Another 2-D Motion
Uniform Circular Motion
Uniform Circular Motion
• The radius is perpendicular to
any tangent to the circle
• Trajectory--circle
• Speed--constant
Qualitative Aspects
r
90°
What is the velocity and acceleration of
an object in uniform circular motion ?
Quantitative Relationships
From definitions
velocity
acceleration
What do we know about a circle
• Every point on the circle is the
same distance from the center.
radius
r
Vector Components
• The motion repeats (periodic)
Time to make one full revolution
period
T
Period
• Average speed for one period
C
T
Angular speed
• Average velocity for one period
r
Some jargon
0
• Distance around the circle is 2π
πr
circumference
C
+y
Quantitative
Describing a position
vf
vf
vf
vo
vo
vo = vf
+x
∆v +y
vf
vo
∆v
a
r
• Magnitude of instantaneous velocity
Change of Velocity
Describe the motion of an object:
velocity, acceleration
vo
• Instantaneous speed
BUT
r
+y
+x
r
r
θ
vo NOT = vf
θ
ry
+x
rx
Velocity
• Same magnitude
• Changes direction
Acceleration
• Not zero
• Changes direction
Both velocity and acceleration depend on
where you are on the circle
Direction
Magnitude
r
r
∆v
a av =
∆t
rx
= cos θ
r
r r
r
∆v = v f − v i
rY
= sin θ
r
r
r = rx î + ry ĵ
Direction of acceleration is
direction of change of
velocity
r
r = r cos θ î + r sin θ ĵ
1
Velocity Magnitude
How does the velocity of an object
depend on its position?
v
+y
vx =
v
r
vy
r
v = v x î + v y ĵ
vx
= cos φ
v
vy
= sin φ
v
θ
+x
vx
(
)
vy = v cosθ
θ
2
a = a2
x + ay
dv x
ax =
dt
dv y
ay =
dt
+y
ax = −
v
r
θ
+x
velocity
θ
v x = − r sinθ
a=
v
v dθ
θ
=
r dt
a constant
The rate that the angle changes is
constant if the speed is constant
vy =
dy
dt
vy =
d(r sinθ
θ)
dt
note:
Pythagorean Theorem
r2 = x2 + y 2
dθ
θ
dt
v dθ
θ
=
r dt
v
= − v sinθ
θ
r
(
d
dθ
θ
v2
vcosθ) = − vsinθ
θ
=−
sinθ
θ
ay =
r
dt
dt
v2
cosθ
θ
r
4
r2
a=
v
v y = r cosθ = v cosθ
r
d
dθ
θ
v2
ax =
− vsinθ
θ) = − vcosθ
=−
cosθ
r
dt
dt
(
dθ
θ
dθ
θ
sin2 θ + cos2 θ = r
dt
dt
r 2 = r 2 cos θ 2 + r 2 sin 2 θ
r 2 = r 2 (cos θ 2 + sin2 θ )
1 = (cos θ 2 + sin 2 θ )
Should agree with what you got in lab
θ
dθ
θ
v x = − r sinθ
θ
dt
dθ
θ
v y = r cosθ
dt
dθ
θ
dt
v y = r cosθ
Acceleration
2
v= r
Do the same for vy
θ + 90° + φ = 180°
φ = 90° - θ
r
v = − v cos 90 o − θ î + v sin 90 o − θ ĵ
vx = -v sinθ
θ
r is a constant
d(cos θ )
dt
v x = − r sin θ
r
r = r cos θ î + r sin θĵ
r
v = − v sin θ î + v cos θ ĵ
2
2
 dθ
 dθ
θ
θ
v = r 2 sin2 θ  + r 2 cos2 θ 
 dt 
 dt 
d(r cos θ )
dt
vx = r
What does φ have to do with position?
)
dx
dt
vx =
r
φ
r
v = − v cos φ î + v sin φĵ
(
v = v2x + v2y
2
v2 = v2
x + vy
ay = −
cos2 θ +
v2
r
v
4
r2
sin2 θ =
v2
sinθ
θ
r
2
v
r
Replay
• Get acceleration into components
r
a = a x î + a y ĵ
cos2 θ + sin2 θ
units of acceleration ?
2
2
[ m / s] / [m] = [m / s ]
correct units for an acceleration
• Use definition of accel. for each component
dv y
dv x
ax =
ay =
dt
dt
• Get definition of velocity for each component
dx
dy
vx =
vy =
dt
dt
• Use
+y
Direction of acceleration
r
a = −a cosθî + −a sin θĵ
compare to
r
r = r cos θ î + r sin θ ĵ
Direction of a is opposite to r
dθ
θ v
=
dt r
• Use Pythagorean Theorem
a 2 = a 2x + a 2y
v
θ
θ
a
r
θ
+x
to get magnitude of acceleration
v2
a=
r
• Use components of acceleration to show
Acceleration is directed inwards
toward the center of the circle
2
Example
Uniform Circular Motion
v
Any object following a
circular path at a constant
speed is accelerating.
a
r
• Direction: along radius
toward center of circle.
The SUM of the
forces on the
object must be
towards the center
of the circle
• Magnitude: a = v2 / r
• The acceleration is
always perpendicular
to the velocity
You have a job at a business which designs
parts for jet engines. Your task, as a member
of a safety team is to compare the motion of
two different parts located on a disk attached
to the turbine shaft. The part furthest from
the turbine shaft is three times the distance
from the shaft as the other part. The turbine
shaft goes through the center of the disk and
is perpendicular to it. Assume the disk is in
uniform circular motion at n rotations/sec.
•C=2πr
Looking down
from above
A
vt
rA
dθ v
=
dt r
Qualitative Analysis
rB = 3 r A
n rotations
per sec
aA or aB
B
rB
aB
∆vB
rB = 3 r A
vA = f(vB )
aA = f(aB )
Quantitative relationships:
θ
vA2
vB2
r
∆v B
r
a av B =
∆t
∆t is same for both objects
aav B> aav A
1
sec
n
Target quantities:
vB1
1
CA
2
rA
3
rB
Find CA
TB = TA = T =
the change of direction of A
is the same as
the change of direction of B
∆vB > ∆vA
vA
C
= A
T
CA = 2 π r A
In any time interval,
vA
Find vA
A
vA
rA
unknowns
Plan:
vB
aA
For constant speed, instantaneous speed
equals average speed
vA or vB
Motion diagram:
Magnitude of instantaneous acceleration
r
∆v A
r
a av A =
∆t
The objects go around circles of different
circumference.
The objects have different speeds
Why?
vB > vA
vA1
∆vA θ
Both objects take the same time to go
around a full circle
Same period
Qualitatively, which is larger
Question: Compare vA and vB
Compare aA and aB
Magnitude of instantaneous velocity
Circle B is larger than circle A
Object B takes same time as object A
to go once around the circle
Both objects have a constant speed
Use relationship between instantaneous
speed and acceleration for uniform
circular motion
B
rB
The time to go once around a circle is the
period ( T ).
Angular speed (rate angle changes) is
Compare means make an equation
involving the two quantities in question.
Use definition of average speed
The distance once around a circle is its
circumference ( C )
If the speed of the object is constant,
•v=C/T
Approach:
distance
sav =
∆t
sav = v for constant speed
a =
v2
r
Find rA
rB= 3 rA
Find rB
CB = 2 π r B
4
CB
Find CB
vB =
CB
T
5
5 unknows, 5 equations
for uniform circular motion
C = 2π
πr
3
5
vB =
Plan to find acceleration
CB
T
v B T = CB
into 4
unknowns
aA
aA
vA
rA
aA =
vBT
= rB
2π
into 3
v BT
= 3r A
2π
vBT
= rA
6π
into 2
v BT
3
vB T
vA = 3
T
CA =
into 1
rA =
1
r
3 B
3
rB
v B2
rB
4 unknowns , 4 equations
200 mi
shuttle
Diagram
1
a
3 B
+y
v
90 min for
1 orbit
A space shuttle typically has a circular
orbit around the earth at a height of 200
miles. It travels with a constant speed and
completes one orbit in 90 minutes. The
radius of the Earth is about 4000 miles. What
is its acceleration?
into 1
vB 2
)
3
v B2
3aB
aA =
v
vA = B
3
Example
(
aA =
4
3
into 1
2 v A = 1 vB
3
Find rB
aB =
into
vA 2
vB2
3aB
aA =
2
vB 2
aB
vB 2
3aB
Find rA
rA =
vB T
6π
vA , rA
1
Find vA
from 1st part
1
vA =
v
3 B
vB2
rB
rB =
2
v BT = 2 πrB
CA = 2 π
Find
aB =
4
r = 4200 mi
a
r
T = 90 min
+x
4000 mi
Earth
Target quantity:
Question: What is acceleration of shuttle?
Approach:
Shuttle travels in a circle around the Earth
a
Quantitative relationships:
a=
v2
r
C=2πr
s=
distance
∆t
v = s for constant s
Uniform circular motion
Use expression for centripetal accel.
4
unknowns
PLAN:
a
Find a
v2
a=
1
r
Find v
2 πr
2
v=
T
2 unknowns, 2 equations
v
Evaluate:
[distance]
[time] 2
Are correct units for accel.
The question is answered since the
acceleration of the shuttle is calculated
Does a make sense? Compare it to g.
2 πr 2
)
(
a= T
r
a=
4 π2r
T2
a=
4 π2 (4200 mi)
(90 min)2
a = 20. 5
mi
1 min 2 5280 ft
(
) (
)
min2 60 sec
1 mi
ft
a = 30
sec2
a = 20. 5
check units
mi
min2
slightly less than the
acceleration if you drop it
on the surface of the Earth
This makes sense since 200 miles
is very close to the surface of the
Earth compared to 4000 miles.
5