# Circular Motion - Ferris State University

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31 Οκτ 2013 (πριν από 4 χρόνια και 6 μήνες)

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Circular Motion

Examples: Planets, atoms, cars
on curves, CD
-
ROM’s, propellers,
etc.

Rotational Kinematics

How do we
describe an
object moving in
a circle?

Measure angles relative to reference line:

Δθ

Θ

= 0
0

Θ

= 60
0

1 revolution = 360
0

360
0

= 2
π

0

Position in
circular motion
is expressed as
an angle,
preferably in

‘x’ in linear motion, ‘
θ
’ in circular
motion.

Example

A fan blade, with a diameter of 1 m, is
rotating at 1 revolution every second.
Through what angle does the tip of the

1 rev in 1 second

0.5 rev in 0.5 sec

0.3 rev in 0.3 sec, so

0.3 rev = 108
0

Speed in Circular Motion

Since we’re not using ‘x’ to measure position, we
can’t define speed as
Δ
x/
Δ
t.

But, since ‘
θ
’ replaces ‘x’, why not define angular
speed by

Δ
θ
/
Δ
t = (
θ
f

θ
o
)/(t
f

t
o
)

This is called angular speed and is given the
symbol (omega),
ω
.

ω
=

Δ
θ
/
Δ
t

Angular Velocity

ω
=

Δ
θ
/
Δ
t

Connection between Linear and
Circular Motion

q

= x/r

= v/r

a

= a/r

Example

What is the angular speed of a 33 1/3 rpm
record?

ω

=33.33 rev/min = 33.33 rev/60 sec

ω

= 33.33 ( 2
π

Acceleration in Circular Motion

Consider your rotating car tires as you
accelerate from 25 mph to 55 mph. What
is happening to the rotational speed of the
tires? R = 33 cm.

V = 11 m/s

V=24.5 m/s

ω
o

ω
f

Linear and Angular Connection

ω

= v/r

So,
ω
o

= (11 m/s)/0.33 m = 33 rad/s

And
ω
f

= (24.5 m/s)/0.33 m = 73.5 rad/s.

Therefore the change in angular speed,

ω
f

ω
o

Δ ω

Angular Acceleration

When you have changing angular speeds,
this means the object has an angular
acceleration,
α

(alpha),

which is calculated
by

α

=
Δ ω
/
Δ
t

2

2

Characterizing Circular Motion

Angular position,
θ

Angular displacement,
Δ
θ

Angular speed,
ω
=
Δ
θ
/
Δ
t

Angular acceleration,
α
=
Δ
ω
/
Δ
t

Kinematics of Circular Motion

ω
=
Δ
θ
/
Δ
t

α
=
Δ
ω
/
Δ
t

ω
ave
=(
ω
f

+
ω
o
)/2

Θ

=

ω
ave
t

ω
f
=
ω
o
+
α
t

Θ

=

Θ
o

+
ω
o
t + ½
α
t
2

ω
f
2

=
ω
o
2

+ 2
αΔ
θ

Kinematics Example

A flywheel of a machine is rotating at 12
rev/s. Through what angle will the wheel
be displaced from its original position after
5 seconds?

Angular speed,
ω

= 12 rev/s = 75 rad/s

Θ

=

ω
ave

= 214875
0
. 59.6875 revolutions, so .6875
revolutions from start position = 247
o
.

A turntable revolves at 33 1/3 rpm. It is shut
off and slow to a stop in 6.3 seconds.
What is the angular acceleration?

Through what angle did it turn as it slow to a
stop?
ω
f
=0,
ω
o

t = 6.3 s

ω
f
=
ω
o
+
α
t

Θ

=

Θ
o

+
ω
o
t + ½
α
t
2

Dynamics of Rotation

Examine circular
motion taking
Newton’s Laws into
consideration.

1
st

Law
-

2
nd

Law
-

3
rd

Law
-

Dynamics of Rotation

1
st

Law

Is Moon at rest?

Is Moon moving in a
straight line?

Conclusion

MOON

EARTH

Dynamics of Rotation

1
st

Law

Objects executing
circular motion have
a net force acting on
them…even if you
can’t see the agent
of the force.

What force acts on the
Moon?

MOON

EARTH

Dynamics of Rotation

2
nd

Law

F
NET

=
m
a

a

is a vector defined
by

a =
Δ
v/
Δ
t

Δ
v

=
v
f

v
o

For circular motion the
speeds are the
same, but the
directions aren’t.

MOON

EARTH

Dynamics of Rotation

MOON

EARTH

-
v
o

v
f

Δ
v=

v
f
-
v
o

Let’s visually
examine the change
in velocity

Dynamics of Rotation

MOON

EARTH

v
o

v
f

Δ
v=

v
f
-
v
o

So, a
Δ
v
exists because
the direction is changing,
not the magnitude. How
do we find the
acceleration?

D
v/v =
D
r/r

D
v/v = v
D
t/r

D
v/
D
t = v
2
/r

a
c

=
v
2
/r

Dynamics of Rotation

MOON

EARTH

The acceleration
toward the center is
called ‘centripetal’
acceleration, a
c
,
given by

a
c

= v
2
/R

In magnitude and
directed toward
center.

R

v

v

Dynamics of Rotation

MOON

EARTH

So, Newton’s 2
nd

Law for rotation
becomes,

F =ma
c

= mv
2
/R

In magnitude and
directed toward
center.

R

v

v

Dynamics of Rotation

MOON

EARTH

A physical statement
that relates cause and
effect. Cause = F,
effect = mv
2
/R

F = mv
2
/R

The right side is ‘what
you see’, the left side is
‘why’.

R

v

v

Dynamics of Rotation

MOON,m

EARTH,M

What is the ‘cause’ for
the Moon’s motion?

F =GMm/R
2

Newton’s Universal Law
of Gravity.

G = 6.67 x 10
-
11

Nm
2
/kg
2

R

v

v

ULG
-
two objects of given masses separated
by known distance exert a gravitational force
of attraction on each other whose size is
determined from

F =GMm/R
2

You are sitting next to a person whose mass
is 55 kg. Your mass is 75 kg. What is the
force of attraction between you if you are 0.8
m apart (center to center)?

F =(6.67 x 10
-
11
Nm
2
/kg
2
)(75kg)(55kg)/(0.8m)
2

F=0.00000043 N = 0.43 microNewtons

How do we know the mass of the
Earth?

Using the ULG,

F =GMm/R
2

And 2
nd

Law,

F = mv
2
/R,

Combine,

GMm/R
2
=

mv
2
/R

M = v
2
R/G

M = (
ω
R)
2
R/G

M =
ω
2
R
3
/G

M =
ω
2
R
3
/G

R = 380,000,000 m

ω

= 2
π

rad /27.3 days =2.664 x 10
-
6

So, M = (2.664 x 10
-
6

2
(380,000,000
m)
3
/(6.67 x 10
-
11
Nm
2
/kg
2
)

M = 5.84 x 10
24

kg.

True value 5.98 x 10
24

kg.

Earth and Moon orbit
the center of mass of
the system.

Located 1070 miles
below the Earth’s
surface or 2880
miles from center of
Earth.

Problem Solving Strategy for
Circular Motion Problems

Is it Kinematics or Dynamics

Kinematics
-
You are trying to characterize
the motion by its position, speed or
acceleration.
.

Dynamics
-
You are trying to relate the
motion to its causes.
.

A tire of diameter 26 inches is spinning with a
constant angular velocity of 2 rad/s. What is the
centripetal acceleration of a point on the rim of
the tire?

R = 0.33 m

ω

Centripetal Accel. a
c

= v
2
/R

v =
ω
R

a
c

= v
2
/R =
ω
2

R

2
*0.33 m

a
c

=1.32 m/s
2
, directed toward axle.

Another example,
.

A dentist’s drill spins at 1800 rpm. If it takes
6 seconds to stop when turned off, what is
the angular acceleration of the drill?

Initial angular speed,

ω
o

Final angular speed =
ω
f

Time, t = 6 s.

Angular Accel,

α

=
Δω
/
Δ
t=(0
-

=
-
2

A car moving at 25 m/s rounds a curve of radius
100 m and is just on the verge of slipping. So if
that is the fastest that it can round this curve,
what is the maximum speed it can travel on a

In both cases the force keeping that car from
slipping will be the same, i.e. static friction. So,
F = mv
2
/R is the equation we will apply to each
case.

v
1
2
/R
1

= v
2

2
/R
2

v
2

= v
1
(R
2
/R
1
)
1/2

Next Example

A 0.50
-
kg mass is attached to the end of a 1.0
-
m string. The system is whirled in a horizontal
circular path. If the maximum tension that the
string can withstand is 350 N. What is the
maximum speed of the mass if the string is not
to break.

M = 0.5 kg, R = 1.0 m, F(max) = 350 N

F = mv
2
/R

V = ( FR/m)
1/2

= (350*1.0/0.5)
1/2

= 26.5 m/s

Next Example

A car goes around a flat curve of radius 50 m at a
speed of 14 m/s. What must be the minimum
coefficient of friction between the tires and the
road for the car to make the turn?

V = 14 m/s, R = 50 m.

F = mv
2
/R and f =
µN = µmg, so

µmg =
mv
2
/R

µ =
v
2
/gR

= (14 m/s)
2
/(9.8 m/s
2
*50m)

µ

=0.4

Next Example

The hydrogen atom consists of a proton of
mass 1.67X10
-
27

kg and an orbiting
electron of mass 9.11X10
-
31

kg. In one of
its orbits, the electron is 5.3X10
-
11

m from
the proton. What is the mutual
gravitational attractive forces between the
electron and proton?

F = GM
1
M
2
/R
2

=(6.67e
-
11*1.67e
-
27*9.11e
-
31)/(5.3e
-
11)
2

F =3.6 x 10
-
47

N

3
rd

Law:

F(earth on moon) =
-
F(moon on Earth)

F
EM

=
-
F
ME

F
EM

-
F
ME

Energy of Orbiting Objects

Consider Moon.

It has velocity, so it has Kinetic Energy.

E = K + U.

What is the potential energy of a bound
object?

K

U

E

K is pos, U is neg, E is neg.