Transcript Ch33 - Wells College

Chapter 10: Rotation of a Rigid Body
about a Fixed Axis
Chapter 10 Goals:
• to reintroduce ourselves to the rotational analogues of position,
velocity and acceleration
• to understand the limitations inherent in the fixed axis assumption
• to remind ourselves of the connections between the rotation of a
body, and the translational kinematics of a point of the body
• to define the kinetic energy of rotation and thereby introduce the
rotational analogue of the mass: the moment of inertia
• to utilize the parallel axis theorem
• to define the (axis-dependent) torque associated with a force and
to understand the relationship between torque and angular
acceleration
• to appreciate the kinematics of rolling motion
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Rotational Kinematics about a Fixed Axis
• here is a body which we want
to be turning about an axis
• rotation axis is a line
“normal” to page through
the origin O
• we overlay an xy
coordinate system
• a spot (fiducial point) on
the body moves from A to B
•
the line from O to the fiducial point will move
through an angular displacement Dq : = qf – qi in
a time Dt = tf – ti
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Rotational Kinematics about a Fixed Axis
3 Warnings
(1) if the axis’s direction is fixed in space but the axis
moves (e.g. rolling motion), then the body is
translating as well as rotating!!
(2) in fact, if the axis is fixed, only points ON the axis
are not translating
(3) if the axis’s direction is not fixed in space then one
cannot even DEFINE Dq as a simple number!! It’s
even more strange than a vector!! It is a tensor and you
need nine numbers in 3d to express it!!
r
arc length
s

• The radian: q :
s
circle radius
r
• To convert: p rad = 180 °
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q
Angular Velocity and Acceleration
• given Dq : = qf – qi in a time Dt = tf – ti
• define average angular velocity:
 avg :
angular
displaceme
nt

time interval
Dq
Dt
  
rad
 sec
1
sec
• of course, this is the slope of the line on a
graph of q (t) that connects (qi , ti) to (qf , tf )
• now we define instantaneous angular velocity:
 : lim
Dt  0
Dq
Dt
• Angular
:
dq
 slope of the tangent t o q (t) graph
dt
 avg :
D
and
 :
d
d q
2

2
Dt
dt
dt
accelerations:
• obviously, these are slopes of the (t) graph
• units: [||] = rad/sec2 = sec–2
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The Analogy Is Pretty Darn Good
• with a fixed axis, it is as if it is a 1d kinematics problem
• quantities can be positive or negative, of course
• a useful alternative to radians as the angle unit is the
cycle, or revolution: 2p radians = 1 revolution
• 1 cycle per second := 1 Hz [Hertz]
• 1 Hz = 60 revolutions per minute = 60 rpm
• angular speed := magnitude of angular velocity = ||
• with the period T as the time for one cycle,  =2p/T
The Case of Constant Angular Acceleration
• needless to say, we recover the ‘big five’:
  constant
  i  t
q  q i   it  2  t
1
2
q  qi 
   i  2 q  q i 
2
2
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1
2
 i   t
Connection between Linear and Angular Velocities for
a Fiducial Point on a Turning Object
• If a point is a distance r from the
v
2d axis line, the point moves in a

circle of that radius
r
• and since the object rotates at (t),
the average speed of the point is
v avg 
circumfere
nce
time to go around once

2p r
T
  avg r
• notice how the rad unit sort of disappears…
• this is also true on an instantaneous basis, even if  is a
function of time: v(t) = (t) r
• velocity vector is tangent to the circular path in space
• we can say that
v   r tˆ
or
v t   r and v r  0
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Connection between Linear and Angular Accelerations
for a Fiducial Point on a Turning Object
• velocity vector is tangent to the path
v

• how does it change with time? In two
at
ways!! Its length may be changing,
and its direction is obviously changing
too!!
• the acceleration of that point has both
ar
components, tangential and radial:

length of v changes
 tangential
direction
 changing
speed (  angular
dv
accelerati on is a t :
of v changes

d  r 
r
accelerati on)
d
dt
dt
dt
 centripeta l accelerati on
 radial accelerati on is a r    r
2
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:  r
Kinetic Energy of Rotation
• here is a rotating body, turning about a fixed
axis (along z) at angular speed 
• the axis does not have to pass through the
body, nor must it be along z!!
• if the axis passes through the CM, that’s fun
• think of the body as being made up of a
collection of point masses.. a system
• ri is the distance of mi from the AXIS, not
from the origin (‘tho figure implies
otherwise)
• vi is the speed of mi so vi =  ri
rotational
kinetic
energy is K R 

1
2
m i vi 
2

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1
2
m i  ri 
2
2
2 2

m
r


2 
i i 

1
Moment of Inertia
rotational
kinetic
that summation
so we can write
•
•
•
•
energy is K R 

1
2
m i vi 
2

1
2
2
is called the moment of inertia ... I :
KR 
1
2
I  in the form like
2
1
2
2 2

m
r


2 
i i 

2
m r
i i
m i  ri 
mv
2

1
2
for a collection of point masses one can do the sum pretty easily
remember: r is the distance from the 1d axis, not from an origin!!
if the system is an extended body (an object) then we have to think!
break the body up into a very large number of very tiny masslets
I :

2
m r 
i i
 dm r where the big challenge
2
is to write dm
body
• To express dm we need to define some kind of ‘mass density’
which may be 1d, 2d or 3d
• examples will clarify the ideas…
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The Parallel-Axis Theorem
• imagine a second axis, parallel to the first, through the CM,
separated by a ‘distance vector’ D (a ‘line-to-parallel-line’distance)
• let the vectors {ri} be expressed as ri = r’i + D where r’I is
measured from the CM axis to mi
 I 

2
m r 
i i

2
2
m r ' i  D   r ' i  D    m r '  D  m  2 D   m r '1
i
i i
i
i
 I CM  MD
2
 2 D   the first moment
 I CM  MD
2
 0  the parallel - axis theorem
of the mass as seen from CM
• one implication is that the CM moment of inertia is the
smallest possible one for a given axis direction
• Another cool fact is that we can ‘split up’ KR:
KR 
1
2
I
2

1 

2 
I CM  MD
2 


but here v CM   D  K R 
 
2
1
2
1
2
I CM  
2
I CM  
2
1
2
2
1
2
Mv CM
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MD 
2
2

How to Calculate a Moment of Inertia
Find the moment of inertia for a stick of length L and mass M for an axis normal to
the stick and through the CM
Solution
Place the stick on the x-axis with origin at CM
The stick has a linear mass density l = M/L and
so a thin ‘slice’ of stick of infinitesimal thickness
dx has an infinitesimal mass dm = l dx
The thin slice’s distance from the axis is x, so
L/2
I CM :
 dm r 
 l x dx 
2
body
2
 L /2
M
L
L/2

x dx 
2
Mx
L/2
3

3L
 L /2
2
3
M
 
L
x
dx
O
2

8
1
12
ML
2
 L /2
Suppose the axis is normal to the stick but at one end… Then D = L/2 and
by P.A.T we get
I  I CM  MD
2

1
12
ML  M
2

L 2
2
 ML
2

1
12
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
1
4

1
3
ML
2
Example of calculation of the moment of inertia
Find the moment of inertia for a hoop of mass M and radius R about an axis
through its c.o.m. and along the hoop’s axis
 r dm 
I CM 

2
body
R dm  R
2
body
2
 dm  MR
2
body
• This one is trivial since all of the matter is at the same distance from the axis:
hoops are special and have the greatest possible ICM!!
• Sometimes, even though the body is (say) 3d, you can get away with a 1d (or
2d) integral, by exploiting simpler results
Example of calculation of the moment of inertia
Find the moment of inertia for a solid cylinder of mass M and
radius R and length L, about an axis through its CM and along
the cylinder’s symmetry axis. Here, we take dm to be a very thin
cylindrical shell of radius r and thickness dr and length L.
The volume mass density of the stuff is
r = Mass/Volume = M/(pR2L), and the infinitesimal mass has a
volume 2prL so dm = 2prLr dr which becomes dm = 2Mr dr/R2
I CM :
 dm r 
2
body
2M
R
2
R
 r dr 
3
0
2 Mr
4R
4
R

2
1
2
MR
2
0
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New Concepts: the Line of Action, and the Moment
Arm, of a Vector
• an origin-dependent quantity: given a vector, we
have the line of action of the vector in space
• the distance d from the origin to the
line of action of the vector quantity is
called the moment arm of the vector
• we are here applying the notion to the
force
• we will see soon enough how to
apply it to the momentum
• it might be ‘obvious’ that if you
want to turn something about an
axis through O, you’ll succeed
better for a given force F if you
maximize the associated d.
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We Want to Maximize the ‘Force of Twist’ by Using the
Wrench as Wisely as Possible
• r is the position vector from the origin O to the point of
application of the force F
• the line of action of F is also shown
• any component of the force ║ to r
will do nothing to help turn the nut
• this component is F cos f, where
f is the angle between F and r tailtail
• the component of the force ┴ to r
is what really does the job
• this component is F sin f
• notice that the moment arm is d,
where d = r sin f
• conclusion: the ‘force of twist’ is force x the moment arm,
or force x distance x sin f
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A Definition: the Cross Product of Two Vectors
• the vector product of a vector with a vector: the
cross(vector(outer(wedge))) product A × B
• the cross product of two vectors is itself a vector!!
magnitude
direction
of A  B is A  B  A B sin q
of A  B is normal to plane of A and B
• which normal? Use Right Hand Rule: swing A into B through
the smaller tail-tail angle, according to how your right hand
operates; its thumb points along A x B
•so we have a completely new ‘thing’, in a completely new
direction, derived from A and B, which couldn’t be captured by
the dot product
• A x B = sort of ‘signed’ area of parallelogram spanned by A
and B; a measure of their ‘perpendicularness’
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More about the cross product of two vectors
A
|A|sin q
q
B
• magnitude of cross product is
magnitude of B times ‘moment’ of
A [ |A| sin q ] from perpendicular
projection of A along line of B
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The Definition of the Torque of a Force
F
d = |r| sin q
r
r
d = |r| sin q
q
tip-to-tail,
where F lives
q
F
tail-to-tail,
in standard position
• the moment arm is d = |r| sin q, with units of length
• t = r × F [|t|] = N-m [NOT Joules! USA: foot-pounds]
• called the torque ; greek letter tau
• also called the moment of the force
• note: the moment of the force is perpendicular to the force
• the magnitude is t = (force)(moment arm) = Fd
• direction is given by R.H.R.
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O
Net Torque about an Origin for a Fixed Axis
• the net torque about an origin is the moment of the net force
• tnet = S ri × Fi in an obvious defintion
• what we care about here is the component of the vector net
torque along the fixed axis, which is associated with the tangential
component of the net force (the radial component of the forces
{Fi} contribute no torque since they are parallel to ri)
• for now focus on a single point mass ...
• so since at = r and Fnet,t = mat, (recall: r = distance from axis to
point that is accelerating tangentially), we get for the component of
the vector torque along the axis as follows:
t net  rF net , t  rma t  r m  : I 
2
• looks for all the world like Fnet = ma, doesn’t it?
• if the axis is fixed (or not changing direction, more generally),
the analogy is pretty darned good..
• so net torques cause angular accelerations about a fixed axiss!!
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What if the Net Torque
about an Origin for a
Fixed Axis is Zero?
F
13 mm
F
• these are the forces applied to a
bolt head by a wrench
• note that the net force is zero: a
so-called force couple
• there are other forces applied to
the bolt by the thing that the bolt is
being tightened into
• suppose the specification says to
‘torque’ the 13 mm bolt to 70 N-m
• the 13 mm bolt has a ‘face dimension’ of 13mm(1/√3) ≈ 7.5 mm
• put the origin at the center of the bolt’s face
• each force therefore has a moment arm of half of that: 3.75 mm
• for the torque to be 70 N-m, we get
t
70 N - m
t  2 Fs  F 

 9300 N
-3
2s
(2)(3.75 x 10 m )
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Example
A heavy bank vault door of mass M requires a
force applied at its edge of 200 N to accelerate
it so that it slams shut in 4 sec after turning
F
through an angle of 90°. The door is w = 1.5
r
meters wide and h = 2.5 meters tall.
a) Find the needed angular acceleration
b) Find the moment of inertia of the door about an axis through the hinge in
terms of the unknown mass
Solution
c) Find that unknown mass
hint: put the axis at the hinge line!! [Why??]
Note that F ┴ r
2q
p rad
1
1
2
2
q  q i   it  2  t  q  0  0  2  t    2 
2
F
t
P.A.T.  I  I C M  MD
•
•
•
•
2

1
12
Mw
2
M

w 2
2
16 s

1
3
Mw
2
Now, what are the forces on the door as seen from the top?
F is the push on the edge, with w = moment arm from the axis
There is also a horizontal force at the hinge! It turns out to not be equal to
F (because the CM of the door has to accelerate too)
But the torque associated with the hinge force is zero!!
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Solution Continued
t ne t  I   I 
 M 
•
•
•
•
•
•
3 F w/
w
2/

t ne t


Fw sin 90



p
16  rad/s
1.5
3
Mw
2
2
3 ( 200 N)
2
1
m

( 600 N)(16 s )
p rad 1.5 m 
 2040 kg
can we find the size of the other force, applied at the hinge?
yes! The CM accelerates at at =  r, where r = w/2
 at = (p/16)(rad/s2)(.75 m) = .147 m/s2
therefore, since Fnet = MaCM = 300 N = 200 N + Fhinge
amazingly, the hinge pushes on the door in the SAME direction as F, with
a force of 100 N (!!)
This actually makes some sense since the attempt to close the door in part
makes the door want to rotate about its CM which means the hinge edge
of the door pushes against the hinges, so by N3 the hinges push back on
the door
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Examples that Synthesize All This Stuff: 10.8&10.11
Fp
r
F
Fg
A bar of length L and mass M is initially
horizontal and supported at its left end by a
frictionless pivot. The end of the bar is released
from rest. Here is the FBD.
a)
Find the initial angular acceleration of the bar, and the initial linear
acceleration of the CM, and the initial angular acceleration of the right tip
of the bar.
b) When the bar is vertical, what is its angular velocity, and what is the
velocity of its CM?
Solution to (a)
Put the origin/axis at the pivot, so the pivot force has a zero moment arm
The weight force’s moment arm is L/2
t net  I   mg
 a CM  
L
2

L
 
2
3
4
mgL
2I
g
but

m g L/

2 mL
2
3


a TIP   L 
3g
2L
3
2
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g !!!
Solution to (b)
Bar is initially motionless and its height is zero, so
initial energy is Emech = U + K = 0
vCM
At bottom, bar is turning at  and its CM has speed vCM, and its
CM height is yCM = − L/2
E mech  
MgL
but v CM 

2
1
2
L
Mv CM 
2
2
2 v CM
or  
2
 E mech  
1
L
MgL
2

2
3
I CM   
2
MgL
 E mech  
Mv CM  0  v CM 
2
2

2
1
2
MgL
I PIVOT 

MgL
2
3
4
2
gL  v CM 
2
11

2  2 v
  ML   CM 
23
 L 
3
gL
2
A third question: what point (distance d from the pivot) on the
bar is moving at √(2gL), which is the speed of a point mass that
has fallen a distance L?
so since
 
2 v CM
  
3g / L
L
therefore
we must have
2 gL  d
3g / L  d 
2
3
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L  . 816 L
2
Example: the Atwood Machine
I
m1
Solution Based on Energy Methods
m2



• Assume system starts from rest and denote the heights of
rising/falling masses to be zero initially
• obviously, the pulley does not rise or fall
• so initially Emech,i = Ki + Ui = 0
• now release masses.. 1 drops and 2 climbs a distance h,
while system picks up linear/angular speed too
mechanical
energy is now
E mech,
f
  m 2  m 1  gh 
1
2
 m 1  m 2 v 2 
with pulley mass M and radius R , we have I   MR
mechanical
energy conservati on  0   m 2  m 1  gh 
1
2
1
• one can also analyze using Newtonian methods
• put an origin at the axis of the pulley and note  = a/R
• there are two tensions, T1 and T2 now, as well
• three FBDs  three equations relating T1 and T2 and a
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I
2
2
and   v / R
M 2
2
 m 1  m 2 v
 2  m 1  m 2  gh  2
m 1  m 2      m 2  m 1  gh  v  


2
 m1  m 2   M 
v
2
2
1

v
2
  m 1  m 2  gh 
a  

 m1  m 2   M 
 2m2  M 
T1  m 1 g 

 m1  m 2   M 
 2 m1   M 
T2  m 2 g 

 m1  m 2   M 
Work Considerations in Rotation I
• we can work by analogy, almost.. work = force × distance so this
smells like work = torque × angular displacement
• more generally, dW = F·ds where ds is an infinitesimal
displacement
• assume the body is pivoting about a fixed axis, so ds = r dq
[where r is distance to point of application of F from axis, again,
and dq is an infinitesimal rotation angle of the body.

ds
r
F
dq
• let the angle between F
and ds (tail-tail) be 
dW = F cos  ds by the
usual usage of the dot
product
• since ds ┴ r the angle
between r and F and r is
f = 90° − 
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Work Considerations in Rotation II
 F
ds
r
90° − 
dq
• but sin(90° − ) = cos 
• so dW = F ds sin(90° − ) =
F r sin(90° − ) dq
• so dW = F (moment arm) dq
•  dW = t dq
• the analogy looks good and
we can push it much further
 W   t d q but only for a fixed axis! !
• if the torque is the net external torque then we can say more!
• so, summing over the system, tnet = S text = I   I d/dt
 W net   t net d q 

I
d
dt
dq  I  d
dq
1
2 
 I   d   D  I   K
dt
2

f
 Ki
• thus we have the rotational analog of the work-energy theorem
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Kinematics of rolling motion I
• Any object of circular cross-section
(t)
• CM is at the center for smooth rolling
• no sliding (so if  ≠ 0, need static friction)
• let v = CM speed
• in one rotation, CM and point of contact (p.o.c.) move a distance
of one circumference 2pR
• time is rotation period: T = distance/speed = 2pR/v
•  v = 2pR/T = R: same formula as for linear speed as related to
angular speed for a turning object, where R is distance from axis!!
• p.o.c appears to move… but that bit of roller can’t slide so instead
we realize that there is no relative motion between the two surfaces
(if there were, there’d be a lot of grinding and smoke!!)
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
(t)
Kinematics of rolling motion II
• CM moves at speed v = R, as established
• p.o.c. cannot move unless slippage: vbot = 0
• thus p.o.c. acts as an instantaneous axis of rotation, that
slides along at speed v
• top of roller is distance 2R from the instantaneous axis, so it
moves forward at speed 2v (!!)
• rotation (left)
+
=
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
• translation
(mid)
• both t+r (right)
(t)
Kinematics of rolling motion III
• c.o.m. moves forward at v in steady motion
• a dot on the edge moves on a path cycloid
• for an instant at the very bottom, it stops
• at the top, it is moving forward at 2v
• top of roller is distance 2R from the instantaneous axis, so it
moves forward at speed 2v (!!)
• what does a flanged wheel do??
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Does it matter where the origin is placed?
• a trivial example: a single force F acts on a body
• let t and t’ be referred to origins O and O’
r’
F
O’
r
τ  r  F  τ : t  ( F )( moment arm )  ( F )( b )  Fb
b
O
τ '  r ' F  τ ' : t '  ( F )( moment arm )  ( F )( 0 )  0
• clearly here the torque depends on the choice of origin
• But… suppose the net force on the body is
zero.. so there MUST be at least two forces
acting on the body if there is any torque at all
F1
• So take the case of a pair of forces that satisfy
F1 = − F2 , and let D be the displacement vector
from the point of application of F1 to that of F2
• The net force on the body is zero  aCM = 0
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
D
F2
What Controls whether Torque is Origin-independent
•let t1 and t2 be referred to origin O
τ net, ext  r1  F1  r 2  F 2
F1
r1
 τ 1  τ 2  r1  F1  r1  D     F1 
D
r2
F2
 r1  F1  r1  F1  D  F1
  D  F1 and that is origin - independen t! !
O
• The force couple seems to contain only one force, but the
other one is lurking in the pair’s torque
• the torque is independent of the origin   the net force
on the body is zero   the CM obeys N1 so body must
be rotating about the CM, while the CM may be in steady
translational motion (or may be stationary)
• if the net force on the body is not zero, then we know how
the CM moves, and we know how it rotates about the CM:
F net, ext 

Fi , ext 
dP
where
dt
P 
p
i
dL
where L   r  p

dt
Copyright © 2008 Pearson Education, Inc., publishing
as Pearson Addison-Wesley.
τ net, ext 
ri  Fi , ext 
i
i