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Section 9 - Uniform Circular Motion
Section Outline
The Speed and Acceleration
The Equations of Motion
A Summary of the Course to Date
What do objects do? They move. In this section we’ll look at uniform circular motion such as the orbit
of the moon. We really only have the definitions of position, displacement, velocity, and acceleration to
guide us. However, that is enough to describe all motion, even motion in circles.
1. The Speed and Acceleration
Circular motion means constant radius,
dr
dt
= 0. That is, the position vector has a constant
length. Uniform means constant speed,
dv
dt
Let’s think a bit more about this constant speed. The definition of speed is the distance covered per
time. In this case the distance is around the circle, so it is just the circumference, 2πr. The time to go
around is called the period usually represented by T. So,
v ≡
∆ s
∆ t
⇒ v =
2 π r
T
Since the direction of the velocity is tangent to the circle, this is called the “tangential speed.”
There are two other ideas associated with circular motion that we should mention just for completeness.
Frequency is the number of orbits per unit time, so it is the reciprocal of the period,
f =
T
Angular frequency is the angle covered per unit of time. Angular frequency is often called “angular
speed” just to add to the confusion. Since it is angle per time, it must the 2π divided by the period,
ω ≡
∆ θ
∆ t
⇒ ω =
2 π
T
= 2 π f.
Tangential Speed v t
2 π r
T
With those definitions established, let’s move on and look back at our
earlier study of the acceleration of the tip of the second hand on a clock.
For the second hand, the velocity vectors are tangent to the circle at any
point. The velocity just before the top and just after the top are moved
into standard position in the lower sketch, so that their components are
easier to find.
v i
= vcos
Δθ
2
i + vsin
Δθ
2
j and
v f
= vcos
Δθ
2
i − vsin
Δθ
2
j
where v is the speed of the tip of the second hand.
The change in velocity is,
v ≡
v f
v i
= v cos
Δθ
2
i − vsin
Δ θ
2
j
− vcos
Δθ
2
i + vsin
Δθ
2
j
v = − 2 vsin
Δ θ
2
j ⇒ d
v = − vdθ
j.
Using the definition of acceleration,
a ≡
d
v
dt
= −v
dθ
dt
j = −
v
r
rdθ
dt
j = −
v
r
ds
dt
j = −
v
2
r
j.
That was a rather formal way to get the acceleration. A simpler
method depends upon seeing the relationship between the two
triangles shown above redrawn at the right. The first triangle is
formed from the two position vectors for the second hand and the
displacement vector. The second triangle is the two velocity
vectors and the change in the velocity. Since both triangles are
isosceles and have the same angle between the equal sides, they
are similar. So,
∆ v
v
∆ r
r
⇒ ∆ v =
v
r
∆ r.
Dividing both sides by the time to change the position (or velocity),
∆ v
∆ t
v
r
∆ r
∆ t
⇒ a =
v
r
v ⇒ a =
v
2
r
In summary, the speed is constant, but there is acceleration toward the center called the “centripetal
acceleration.” The magnitude of the centripetal acceleration is,
For the motion to remain uniform and circular there must be acceleration toward the center of precisely
this magnitude.
Centripetal Acceleration
a c
v
2
r
r r
∆r
∆θ
∆θ ∆v
v
v
The position, velocity, and acceleration vectors are redrawn in
standard position at the right. The acceleration vector as a function of
time can be written as,
a = −a cosωt
i − a sinωt
j = −ω
2 r cosωt
i − ω
2 r sinωt
j.
Let’s work our way through the x-components first,
a x
= −ω
2 r cosωt.
Note that the x-component is negative consistent with the sketch at the
right. To find the x-component of the velocity, use the definition of
acceleration,
a x
dv x
dt
dv x
dt
= −ω
2 r cosωt ⇒ dv x
= −ω
2 r cosωtdt.
Now integrate keeping in mind the ω and r are constant,
dv x
0
v x
∫
= −ω
2 r cosωtdt
0
t
∫
⇒ v x
= −ω
2 r
ω
sinωt
⇒ v x
= −ωrsinωt.
Again, it matches the sketch. To find the x-component of the position, use the definition of velocity,
v x
dx
dt
dx
dt
= −ωrsinωt ⇒ dx = −ωr sinωtdt.
Now integrate being careful with the limits,
dx
r
x
∫
= −ωr sinωtdt
0
t
∫
⇒ x − r = −ωr −
ω
[cosωt − 1 ]
⇒ x − r = r[cosωt − 1] ⇒ x = r cosωt.
To find the acceleration as a function of position, notice that,
a x
= −ω
2 r cosωt ⇒ a x
= −ω
2 x.
To get the velocity as a function of position, square the velocity as a function of time,
v x
2 = ω
2 r
2 sin
2 ωt = ω
2 r
2 1 − cos
2
( ωt) =^ ω
2 r
2 − r
2 cos
2
( ωt) =^ ω
2 r
2 − x
2
( ) ⇒^ v x
= ±ω r
2 − x
2
where the sign is chosen by the coordinates.
The equations for the y-components are derived in a similar manner. In summary,
a x
(t) = −ω
2 r cosωt a y
(t) = −ω
2 rsin ωt
v x
(t) = −ωrsinωt v y
(t) = ωr cosωt
x ( t ) = r cos ω t y(t) = r sinωt
a x
(x) = −ω
2 x a y
(y) = −ω
2 y
v x
(x) = ±ω r
2 − x
2 v y
(y) = ±ω r
2 − y
2
r
v
a
θ
θ
θ
θ
θ
θ
Example 9.2: Using the coordinates at the right where the moon
crosses the x-axis at t = 0. (a)Find the angular frequency. When
the moon is at the position shown, find (b)the velocity, (c)the
acceleration, and (d)the time. The moon’s radius of orbit is
3.82x
8 m and its period is 27.3 days.
Given: r = 3.82x
8 m, θ = 60˚, and T = 27.3days
Find: ω = ?, v = ?, a =? and t =?
(a)Using the angular speed,
ω =
2 π
T
2 π
⇒ ω^ =^ 2.66x^10
− 6 rad / s (^).
The components of the position vector can be found from the equations of motion,
x = r cosωt = (3.82x 10
8 )cos60˚= 1.91x 10
8 m and
y = r sinωt = (3.82x 10
8 )sin 60 ˚= 3.31x 10
8 m.
(b)The components of the velocity at this position can be found using the equations of motion,
v x
(x) = ±ω r
2 − x
2 = −(2.66x 10
− 6 ) (3.82x 10
8 )
2 − (1.91x 10
8 )
2 ⇒ v x
= − 880 m / s and
v y
(y) = ±ω r
2 − y
2 = +(2.66x 10
− 6 ) (3.82x 10
8 )
2 − (3.31x 10
8 )
2 ⇒ v y
= 507m / s.
The velocity is,
v = −(880m / s)
i + (507m / s)
j.
(c)The components of the acceleration at this position are again from the equations of motion,
a x
(x) = −ω
2 x = −(2.66x
− 6 )
2 (1.91x 10
8 ) ⇒ a x
= −1.35mm / s
s and
a y
(y) = −ω
2 y = −(2.66x
− 6 )
2 (3.31x 10
8 ) ⇒ a y
= −2.34mm / s
s .
The acceleration is,
a = −(1.35mm/ s
2 )
i − (2.34mm / s
2 )
j (^).
(d)Using the equation of motion for position as a function of time,
x(t) = r cosωt ⇒ ωt =
π
⇒ t =
π
3 ω
π
3 (2.66x 10
− 6 )
⇒ t^ =^ 3.94x^10
5 s = 4.56days (^).
To summarize, an object in uniform circular motion has a tangential speed given by,
v t
2 π r
T
The object is accelerating even though it is moving at a constant speed. The acceleration is due to the
changing velocity vector. The magnitude of the acceleration required to maintain the uniform circular
motion is given by,
a c
v
2
r
The equations of motion for uniform circular motion are:
a x
(t) = −ω
2 r cosωt a y
(t) = −ω
2 rsin ωt
v x
(t) = −ωrsinωt v y
(t) = ωr cosωt
x ( t ) = r cos ω t y(t) = r sinωt
a x
(x) = −ω
2 x a y
(y) = −ω
2 y
v x
(x) = ±ω r
2 − x
2 v y
(y) = ±ω r
2 − y
2
!^ y
v
!=60˚
r
x
a
t > 0
