Uniform Circular Motion
Uniform Circular Motion
• Physics of motion of a mass in a circle at Constant Speed.
Constant Speed  The Magnitude (size) of the
velocity vector v is Constant.
BUT the DIRECTION of v changes continually!
v  r
r
r
v = |v| = constant
Uniform Circular
Motion is Circular
Motion at Constant
Speed.
•The direction of the
velocity is continually
changing.
The velocity vector
is always tangent to
the circle.
• Consider details of the motion
of a mass m in a circle at
v = |v| = constant
Constant Speed.
Question:
Is there an acceleration?
• To answer this, consider both
r
Newton’s 1st Law
&
Newton’s 2nd Law!
• Recall that by Definition,
Acceleration  Time Rate
of Change of Velocity
• That is: a = (Δv/Δt)
r
• Acceleration 
Time Rate of Change of Velocity
a = (Δv/Δt)
• Also, recall that both a & v are vectors.
Constant Speed  The Magnitude (size) of
the velocity vector v is Constant.
BUT the DIRECTION of v changes
continually!
 An object moving in a circle
undergoes an acceleration!!
Centripetal (Radial) Acceleration
Look at the vector velocity change Δv in the limit that the
time interval Δt becomes infinitesimally small & get:
Similar Triangles
 (Δv/v) ≈ (Δℓ/r)
As Δt  0, Δθ  0, A B
As Δt  0, Δv   v &
Δv is in the
radial direction
 a  aC is radial!
Close view of the
fact that, in the
infinitesimal limit,
Δv  0
& its direction is
towards the center
of the circle.
• This type of acceleration is sometimes called the
Centripetal Acceleration & sometimes
called the Radial Acceleration.
Its vector direction is Radially Inward!
• The word “Centripetal” is from Greek. It means
“Towards the Center”
“Centripetal Acceleration”  Center
Directed or Center Seeking Acceleration
• Below is a typical figure for a particle moving in
uniform circular motion, radius r (speed v = constant):
• The velocity vector v
is always tangent
to the circle!
• The centripetal acceleration
vector aC
is always Radially Inward!
 aC  v always!!
Period & Frequency
• Consider again a particle moving in uniform
circular motion of radius r (speed v = constant).
• One common way to describe this motion is
in terms of the Period T & the frequency f.
Period T  The time for one revolution
(time to go around once!)
Frequency f  The # of revolutions
per second.
• They are obviously related by:
T = (1/f)
• Consider again a particle moving in uniform
circular motion, radius r (speed v = constant)
• Circumference = Distance Around= 2πr
 Speed: v = (2πr/T) = 2πrf
 Centripetal Acceleration:
aC = (v2/r) = (4π2r/T2)
Example: Acceleration of a Revolving Ball
A 150-g ball at the end of a string is revolving
uniformly in a horizontal circle of radius 0.600 m.
The ball makes f = 2.0 revs/second (period T = 0.5 s).
Calculate its centripetal acceleration.
r
r
Example: Acceleration of a Revolving Ball
A 150-g ball at the end of a string is revolving
uniformly in a horizontal circle of radius 0.600 m.
The ball makes f = 2.0 revs/second (period T = 0.5 s).
Calculate its centripetal acceleration.
Solution
r
v = (2πr/T) = 7.54 m/s
r
Example: Acceleration of a Revolving Ball
A 150-g ball at the end of a string is revolving
uniformly in a horizontal circle of radius 0.600 m. The
ball makes f = 2.0 revs/second (period T = 0.5 s).
Calculate its centripetal acceleration.
Solution
r
r
v = (2πr/T) = 7.54 m/s
aC = (v2/r) = 94.7 m/s2
Example: Moon’s Centripetal Acceleration
The Moon’s nearly circular orbit about the Earth
has a radius of about 384,000 km (3.84  108 m)
and a period T of 27.3 days (2.36  106 s). Calculate
the acceleration of the Moon toward the Earth.
Example: Moon’s Centripetal
Acceleration
The Moon’s nearly circular orbit about the Earth
has a radius of about 384,000 km (3.84  108 m) &
a period T of 27.3 days (2.36  106 s). Calculate the
acceleration of the Moon toward the Earth.
Solution
v = (2πr/T), aC = (v2/r) = (4π2r/T2)
Example: Moon’s Centripetal Acceleration
The Moon’s nearly circular orbit about the Earth
has a radius of about 384,000 km (3.84  108 m)
and a period T of 27.3 days (2.36  106 s). Calculate
the acceleration of the Moon toward the Earth.
Solution
v = (2πr/T), aC = (v2/r) = (4π2r/T2)
aC = 2.72  10-3 m/s2
Examples 5.1 & 5.2: A Bug on a CD
A bug sits on the edge of a CD, of
radius r = 6 cm (0.06m), as in the
figure. It undergoes uniform
circular motion as the CD spins. It
goes around the CD 6 times/sec.
Calculate
a. The period of the motion.
b. The speed of the bug.
c. The centripetal acceleration of the bug.
Examples 5.1 & 5.2: A Bug on a CD
A bug sits on the edge of a CD, of
radius r = 6 cm (0.06m), as in the
figure. It undergoes uniform
circular motion as the CD spins. It
goes around the CD 6 times/sec.
Calculate
a. The period of the motion.
Because f = 6 rev/s, T = 1/f = 0.17s
b. The speed of the bug.
c. The centripetal acceleration of the bug.
Examples 5.1 & 5.2: A Bug on a CD
A bug sits on the edge of a CD, of
radius r = 6 cm (0.06m), as in the
figure. It undergoes uniform
circular motion as the CD spins. It
goes around the CD 6 times/sec.
Calculate
a. The period of the motion.
Because f = 6 rev/s, T = 1/f = 0.17s
b. The speed of the bug.
v = (2πr/T) = 2.3 m/s
c. The centripetal acceleration of the bug.
Examples 5.1 & 5.2: A Bug on a CD
A bug sits on the edge of a CD, of
radius r = 6 cm (0.06m), as in the
figure. It undergoes uniform
circular motion as the CD spins. It
goes around the CD 6 times/sec.
Calculate
a. The period of the motion.
Because f = 6 rev/s, T = 1/f = 0.17s
b. The speed of the bug.
v = (2πr/T) = 2.3 m/s
c. The centripetal acceleration of the bug.
aC = (v2/r) = 88.2 m/s2
Newton’s Laws + Circular Motion
• It’s straightforward to see how Newton’s
2nd Law can be applied to circular motion:
• Since the acceleration is directed toward
the center of the circle, the net force must
be in that direction also!
• This “Centripetal Force” can be supplied
by a variety of physical objects or forces
• Also, the “circle” does not need to be a
complete circle.
Uniform Circular Motion; Dynamics
• Consider a particle moving in
uniform circular motion at radius
r & speed v = constant.
• Centripetal Acceleration is:
aC = (v2/r) , aC  v always!!
aC is radially inward always!
Newton’s 1st Law 
There must be a force acting!
Newton’s 2nd Law 
∑F = ma = maC = m(v2/r)
Direction:
The total force must be radially inward always!
A force is required to keep an
object moving in a circle. If
the speed is constant, the force
is directed toward the center of
the circle. The direction of the
force is continually changing
so that it is always pointed
toward the center of the circle.
∑F = ma = maC =
2
m(v /r)
Example: A ball twirled on the end of a string.
In that case, the force is the tension in the string.
“Centripetal Force”
Newton’s 2nd Law:
∑F = ma = maC= m(v2/r)
• The total force ∑F must be radially inward always!
• The force which enters Newton’s 2nd Law in this
case is often called a “Centripetal Force”.
(It is a center directed force)
• The “Centripetal Force” is NOT a new kind of
force! It could be string tension, gravity, etc.
• It’s the right side of ∑F = ma, not the left side!
• It’s the form of “ma”, for circular motion!
Centripetal Force
You can understand that
the centripetal force
must be inward
by thinking about the ball on
a string. Strings only pull;
they never push.
MISCONCEPTION!!
The force on the ball is
NEVER outward
(“Centrifugal”). It is
ALWAYS inward
(Centripetal) !!
An outward (“centrifugal”)
force ON THE BALL
is NOT a valid concept!
The string tension force
F
F
ON THE BALL
is Inward (centripetal).
The string tension force F on the ball is
INWARD toward the center of the circle!
What happens if the cord on the
ball is broken or released?
For the ball to move in a circle, there
must be an inward (Centripetal) force
pointed towards the circle center so that
the natural tendency of the object to
move in a straight line (Newton’s 1st
Law!) will be overcome. If the centripetal
force goes to zero, the ball will fly off in a
direction tangent to the circle
(Newton’s 1st Law again!)
There is no centrifugal force pointing outward on the ball!
Example (Estimate)
Estimate the force a person must exert on a string attached
to a 0.15 kg ball to make the ball revolve in a horizontal
circle of radius 0.6 m. The ball makes 2 rev/s.
m = 0.15 kg, r = 0.6 m, f = 2 rev/s  T = 0.5 s
Assumption: Circular path is  in horizontal plane, so
φ  0  cos(φ)  1
Newton’s 2nd Law:
∑F = ma  FTx = max= maR = m(v2/r)
v = (2πr/T) = 7.54 m/s
So, the tension is (approximately) FTx  14 N
Example: Revolving Ball (Vertical Circle)
A ball, mass m = 0.15 kg on the
end of a (massless) cord of length
r = 1.1 m cord is swung in a
vertical circle. Calculate:
a. The minimum speed the ball must
have at the top of its arc so that the
ball continues moving in a circle.
b. The tension in the cord at the
bottom of the arc, assuming that
there the ball is moving at twice the
speed found in part a.
Hint:
The minimum speed at the top will
happen for the minimum tension FT1
Note: Here, string tension &
gravity are acting together
(both enter Newton’s 2nd
Law!) to produce centripetal
acceleration.
Problem
r = 0.72 m, v = 4 m/s, m = 0.3 kg
Use: ∑F = maR
• At the top of the circle,
Newton’s 2nd Law is:
(down is positive!)
FT1 + mg = m(v2/r)
FT1 = 3.73 N
• At the bottom of the circle:
Newton’s 2nd Law is:
(up is positive!)
FT2 - mg = m(v2/r)
FT2 = 9.61 N
Example: Back to a Bug on a CD
From Examples 5.1 & 5.2
A bug sits on the edge of a CD, of
radius r = 0.06m, as in the figure. It
undergoes uniform circular motion
as the CD spins. It goes around the
CD 6 times/sec. Results from before:
T = 0.17s, v = 2.3 m/s, aC = 88.2 m/s2
Given: Mass m = 5  10-3 kg
Calculate: The force which keeps the bug on the CD
F = maC = 0.1 N
Conceptual Example
A Ferris wheel rider moves in a
vertical circle of radius r at
constant speed v. Is the normal
force FN1 that the seat exerts on
the rider at the top of the wheel
1
a. less than, b. more than, or
c. equal to the normal force FN2
that the seat exerts on the rider at
the bottom of the wheel?
Use
Newton’s 2nd Law:
∑F = maC at top & bottom.
Solve for normal force & compare.
2
Conceptual Example
• A tether ball is hit so that it
revolves around a pole in a circle
of radius r at constant speed v.
• In what direction is the
acceleration?
• What force causes it?
Newton’s 2nd Law: ∑F = ma
x: ∑Fx = max
 FTx = maC = m(v2/r)
y:
∑Fy = may = 0
 FTy - mg = 0, FTy = mg
Centripetal Force Example
Outside the International
Space Station
• Whirls a ball on a string in a
perfect circle.
• The centripetal acceleration is
produced by the tension in the
string.
• If the string breaks, the object
would move in a direction
tangent to the circle at a constant
speed.