# Characterizing Circular Motion

Mécanique

14 nov. 2013 (il y a 5 années et 4 mois)

94 vue(s)

Characterizing Circular Motion

Angular position,
θ

Angular displacement,
Δ
θ

Angular speed,
ω
=
Δ
θ
/
Δ
t

Angular acceleration,
α
=
Δ
ω
/
Δ
t

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

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