Chapter 5: Circular Motion
Rotating Objects:
Wheels, moon, earth, CDs, DVDs etc.
Rigid bodies.
Description of circular motion.
Angular Position, Angular Displacement
s
θ
r
Angle (in radians) θ = arc length/radius = s/r.
z Angular displacement ∆θ = θ2-θ1
ÎHow far from initial angular position it has
rotated.
z ∆θ = positive going CCW, negative going CW.
z Magnitude ∆θ = ∆s/r. If ∆s = r, then ∆θ = 1 radian
z
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Angular Displacement
360o
2π rads
180o
π rads
90o
π/2 rads
In 1 complete turn,
∆s = circumference of the circle = 2πr.
z In 1 turn, ∆θ = ∆s/r = (2πr)/r = 2π rads.
z 360o = 2π rads, 180o = π rads, 90o = π/2 rads.
z
Angular Velocity (ω)
ω
z
Average Angular velocity ω = ∆θ/∆t
Î How fast it is rotating
Î It is a vector quantity.
Î Direction: Up for CCW rotation, down for CW
rotation. [ Right Hand Rule].
Î Units = radians/second.
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Examples
1.Change 1 rpm to rad/sec.
2. What is the angular speed of
the earth’s rotation?
3. What is the angular speed of
the minute hand of an analogue
clock?
Circular to Linear
z Linear
distance = Arc length ∆s = r ∆θ
(θ in radians)
z Speed
|v| = ∆s/∆t = r ∆θ/∆t = rω
z Direction of v is tangent to circle
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Example:
1. A CD spins with angular speed
20 radians/second. What is the
linear speed 6 cm from the center
of the CD?
V = r ω = 0.06 x 20 = 1.2 m/s
Example:
2. If you were seated at the equator,
what would be your linear speed due
to the rotation of the earth? (radius
of the earth = 6.371 x 106 m.
v = rω
ω = ∆θ/∆t = 2π/(24x3600) = 7.27 x 10-5 rad/s
V = 6.37 x 106 x 7.27 x 10-5
= 463 m/s
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Circular Motion Act
b
a
c
v
Answer: b
A ball is going around in a circle attached to a
string. If the string breaks at the instant shown,
which path will the ball follow?
Frequency (f)
¾ Frequency (f) = number of turns a rotating
object makes in one second.
¾ f = # of turns/time taken to make the turns.
¾ Unit: Cycles per second or hertz (Hz)
¾ 1 cycle/second = 1 Hz
¾ Eg 1. The frequency of the second hand of an
analogue clock = 1 turn/60s = 1/60 Hz.
¾ The frequency of the moon’s rotation around
the earth = 1 turn/28 days = 4.13 x 10-7 Hz.
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Period (T)
¾ Period (T) = Time taken to make one cycle.
¾ T = Time to make one turn.
¾ Unit: seconds
¾ Since f = # of cycles per second, Period and
frequency are related by the expression
¾ f = 1/T or T = 1/f
¾ Eg 1. The period of the second hand of an
analogue clock = 60 seconds.
¾ The period of the moon’s rotation around the
earth = 28 days = 2.42 x 106 s.
Relation between ω, v, f and T
r
v
In one complete rotation:
Distance moved = 2πr.
Time taken = period, T
Speed v = dist/time = 2πr/T
ie v = 2πr/T = 2πrf
Since v = rω, we have ω = 2π/T = 2πf
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Uniform Circular Motion
• Uniform Circular Motion is motion of an
object in a circular path with constant
(uniform) speed.
• Its linear speed is constant. Its angular
velocity (ω) is constant.
• Direction constantly changing.
• Hence linear velocity v is not constant.
• The instantaneous direction of v is
tangential to the circular path.
• Since velocity v is not constant, an object in
uniform circular motion must have an
acceleration.
Uniform Circular Motion
v
r
ω (out of page)
v = rω
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Acceleration in Uniform Circular Motion
The acceleration is due to change in direction of the
velocity, not its magnitude (since magnitude of the velocity
is constant).
∆v
v2
R
∆R
v1
aave= ∆v / ∆t
-v1 Direction of a =
v2
direction of ∆v
∆v = v2 – v1
As ∆t →0, v1 and v2 will be almost parallel to each other
with ∆v being perpendicular to them.
∆v will point towards center of circle.
Therefore, acceleration a is directed towards the center
of the circle.
Uniform Circular Motion
(circular motion with constant speed)
• Instantaneous velocity is
tangent to circle.
aC
v
• Instantaneous acceleration
is radially inward. Hence it is
called radial or centripetal
acceleration (aC).
• The magnitude of aC = v2/r
• There must be a net force
to provide the acceleration.
aC = v2/r = rω2 = (2πr/T)2/r = 4π2r/T2 = 4π2rf2
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Net Force in Uniform Circular Motion
FC
Conclusion:
v
¾There must be a net force to
provide the acceleration.
¾ This net force must also be
directed radially towards the
center. Hence it is called radial
or centripetal force (FC).
¾ Fnet = FC = maC = mv2/r = mrω2
9Any object moving in a circular path must have a
net force exerted on it directed towards the
center of the circle.
9The magnitude of this force = mv2/r = mrω2
A spider sits on a turntable that is rotating
at a constant 33 rpm.
The centripetal acceleration of the spider (aC) is
(A)Greater the closer the spider is to the central
axis
(B)Greater the farther the spider is from the
central axis
(C)Zero
(D)Non zero and independent of the spider’s
location on the turn table
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Examples of Circular Motion
1. A ball, mass m, swung in a vertical circle by a
string of length L
A
¾Fnet = maC
mg
T
L
¾-TA-mg = -mv2/L
¾ TA = mv2/L - mg
If T = 0, the ball will not
move in the circular path.
For circular motion, minimum swing speed is
obtained by setting T = 0
0 = mv2/L – mg or v = √(gL)
At the bottom of the circular path:
A
¾Fnet = maC
¾TB - mg = mv2/L
T
B
mg
¾ TB = mv2/L + mg
In uniform circular motion, TB > TA
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2. Water in a bucket swung in a vertical circle
¾Fnet = maC
r
mg
N
¾-N-mg = -mv2/r
¾ N = mv2/r - mg
If N = 0, the water will
pour out.
For water not to pour out, minimum speed is
obtained by setting T = 0
0 = mv2/r – mg or v = √(gr)
What is the minimum speed you must
have at the top of a 20 meter roller
coaster loop, to keep the wheels on the
track.
A 20 m roller coaster loop;
r = 10 m
v = √(gr) = √(9.8 x 10) = √ (98)
= 9.9 m/s
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3. Small ball swung in a horizontal circle
¾Fnet = maC
¾ Small mass,T >> mg
r
¾ T = mv2/r
T
mg
4. Car driven on valley/hill
N
v
r
N
mg
v
r
mg
¾Fnet = maC
¾ Valley: NV – mg = mv2/r and NV = mg + mv2/r
¾ Hill: NH – mg = -mv2/r and NH = mg – mv2/r
¾ On flat road, NF = mg
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5. Car rounding a curve on a flat road
¾Fnet = maC
r
¾Fy: N – W = 0
¾ Fx: fs = mv2/r
N
W
¾ If v is increased, fs will
increase up till fs max (= µsN)
fs
¾ mv2/r ≤ µsN (= µsmg)
¾ v ≤ √(µsrg) OR µs ≥ v2/rg
6. Car rounding a curve on a banked road
¾To reduce chances of skidding, roads
are banked.
¾Centripetal force will then come from
a component of the normal force,
reducing total reliance on friction.
¾ At a particular speed, all of the
centripetal force Fc will be
contributed by the normal force N.
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Nx
y
N
mg
Ny
θ
θ
Nx = N sinθ
N
x
Ny = N cosθ
With no dependence on friction:
Along y-axis: Ny = Ncosθ - mg = 0 ie, Ncosθ = mg
Along x-axis: Nsinθ = mv2/r
OR: tan θ = v2/rg, ie Banking angle θ = tan-1(v2/rg)
Speed v ≤ √(rg tan θ)
Example 1
A 1,000 kg car rounds a curve on a flat
road of radius 50 m. What should be the
speed limit if the coefficient of static
friction is 0.3?
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Example 2
A car rounds a curve on a flat road of
radius 25 m at a speed of 13 m/s. What
should be the banking angle of the road
for the car to safely negotiate the turn if
the road is icy?
Circular Orbits
Satellites orbit the earth in circular path
due to gravitational force. Its speed
must be just right.
Too low – falls to the earth.
Too high – escapes into space
m
v
Fc = GmM/r2 = mv2/r
v2 = GM/r = GM/(rE + h)
r
M
For Geosynchronous satellites,
V = 2πr/T, where T = 24 hours
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Example 3
The Hubble telescope moves in a circular
orbit 613 km above the earth’s surface.
The radius of the earth is 6.37 x 103 km
and the mass of the earth is 5.98 x 1024
kg.
(a) What is the speed of the satellite in this
orbit?
(b) Is the telescope geosynchronous?
G = 6.67 x 10-11 N.m2/kg2
Non-Uniform Circular Motion
• Uniform circular motion – circular motion in
which speed stays constant and direction
changes.
• Centripetal acceleration – changes only
direction of velocity, not its magnitude.
• ac – points along radius towards the center.
• Non-uniform circular motion – when both
magnitude and direction of the velocity
changes.
• The acceleration is not along the radius.
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Non-Uniform Circular Motion
Non-uniform circular motion – when both
magnitude and direction of the velocity changes.
The acceleration is not along the radius. This
acceleration (a) can be resolved into tangential
(at) and centripetal (aC) components
a
at
ac
a = ac + at
ac = changes direction of v.
at = changes magnitude of v
Angular Acceleration
z Angular
acceleration is the change in
angular velocity ω divided by the
change in time.
= ∆ω/∆t = (ω - ωo)/∆t
z Units: rad/s2
z Note: tangential acceleration at = rα
zα
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Example
If the speed of a roller coaster car is 15
m/s at the top of a 20 m loop, and 25
m/s at the bottom. What is the car’s
average angular acceleration if it takes
2 seconds to get from the top to the
bottom.
α = ∆ω/∆t = (ω - ωo)/∆t
= vo/r = 15/10 = 1.5 rad/s
z ω = v/r = 25/10 = 2.5 rad/s
z α = ∆ω/∆t = (2.5 – 1.5)/2 = 0.5 rad/s2
z ωo
Kinematics for Circular Motion with constant α
Linear Variables
Angular Variables
When a = constant
When
v = vo + at
ω = ωο + αt
v2 = vo2 + 2a (x – xo)
ω2 = ωο2 + 2 α(θ − θο)
x = x0 + vot + ½ at2
θ = θ ο + ω ot + ½ α t 2
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α = constant
CD Player Example
The CD in your disk player spins at
about 20 radians/second. If it
accelerates uniformly from rest
with angular acceleration of 15
rad/s2, how many revolutions does
the disk make before it is at the
proper speed?
ω2 = ωο2 + 2 α(θ − θο)
1 Revolution = 2 π radians
Example
A small ball of mass 200 g is rotated in
a horizontal circle of radius 0.5 m. If
the ball makes 10 revolutions in 5
seconds, what is its centripetal
acceleration?
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Example
The moon orbits the earth in a nearly
circular path once every 27.3 days. It
the distance from the moon to the earth
is 384,000 km, what is the moon’s
acceleration?
Example
Calculate the velocity of a
satellite moving in a stable
circular orbit about the Earth at
a height of 3600 km.
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Example
A hypothetical planet has a
radius 2.5 times that of earth but
has the same mass. What is the
acceleration due to gravity near
its surface?
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