Physic 231 Lecture 16
•
•
Main points of last lecture:
Description of circular motion in
cylindrical coordinates (r,θ):
•
•
∆θ
∆θ
, ω = lim ∆t →0
ω=
∆t
∆t
∆ω
∆ω
α =
, α = lim ∆t →0
∆t
∆t
∆s = ∆θr; vt = ωr; at = αr
 y
r = x + y ;θ = tan  
x
x = r cos(θ ); y = r sin (θ )
2
2
∆s = ∆θr
Main points of today’s lecture:
Rotational motion definitions:
−1
•
Rotational kinematics equations:
∆θ = ω t
ω = ω0 + αt
1
1
ω = (ω + ω 0 ) ∆θ = (ω + ω 0 )t
2
2
1
∆θ = ω 0t + αt 2 ω 2 = ω 02 + 2α∆θ
2
• Rolling motion:
∆x = s = r∆θ
v = rω; a = rα
Example
•
A wheel has a radius of 4.1 m. How far (path length) does a point on
the circumference travel if the wheel is rotated through angles of 30°,
30 rad, and 30 rev, respectively?
if θ = 30o , then we must first convert to radians :
1rad
= 0.523 rad
θ = 30
0
57.3
s = rθ = 4.1m ⋅ 0.523 = 2.15m
if θ = 30 rad, then :
s = rθ = 4.1m ⋅ 30 = 123m
if θ = 30 rev , then we must first convert to radians :
2π rad
= 188.5 rad
θ = 30 rev
rev
s = rθ = 4.1m ⋅ 188.5 = 723m
o
Angular velocity
•
•
The rate of change of the angular displacement is defined to be the
angular velocity ω.
In term of the angular displacement, the average angular velocity over
a time interval ∆t is:
∆θ
ω=
•
∆t
The instantaneous angular velocity can be
obtained by making the time interval very
short:
∆θ
ω = lim ∆t →0
•
∆t
Note: if ∆θ measured in radians, ∆s=r ∆θ
is the arc length covered in time ∆t. Thus:
∆s r∆θ
vt =
=
= rω
∆t
∆t
•
is the average tangential speed. And
vt = lim ∆t →0
∆s
r∆θ
= lim ∆t →0
= rω
∆t
∆t
is the instantaneous tangential speed of the
object.
Fig. 7.4, p. 189
Slide 5
Example
•
Two people start at the same place and walk around a circular lake in
opposite directions. The first has an angular speed of 1.7x10-3 rad/s,
while the second has an angular speed of 3.4x10-3 rad/s. How long will
it be before they meet?
∆θ
we want ∆θ1 − ∆θ 2 = 2π :
∆θ1 = ω1t; ∆θ 2 = ω 2t
1
∆θ2
2π = ∆θ1 − ∆θ 2 = ω1t − ω 2t = (ω1 − ω 2 )t
2π
6.28
t=
=
= 1231s
-3
-3
(ω1 − ω2 ) 1.7x10 rad/s − (-3.4x10 rad/s)
•
If first person is walking at a speed of 0.25 m/s, what is the radius of
the lake?
– a) 150 m
– b) 15 m
v
.25m / s
ω
v
=
r
⇒
r
=
=
= 147m
– c) 45 m
ω .0017rad / s
– d) 1200 m
Angular acceleration
•
•
Many times, the angular velocity changes with time. We quantify this
by the angular acceleration α.
In term of the angular velocity , the average angular acceleration over a
time interval ∆t is:
∆ω
α =
•
Note: if ∆ω measured in rad/s, ∆vt=r∆ω is the change in tangential
speed during time ∆t. Thus:
at =
•
∆t
∆vt r∆ω
=
= rα
∆t
∆t
The instantaneous angular acceleration can be obtained by making
the time interval very short:
α = lim ∆t →0
•
∆ω
∆t
is the average tangential acceleration. And at=rα is the
instantaneous tangential speed of the object.
at = lim ∆t →0
r∆ω
∆vt
= lim ∆t →0
= rα
∆t
∆t
Analogy between linear and angular kinematics for
constant acceleration
•
If the acceleration is constant, we obtained a set of equation on the left
side of the table relating ∆x, v, a and t. If the angular acceleration is
constant and we go through the same steps, we obtain the analogous
set of equation on the right side of the table relating ∆θ, ω, α, and t.
∆x = v t
v = v0 + at
1
v = (v + vo )
2
1
∆x = (v + vo )t
2
1
∆x = v0t + at 2
2
v 2 − v02 = 2a∆x
∆θ = ω t
ω = ω 0 + αt
1
(ω + ω0 )
2
1
∆θ = (ω + ω 0 )t
2
1
∆θ = ω 0t + αt 2
2
ω 2 = ω02 + 2α∆θ
ω=
Conceptual question
•
A ladybug sits at the outer edge of a merrygo-round, and a gentleman
bug sits halfway between her and the axis of rotation. The merry-goround makes a complete revolution once each second.The gentleman
bug’s angular speed is
–
–
–
–
a) half the ladybug’s.
b) the same as the ladybug’s.
c) twice the ladybug’s.
d) impossible to determine
Example
•
After 10 s, a spinning roulette wheel has slowed to an angular speed of
1.88 rad/s. During this time, the wheel rotates through an angle of 44.0
rad. Determine the angular acceleration of the wheel.
∆θ = ω t
ω = ω 0 + αt
1
(ω + ω0 ) ∆θ = 1 (ω + ω0 )t
2
2
1
∆θ = ω 0t + αt 2 ω 2 = ω 02 + 2α∆θ
2
ω = ωo + αt ⇒ ωo = ω − αt
ω=
ω
1.88 rad/s
t
10 s
∆θ
44.0 rad
1
1
∆θ = ωot + αt 2 = (ω − αt )t + αt 2
2
2
1 2
⇒ ∆θ = ωt − αt
2
ωt − ∆θ 1.88rad / s ⋅ 10s − 44rad
2
rad
s
⇒α =
=
=
−
0
.
5
/
2
1 2
(
)
s
0
.
5
10
t
2
Quiz
•
A centrifuge decelerates to rest over a time interval of 10 seconds from
an initial angular velocity of 1000 rad/s. Over this time interval, the
angular displacement is about:
– a) 5 rad
– b) 5000 rad
– c) 10 rad
– d) 10000 rad
ω − ωo
− 1000rad / s
= −100rad / s 2
t
10s
1 2
1
2
∆θ = ωot + αt = (1000rad / s )(10s ) + (− 100rad / s 2 )(10s )
2
2
∆θ = 5000rad
ω = ωo + αt ⇒ α =
=
Rolling motion
•
Consider a wheel of radius r rolling on
the ground. The relationship between the
distance traveled and the angle is
obtained by considering the arc length s:
d = s = rθ
•
Similarly there are corresponding
relationships between v, a and ω, α.
v = vt = rω
a = at = rα
∆θ