the centripetal acceleration of the Earth as it moves in its orbit
around the sun?
For the solution of this problem, we assume that the Earth is moving on a circle
around the sun, and that the speed of the Earth is constant on its orbit. We know:
h
min
s
T = 1a = 365.25d × 24 × 60
× 60
= 3.154 × 107 s, R = 1.496 × 1011 m
d
h
min
So, we get for the cetripetal acceleration:
2πR 2
v2
4π 2 R
4π 2 1.496 × 1011 m
m
ac =
= T
=
=
= 5.937 × 10−3 2
2
2
R
R
T
s
(3.154 × 107 s)
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2015
Department of Physics, Eastern Mediterranean University
Page 3 of 5
PHYS101
Uniform Circular Motion
Tangential and Centripetal Acceleration
We can also describe the motion of a particle
along an arbitrary curved line. As we have
discussed in the previous section, the centripetal acceleration is only due to the change
of the direction of the velocity vector. But, we
have not discussed the change of the change
of the magnitude of the velocity, which causes
an increasing speed if the acceleration vector
and the velocity vector point into the same
Figure 3: Motion of a particle along an direction, or a deceleration if the acceleration
arbitrary curved line
vector points into the opposite direction of the
velocity vector. Considering the motion along
an arbitrary curved line we can inscribe at every point a circle and determine the centripetal and tangential acceleration. The total acceleration is given as:
(11)
~a = ~ac +~at ,
where the tangential acceleration causes a change in the speed and its magnitude is
dv d
at = =
(12)
|~v(t)| .
dt
dt
As one can see easily from figure 3, it is obvious that the tangential and centripetal
accelerations are perpendicular to each other.
Example 2. A car leaves a stop sign and exhibits a constant acceleration of 0.3m/s2
parallel to the roadway. The car passes over a rise in the roadway such that the top of
the rise is shaped like an arc of a circle of radius 500m. At the moment the car is at the
top of the rise, its velocity vector is horizontal and has a magnitude of 6m/s. What are
the magnitude and direction of the total acceleration vector for the car at this instant?
Figure 4: A car passes over a rise, shaped like a circle
The magnitude of the centripetal acceleration can be easily determined as:
v2
m
(6m/s)2
ac =
=
= 0.072 2
R
500m
s
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2015
Department of Physics, Eastern Mediterranean University
Page 4 of 5
PHYS101
Uniform Circular Motion
The total acceleration is the vector sum of the centripetal acceleration and the tangential acceleration.
~a = ~at +~ac = 0.3
m
m
î
−
0.072
ĵ
s2
s2
Figure 5: Accelerations acting on the car
Then, the magnitude of the acceleration is
r
q
m 2 m 2
m
a = a2t + a2c =
0.3 2 + −0.072 2 = 0.309 2 .
s
s
s
The angle φ with the positive x-axis is then:
φ = tan−1
ac
at
= tan−1
−0.072 sm2
0.3 sm2
!
= −13.5◦
Example 3 (Moon’s centripetal acceleration). The Moon’s nearly circular orbit about
the Earth has a radius of about 384000km and a period T of 27.3 days. Determine the
acceleration of the Moon toward the Earth.
The distance the Moon travels in 27.3 days is
d = 2πR = 2π3.84 × 108 m = 2.41 × 109 m.
The time for one revolution of the Moon about the Earth is
h
min
s
× 60
= 2.36 × 106 s
T = 27.3d = 27.3d × 24 × 60
d
h
min
So the average speed of the Moon around the Earth:
v avg =
d
2.41 × 109 m
m
=
= 1.0211 × 103
6
T
s
2.36 × 10 s
So, finally we get for the centripetal acceleration
1.0211 × 103 ms
v2
=
ac =
R
2.41 × 109 m
2
= 0.266 × 10−3
m
m
= 2.78 × 10−3 2 .
s
s
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2015
Department of Physics, Eastern Mediterranean University
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