Download Chapter 12: Rotation of Rigid Bodies

Chapter 12: Rotation of Rigid Bodies
Center of Mass
Moment of Inertia
Torque
Angular Momentum
Rolling
Statics
Translational
m
x
v  dx / dt
a  dv / dt
F  ma
p  mv
KE  1/ 2mv
Work  Fd
P  Fv
vs
Rotational
Connection
s    r
v  r
ac   r
2
at   r
2
  Fr
L  pr
I

  d / dt
  d  / dt
  I
L  I
KE  1/ 2 I 
Work  
P  
2
Center of Mass
The geometric ‘center’ or average location of the mass.
Rotational & Translational Motion
Objects rotate about their Center of Mass.
The Center of Mass Translates as if it were a point particle.
v CM
drCM

dt
Center of Mass
The Center of Mass Translates as if it were a point particle
and, if no external forces act on the system, momentum is
then conserved. This means:
EVEN if the bat EXPLODED into a thousand pieces, all the
pieces would move so that the momentum of the CM is
conserved – that is, the CM continues in the parabolic
trajectory!!!! THIS IS VERY VERY IMPORTANT!
System of Particles Center of Mass
• A projectile is fired into
the air and suddenly
explodes
• With no explosion, the
projectile would follow
the dotted line
• After the explosion, the
center of mass of the
fragments still follows the
dotted line, the same
parabolic path the
projectile would have
followed with no
explosion!
If no external forces
act on the system,
then the velocity of
the CM doesn’t
change!!
Center of Mass: Stability
If the Center of Mass is above the base
of support the object will be stable. If
not, it topples over.
Balance and Stability
This dancer balances en pointe by having her center
of mass directly over her toes, her base of support.
Slide 12-88
Center of Mass
The geometric ‘center’ or average location of the mass.
Extended Body:
rCM
1

r dm

M
System of Particles:
xCM 
m x
i i
i
M
CM Lecture Problem
1. Identical particles are placed at the 50-cm and 80-cm
marks on a meter stick of negligible mass. This rigid
body is then mounted so as to rotate freely about a
pivot at the 0-cm mark on the meter stick. A) What is
the CM of the system? B) What is the torque acting on
the system? C) What is the rotational inertia of the
system about the end? D) If this body is released from
rest in a horizontal position, what is the angular
acceleration at the release? E) What is the angular
speed of the meter stick as it swings through its lowest
position?
Prelab
A meter stick has a mass of 75.0 grams and has two masses
attached to it – 50.0 grams at the 20.0cm mark and 100.0
grams at the 75.0 cm mark.
(a) Find the center of mass of the system - that is, at what mark
on the meter stick should the fulcrum be placed so that the
system balances?
(b) A fulcrum is then placed at the CM. Sketch and label the
system.
(c) Show that the net torque about the cm is zero.
(d) Calculate the rotational inertia of the system about the CM
axis. Box answers, make it neat.
Center of Mass
The geometric ‘center’ or average location of the mass.
Extended Body:
rCM
1

r dm

M
Center of Mass of a Solid Object
Divide a solid object into
many small cells of mass m.
As m  0 and is replaced
by dm, the sums become
Before these can be integrated:
 dm must be replaced by
expressions using dx and dy.
 Integration limits must be established.
Slide 12-41
Example 12.2 The Center of Mass of a Rod
Slide 12-42
Example 12.2 The Center of Mass of a Rod
Slide 12-43
Example
Extended Body:
1
rCM 
r dm

M
You must generate an expression for the density and the mass differential, dm, from
geometry and by analyzing a strip of the sign. We assume the sign has uniform
density. If M is the total mass then the total volume and density is given by:
1
M
V  abt ,  
1
2
abt
2
Where a, b and t are the width, height and
thickness of the sign, respectively. Then the
mass element for the strip shown is:
M
2 My
dm  dV  ( ytdx ) 
ytdx 
dx
1
ab
abt
2
xCM
a
1
1
2My
2 b
2 x 
2

x
dm

x
dx

x
(
x
)
dx


a


2 


M
M
ab
ab a
a  3 0 3
0
a
3
Newton’s 2nd Law for Rotation
The net external torques acting on an object
around an axis is equal to the rotational
inertia times the angular acceleration.
  I
Force thing
Acceleration thing
Inertia thing
The rotational equation is limited to rotation about a single
principal axis, which in simple cases is an axis of symmetry.
Torque: Causes Rotations
  Fr sin   Fd
lever arm: d  r sin 
The moment arm, d, is
the perpendicular
distance from the axis
of rotation to a line
drawn along the
direction of the force
The horizontal component of F (F cos ) has
no tendency to produce a rotation
Torque: Causes Rotations
  Fr sin   Fd
The direction convention is:
Counterclockwise rotations are positive.
Clockwise rotations are negative.
Newton’s 1st Law for Rotation
If the sum of the torques is zero, the system
is in rotational equilibrium.
 boy  500 N 1.5m  750 Nm
  0
 girl  250 N  3m  750 Nm
Torque
Is there a difference in torque?
(Ignore the mass of the rope)
NO!
In either case, the lever arm is the same!
3m
What is it?
CM Lecture Problem
1. Identical particles are placed at the 50-cm and 80-cm
marks on a meter stick of negligible mass. This rigid
body is then mounted so as to rotate freely about a
pivot at the 0-cm mark on the meter stick. A) What is
the CM of the system? B) What is the torque acting
on the system? C) What is the rotational inertia of the
system about the end? D) If this body is released from
rest in a horizontal position, what is the angular
acceleration at the release? E) What is the angular
speed of the meter stick as it swings through its lowest
position?
Newton’s 2nd Law for Rotation
The net external torques acting on an object
around an axis is equal to the rotational
inertia times the angular acceleration.
  I
Force thing
Acceleration thing
Inertia thing
The rotational equation is limited to rotation about a single
principal axis, which in simple cases is an axis of symmetry.
Torque is a Vector!
=rxF
The direction is given by the right hand rule where the fingers
extend along r and fold into F. The Thumb gives the direction of .
The Vector Product
• The magnitude of C is
AB sin  and is equal to the
area of the parallelogram
formed by A and B
• The direction of C is
perpendicular to the plane
formed by A and B
• The best way to determine
this direction is to use the
right-hand rule
A  B   Ay Bz  Az By  ˆi   Ax Bz  Az Bx  ˆj   Ax By  Ay Bx  kˆ
COMPARE!
  r F
  Fr sin   Fd
Lrp
L  mvr sin 
CROSS PRODUCT
F and d must be
mutually perpendicular!
CROSS PRODUCT
L and p must be
mutually perpendicular!
DOT PRODUCT
W  F  d  Fd cos 
F and d must be
mutually PARALLEL!
A  B   Ay Bz  Az By  ˆi   Ax Bz  Az Bx  ˆj   Ax By  Ay Bx  kˆ
Two vectors lying in the xy plane are
given by the equations A = 5i + 2j and
B = 2i – 3j. The value of AxB is
a. 19k
b. –11k
c. –19k
d. 11k
e. 10i – j
CM Lecture Problem
1. Identical particles are placed at the 50-cm and 80-cm
marks on a meter stick of negligible mass. This rigid
body is then mounted so as to rotate freely about a
pivot at the 0-cm mark on the meter stick. A) What is
the CM of the system? B) What is the torque acting
on the system? C) What is the rotational inertia of the
system about the end? D) If this body is released from
rest in a horizontal position, what is the angular
acceleration at the release? E) What is the angular
speed of the meter stick as it swings through its lowest
position?
Net external torques

Find the Net Torque
  Fr sin   Fd
  F d
1 1
 F2 d2
 (20 N )(.5m)  (35N )(1.10m sin 60)
OR
 (20 N )(.5m)  (35N cos30)(1.10m)
 23.3Nm CCW
F and d must be
mutually
perpendicular!