Physics 2101
Section 3
March 3rd : Ch. 10
Announcements:
• Monday’s
Monday s Review Posted (in
Plummer’s section (4)
• Today start Ch.
Ch 10.
10
• Next Quiz will be next week
Test#2
# (Ch.
( h 77-9)) will
ll be
b at 6 PM, March
h 3, LockettLockett
k -6
Chap. 10: Rotation
Conservation of Angular Momentum
Describe their motion:
θ (t) → ω (t) → α (t)
Iω
I→
→ Torque
2
2
Rotation
In this chapter we will study the rotational motion of rigid bodies about
a fixed axis. To describe this type of motion we will introduce the
following new concepts:
-Angular
g
displacement
p
θ=s/r
- Average and instantaneous angular velocity (symbol: ω )
- Average and instantaneous angular acceleration (symbol: α )
- Rotational
R t ti l inertia,
i ti also
l known
k
as momentt off inertia
i ti (symbol
(
b l I)
- Torque (symbol τ )
We will also calculate the kinetic energy associated with rotation, write
Newton’s second law for rotational motion, and introduce the workkinetic energy theorem for rotational motion.
Rotation Variables
Here we assume (for now):
Rigid body
- all parts are locked together ( e.g. not a cloud)
Axis of rotation - a line about which the body rotates (fixed axis)
Reference line - rotates perpendicular to rotation axis
s arc length
θ (t)
( )= =
r
radius
looking down
Rotation axis
Δθ = θ 2 − θ1
dθ (t)
ω (t) =
dt
dω (t) d 2θ (t)
α(t) =
=
dt
dt 2
Angular position (displacement) of line at time t:θ (t)
(Δθ(t))
Angle
g is in units of radians ((rad)) : 1 rev = 360 =
Angular velocity of line at time t:
ω (t)
2 πr
= 2 π rad
r
(positive if ccw)
All points move with same angular velocity
Angular acceleration of line at time t:
(measured from+ xˆ : ccw)
α(t)
Units of rad/sec
(positive if ccw)
All points move with same angular acceleration
Units of rad/sec2
Analogies Between Translational and Rotational Motion
Translational Motion Rotational Motion
x ↔ θ
v
↔ ω
a
↔ α
v = v0 + at
at 2
x = xo + v0t +
2
v 2 − v02 = 2a ( x − xo )
↔ ω = ω0 + α t
↔ θ = θ0 + ω0t +
↔
2
ω 2 − ω02 = 2α (θ − θ0 )
mv 2
K=
2
m
F = ma
↔
↔
Iω 2
K=
2
I
τ = Iα
F
↔
τ
P = Fv
↔
P = τω
↔
αt 2
Problem 10-8: The wheel in the
picture has a radius of 30cm and is
rotating at 2.5rev/sec. I want to shoot
a 20 cm long arrow parallel to the
axle
l without
ith t hitting
hitti an spokes.
k (a)
( )
What is the minimum speed? (b)
Does it matter where between the
axle and rim of the wheel you aim?
If so what is the best position.
The arrow must pass through the wheel in a time less than it takes for the next spoke to rotate
1
rev
8
Δt=
= 0.05s
2.5rev / s
((a)) The minimum speed
p
is
vmn =
20cm
= 400cm / s = 4m / s
0.05s
(b) No there is no dependence upon the radial position.
Are angular quantities Vectors?
Some! Once coordinate system is established for
rotational motion, rotational quantities of velocity
and acceleration are given by rightright-hand rule
(angular displacement is NOT a vector)
Note: the movement is NOT along the
direction of the vector. Instead the body is
moving around the direction of the vector
Sample Problem 10-1: The disc in the figure below is rotating
about
bou an axiss described
desc bed by:
θ (t) = −1.00
1 00 − 0.600t
0 600t + 0.25t
0 25t 2
t 0 = 1.2s
R write
Re
it the
th equation
ti
ω = 0rad / s
θ (t) = −1.36 + 0.25 (t − 1.2 )
2
θ (−2) = 1.2rad
dθ (t)
= +0.50(t − 1.2)rad / s
θ (0) = −1.0rad ω (t) =
dt
dω (t)
α (t) =
= +0.50rad / s2
θ (1.2) = −1.36rad
dt
Rotation with Constant Rotational Acceleration
Rotational motion
1-D translational motion
ω = ω 0 + αt
θ = θ0 + ω 0 t + 12 αt 2
v = v0 + att
x = x0 + v0 t + 12 at 2
Problem: A disk, initially rotating at 120 rad/s, is slowed down with a constant
acceleration of magnitude 4 rad/s2.
a)
How much time does the disk take to stop?
ω = ω 0 + αt ⇒ t =
b)
(0 −120rad / s)
−4rad / s
2
= 30s
Through what angle does the disk rotate during that time?
θ = θ0 + ω 0 t + 12 αt 2 ⇒ Δθ = (120rad / s)(30s) + 12 (−4rad / s 2 )(30s) 2
= 1800rad = 286.5rev
286 5rev
Equations for constant Angular Acceleration
Remember these? Solve
angular problems the same
way.
Checkpoint 2: In four situations, a rotating body has angular position give by
the following. To which situations do the angular equations for constant
acceleration apply?
(a) : θ (t ) = 3t - 4
(b) : θ (t ) = -5t 3 + 4t 2 + 6
2 4
−
2
t
t
(d ) : θ (t ) = 5t 2 − 3
(c) : θ (t ) =
Relating Translational and Rotational Variables
Rotational position and distance moved
s = θr
( l radian
(only
di units)
i )
Rotational and translational speed
dr ds d θ
v=
= =
r
dt dt dt
v= ωr
Relating period and rotational speed [distance = rate × time]
T=
2πr 2π
=
v
ω
ωT = 2π
units of time (s)
Relating Translational and Rotational Variables
“acceleration” is a little tricky
Rotational
i l andd translational
l i l acceleration
l i
a)
from
v= ωr
dv d ω
=
r
dt
dt
dω
at =
r = αr
dt
from change in angular speed
tangential acceleration
b) from before we know there
there’ss also a “radial”
radial component
v2
2
ar = = ω r
r
radial acceleration
c) must combine two distinct rotational accelerations
a tot = a r + a t
2
2
2
= ω r + αr
2
2
2
Problem 10-30: Wheel A of radius rA=10
cm is coupled by belt B to wheel C of
radius rC=25 cm. The angular speed of
wheel A is increased from rest at a
constant rate of 1.6 rad/s2. Find the time
needed for wheel C to reach an angular
speed of 100 rev/min.
If the belt does not slipp the tangential
g
acceleration of each wheel is the same.
rAα A = rCαC
rA
αC = α A = 0.64
0 64 radd / s 2
rC
The angular velocity of C is.
ωC ωC rC
ωC = αC t so t =
=
αC rAα A
with ω = 100 rev/sec t = 16 s