Download Rotational Motion: Statics and Dynamics

Physics 207 – Lecture 17
Lecture 17 (Catch up)
Goals:
• Chapter 12
Introduce and analyze torque
Understand the equilibrium dynamics of an extended
object in response to forces
Employ “conservation of angular momentum” concept
Assignment:
HW8 due March 17th
Thursday, Exam Review
Physics 207: Lecture 17, Pg 1
Rotational Motion: Statics and Dynamics
Forces are still necessary but the outcome depends
on the location from the axis of rotation
This is in contrast to the translational motion and
acceleration of the center of mass. Here the position
of these forces doesn’t matter (doesn’t alter the
physics we see)
However: For rotational statics & dynamics: we
must reference the specific position of the force
relative to an axis of rotation. It may be necessary to
consider more than one rotation axis!
Vectors remain the key tool for visualizing Newton’s
Laws
Physics 207: Lecture 17, Pg 2
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Physics 207 – Lecture 17
Angular motion can be described by vectors
With rotation the distribution of mass matters. Actual
result depends on the distance from the axis of
rotation.
Hence, only the axis of rotation remains fixed in
reference to rotation.
We find that angular motions may be quantified by
defining a vector along the axis of rotation.
We can employ the right hand rule to find the vector
direction
Physics 207: Lecture 17, Pg 3
The Angular Velocity Vector
• The magnitude of the angular velocity vector is .
• The angular velocity vector points along the axis
of rotation in the direction given by the right-hand
rule as illustrated above.
• As ω increased the vector lengthens
Physics 207: Lecture 17, Pg 4
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Physics 207 – Lecture 17
So: What makes it spin (causes ω) ?
A force applied at a distance from
the rotation axis gives a torque
θ
r
FTangential
=|FTang| sin θ
a
FTangential
F
Fradial
r
Fradial
If a force points at the axis of rotation the wheel won’t
turn
Thus, only the tangential component of the force matters
With torque the position & angle of the force matters
τNET = |r| |FTang|
|r| |F| sin θ
Physics 207: Lecture 17, Pg 5
Rotational Dynamics: What makes it spin?
a
A force applied at a distance
FTangential
from the rotation axis
τNET = |r| |FTang|
F
Fradial
|r| |F| sin θ
r
Torque
is the rotational equivalent of force
Torque has units of kg m2/s2 = (kg m/s2) m = N m
τNET =
r FTang = r m aTang
=rmrα
= (m r2) α
For every little part of the wheel
Physics 207: Lecture 17, Pg 6
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Physics 207 – Lecture 17
For a point mass
τNET = m r2 α
The further a mass is away from
this axis the greater the inertia
(resistance) to rotation (as we
saw on Thursday)
τNET = I α
a
FTangential
F
Frandial
r
This is the rotational version of FNET = ma
of inertia, I Σi mi ri2 , is the rotational
equivalent of mass.
If I is big, more torque is required to achieve a
given angular acceleration.
Moment
Physics 207: Lecture 17, Pg 7
Rotational Dynamics: What makes it spin?
A force applied at a distance from
the rotation axis gives a torque
τNET = |r| |FTang|
a
FTangential
F
|r| |F| sin θ
Fradial
r
A constant torque gives constant angular acceleration
if and only if the mass distribution and the axis of
rotation remain constant.
Physics 207: Lecture 17, Pg 8
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Physics 207 – Lecture 17
Torque, like ω, is a vector quantity
Magnitude is given by (1)
|r| |F| sin θ
(2)
|Ftangential | |r|
(3)
|F| |rperpendicular to line of action |
Direction is parallel to the axis of rotation with respect to the
“right hand rule”
r sin θ
F cos(90°−θ) = FTang. line of action
r
a
90°−θ
θ
F
F
F
Fradial
r
r
r
And for a rigid object τ = I α
Physics 207: Lecture 17, Pg 9
Exercise Torque Magnitude
In which of the cases shown below is the torque provided by
the applied force about the rotation axis biggest? In both
cases the magnitude and direction of the applied force is the
same.
Remember torque requires F, r and sin θ
or the tangential force component times perpendicular distance
L
F
A. Case 1
B. Case 2
F
L
axis
C. Same
case 1
case 2
Physics 207: Lecture 17, Pg 10
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Physics 207 – Lecture 17
Example: Rotating Rod Again
A uniform rod of length L=0.5 m and mass m=1 kg is
free to rotate on a frictionless pin passing through
one end as in the Figure. The rod is released from
rest in the horizontal position.
What is the initial angular acceleration α?
L
m
Physics 207: Lecture 17, Pg 11
Example: Rotating Rod
A uniform rod of length L=0.5 m and mass m=1 kg is free …
What is its initial angular acceleration ?
1. For forces you need to locate the Center of Mass
CM is at L/2 ( halfway ) and put in the Force on a FBD
2. The hinge changes everything!
L
m
mg
Σ F = 0 occurs only at the hinge
but τz = I αz = - r F sin 90°
at the center of mass and
IEnd αz = - (L/2) mg
and solve for αz
Physics 207: Lecture 17, Pg 12
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Physics 207 – Lecture 17
Work (in rotational motion)
Consider the work done by a force F acting on an
object constrained to move around a fixed axis. For an
infinitesimal angular displacement dθ :where dr =R dθ
F
dW = FTangential dr
dW = (FTangential R) dθ
φ
axis of
rotation
R
dθ
dr =Rdθ
dW = τ dθ (and with a constant torque)
We can integrate this to find:
W = τ θ = τ (θf−θi)
Analog of W = F •∆r
W will be negative if τ and θ have opposite sign !
Physics 207: Lecture 17, Pg 13
Statics
Equilibrium is established when
Translatio nal motion
Rotational motion
r
ΣFNet = 0
r
Στ Net = 0
In 3D this implies SIX expressions (x, y & z)
Physics 207: Lecture 17, Pg 14
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Physics 207 – Lecture 17
Example
Two children (60 kg and 30 kg) sit on a horizontal teeter-totter. The
larger child is 1.0 m from the pivot point while the smaller child is
trying to figure out where to sit so that the teeter-totter remains
motionless. The teeter-totter is a uniform bar of 30 kg its moment of
inertia about the support point is 30 kg m2.
Assuming you can treat both children as point like particles, what is
the initial angular acceleration of the teeter-totter when the large
child lifts up their legs off the ground (the smaller child can’t reach)?
For the static case:
Rotational motion
r
Στ Net = 0
Physics 207: Lecture 17, Pg 15
Example: Soln.
r
Use Στ
Net
=0
N
30 kg
30 kg
0.5 m
300 N 300 N
60 kg
1m
600 N
Draw a Free Body diagram (assume g = 10 m/s2)
0 = 300 d + 300 x 0.5 + N x 0 – 600 x 1.0
0= 2d + 1 – 4
d = 1.5 m from pivot point
Physics 207: Lecture 17, Pg 16
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Physics 207 – Lecture 17
Classic Example:
Will the ladder slip?
Not if
r
ΣFNet = 0
r
Στ Net = 0
Physics 207: Lecture 17, Pg 17
Classic Example:
Will the ladder slip?
Physics 207: Lecture 17, Pg 18
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Physics 207 – Lecture 17
EXAMPLE 12.17 Will the ladder slip?
Physics 207: Lecture 17, Pg 19
EXAMPLE 12.17 Will the ladder slip?
Physics 207: Lecture 17, Pg 20
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Physics 207 – Lecture 17
Angular Momentum:
We have shown that for a system of particles,
r
r
p ≡ mv
momentum
r dpr
F=
=0
dt
is conserved if
What is the rotational equivalent of this (rotational “mass” times
rotational velocity)?
r
r
L ≡ Iω
angular momentum
r
dL
dt
r
τ =
is conserved if
= 0
Physics 207: Lecture 17, Pg 21
Angular momentum of a rigid body about a fixed axis:
Consider a rigid distribution of point particles rotating in the xy plane around the z axis, as shown below. The total angular
momentum around the origin is the sum of the angular
momentum of each particle:
L z ≡ I z ω = ∑ m i ri ω = ∑ m i ri v i = ∑ ri p i
2
i
i
i
v1
( ri and vi , are perpendicular)
m2
Using vi = ω ri , we get
r
r
L = Iω
v2
r2
m3
j
ω
i
r3
r1 m1
v3
Even if no connecting rod we can define an Lz
Physics 207: Lecture 17, Pg 22
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Physics 207 – Lecture 17
Example: Two Disks
A disk of mass M and radius R rotates around the z axis
with angular velocity ω0. A second identical disk, initially
not rotating, is dropped on top of the first. There is
friction between the disks, and eventually they rotate
together with angular velocity ωF.
z
z
ω0
ωF
Physics 207: Lecture 17, Pg 23
Example: Two Disks
A disk of mass M and radius R rotates around the z axis with
initial angular velocity ω0. A second identical disk, at rest, is
dropped on top of the first. There is friction between the disks,
and eventually they rotate together with angular velocity ωF.
No External Torque so Lz is constant
Li = Lf I ωi i = I ωf ½ mR 2 ω0 = ½ 2mR 2 ωf
z
z
ω0
ωF
Physics 207: Lecture 17, Pg 24
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Physics 207 – Lecture 17
Example: Throwing ball from stool
A student sits on a stool, initially at rest, but which is free to
rotate. The moment of inertia of the student plus the stool is I.
They throw a heavy ball of mass M with speed v such that its
velocity vector moves a distance d from the axis of rotation.
What is the angular speed ωF of the student-stool system
after they throw the ball ?
M
v
ωF
I
d
I
Top view: before
after
Physics 207: Lecture 17, Pg 25
Example: Throwing ball from stool
What is the angular speed ωF of the student-stool
system after they throw the ball ?
Process: (1) Define system (2) Identify Conditions
(1) System: student, stool and ball (No Ext. torque, L is
constant)
(2) Momentum is conserved
Linit = 0 = Lfinal = -m v d + I ωf
M
v
ωF
I
d
I
Top view: before
after
Physics 207: Lecture 17, Pg 26
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Physics 207 – Lecture 17
Angular Momentum as a Fundamental Quantity
The concept of angular momentum is also valid on a
submicroscopic scale
Angular momentum has been used in the
development of modern theories of atomic, molecular
and nuclear physics
In these systems, the angular momentum has been
found to be a fundamental quantity
Fundamental here means that it is an intrinsic
property of these objects
Physics 207: Lecture 17, Pg 27
Lecture 17
Assignment:
HW7 due March 17th
Thursday: Review session
Physics 207: Lecture 17, Pg 28
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