Transcript No Slide Title

Chapter 9
Rotations of Rigid Bodies
Up to this point when studying the motion of objects we have
made the (implicit) assumption that these are “point objects”
i.e. all the mass is concentrated at one point.
In chapter 9 we will drop this assumption and study the
rotation of rigid bodies. These are objects that do not change
volume or shape. All that is needed to describe the rotations
of rigid bodies is Newton’s laws of motion. From these we
will develop the equations that describe rotational motion
The following new concepts will be introduced:
Moment of Inertia ( I )
Torque (  )
Angular Momentum ( L )
(9-1)
Cartesian coordinates: (x, y)
y
Polar coordinates: (r, )
How to convert Cartesian to polar
coordinates and vice versa
P
r
y

O
x
x
(x, y)
A


(r, )
From triangle OAP we have:
x  r cos  , y  r sin 
y
2
2
r x y
, tan  
x
(9-2)
(9-3)
O
Consider the rotation of a flat rigid
body (the hand of a clock in the
picture) in the xy-plane about the
origin O which is fixed. Any point
P of the rigid point (the tip of the
hand in the picture) remains at a
constant distance from O.
All we need to describe the motion of P (and thus the motion
of the rigid body) is the angle  between the position vector r
of point P and the x-axis. We express the angle  as function
of time (t)
Note: Point O is the only point of the rotating body that does
not move
Rotation
axis
r
In a three dimensional rigid
body that rotates about a fixed
axis all points on the rotation
axis do not move
Any point P on the rigid body rotates on a plane that is
perpendicular to the rotation axis, on a circular orbit of
fixed radius r
(9-4)
t + t

t
 av 
 (t  t )   (t )
t
 d
  lim

t
dt
t  o


t
Consider a point P
on a pot that is
rotating on a
potter’s wheel
(fig.a). The
position of P is
described by the
angle 
Average angular velocity
Instantaneous angular velocity
Units: rad/s
(9-5)
The angular velocity vector
The vector  is defined as follows:
1. Magnitude

d
dt
2. Direction and sense of 
Direction: The axis of  coincides
with the rotation axis
Sense: Curl the fingers of the right
hand in the direction of rotation.
The thumb points along 
(9-6)
Special case: Rotation with constant angular velocity 
d

dt
t

t
t
 d    dt 
0
d   dt
0

 
Integrate both sides from 0 to t
  t 0
t
  - o   t

0
  o  t
This equation is the analog in rotational motion of the equation
that describes motion along the x-axis with constant velocity:
x(t) = xo + vt
Note: In this chapter and the next we will take advantage of
the similarity in form between equations that describe motion
along a line and rotational motion.
(9-7)
(9-8)
(t) = o + t
eqs.1
Rotational motion with
constant angular velocity
Period T is the time required to
complete one revolution
( = 2)
Frequency f is the number of revolutions per second
1
f 
From the definition 
T
Relationship between f and  : In eqs. 1 we set o = 0
and  = 2  2 = T = /f 
  2 f
t + t In general the angular velocity is not
t
 av 
 (t  t )   (t )
t
constant but it changes with time.
For this reason we introduce the
notion of angular acceleration to
describe the rate at which  changes
with t


t
 d  d 2
  lim

 2
t
dt
dt
t  0
Average angular acceleration
Units: rad/s2
Instantaneous angular
acceleration
(9-9)
Motion with constant angular acceleration 
d

dt
t
 d   dt
t
 d    dt 
0
Integrate both sides from 0 to t
 0   t 0

(9-10)
t
t
   o   t 
0
Compare with: v  vo  at
  o   t
d

dt
 d   dt
Integrate both sides from 0 to t
t
 t 
0 d  0  dt   o 0 dt    0 t dt    0  o t 0    2 
0
2
2

t
t
   o  ot 
   o  ot 

2
2
t
t
t
t
t
at 2
Compare with: x  xo  vo t 
2
t
2
Non-uniform acceleration
(9-11)
In general angular acceleration 
is not constant
r
C
In this most general case we
decompose the vector a into two
components:
Rotation axis
1. ac (centripetal acceleration) that points towards the
center C
2. at (tangential acceleration) that points along the tangent
v2
dv d (r )
d
2
ac    r , at 

r
 r
r
dt
dt
dt
at  r
(9-12)
Ri
y
vi
Ri
mi
x
O
Rotational kinetic energy of
a rigid body that rotates about
an axis with angular velocity
. Divide the object into N
elements with masses m1,
m2, …, mN
Each mass element mi moves on a
circle of radius Ri with speed
vi =  Ri. The kinetic energy of mi
is given by:
mi vi2 mi 2 Ri2
Ki 

2
2
(9-13)
Ri
The kinetic energy of mi is:
mi 2 Ri2
K i 
2
The total rotational kinetic energy K
is the sum of all the kinetic energies
K  K1  K 2  ...  K N
K
vi
mi
x
O
2
2
2

m
R


m
R

...


m
R
 11
1 2
1 N 
2
The sum in the brackets is known as
y
Ri
2
the "moment of inertia"
or "rotational inertia"
(symbol I) of the object
I 2
K
2
mv 2
K
in linear motion
2
Moment of Inertia I
(9-14)
I  m1 R12  m2 R22  ...  mN RN2
Ri
Recipe for the determination
of I for a given object
1. Divide the object into N elements with masses
m1, m2, …, mN
2. The contribution of each element Ii = miRi2
3. Sum all the terms I = I1 + I2 + …. + IN
4. Take the limit as N  
I   R 2 dm
(9-15)
I   R 2 dm
The moment of inertia of an object depends on:
a. The mass and shape of the object, and
b. the position of the rotation axis
A
x
x’
Moment of inertia of a rod of
length L and mass M
Linear mass density  = M/L
Divide the rod into elements of
length dx and mass dm = dx
Consider a rotation axis through the center of mass O
dI  x dm   x dx  I CM
2
I CM
2
L/2
x 
   x dx    
 3 L / 2
L / 2
L/2
3
2
 L3
( L) L2 ML2



12
12
12
L
For a rotation axis through the end point A we have: I    x 2 dx 
0
L
 x 
 L3 ( L) L2 ML2
I   


3
3
3
 3 0
3
(9-16)
Parallel axis theorem
I = ICM + md2
d is the distance between the
rotation axis and the parallel
axis that passes through the
center of mass
Moment of inertia I
about any axis of an
object of mass m
=
Moment of Inertia ICM
about a parallel axis
that passes through the
center of mass
+ md2
(9-17)
Torque  of a force acting on a point
P
r

F
r
O
(9-18)
Associated with any force F we can
define a new vector  , known as “the
torque of F” , which plays an
important role in rotational dynamics
Magnitude of 
 = Fr r = “arm of the moment”
r = rsin 
 = Frsin
Direction of 
If the force tends to rotate the object on
which it acts the clockwise (CW) direction (as in the picture)
the torque points into the plane of rotation and has a negative
sign. If on the other hand the rotation is counterclockwise
(CCW) the torque points out of the plane and is positive
(9-19)
The equivalent of the “second law”
in rotational dynamics
Consider the object shown in the
figure which can rotate about a
vertical axis. We divide the object
into element of masses
m1, m2, … mN On each of these
elements we apply a force
F1, F2, ... FN
For simplicity we assume that these forces are perpendicular
to the object. Consider one element of mass mi The force Fi
results in a torque i = Firi , Fi = miai = miri  i = miri2
Total torque  = 1+ 2 + …+ N = (m1r12 +…+ mNrN2) = I
Thus:
  I
Compare this with: F = ma
Example (9-6) page 250. A bucket of mass m = 12 kg is
connected via a rope to a cylindrical flywheel of mass M = 88
kg and radius R = 0.5 m. The bucket is dropped and the
flywheel is allowed to spin. Determine the angular velocity of
the flywheel after the bucket has fallen for 5 s.
(9-20)
System  bucket
R
a
Fynet  T  mg  ma 
T  mg  ma (eqs.1)
System  bucket
  TR (eqs.2)
We substiture T from eqs.1 into eqs.2
  (mg  ma) R (eqs.3)
a MR 2 a MRa
  I  I 

(eqs.4)
R
2 R
2
Compare eqs.3 and eqs.4 
MRa
 (mg - ma ) R 
2
m
12
ag
 9.8
 2.1 m/s 2
M
12  88 / 2
m
2
(9-21)
R
a
The acceleration of the bucket is
also the acceleration of the rope and
therefore the tangential acceleration
at the rim of the flywheel
a  2.1 m/s 2
a 2.1
 
 4.2 rad/s 2
R 0.5
  o   t  4.2  5  21 rad/s
(9-22)
Torque of the gravitational force
R
The torque of the gravitational force
on a rigid body Fg is equal to the
torque of Fg acting at the center of
mass of the object
Thus  = mgR
R = Rsin
 = mgRsin
(9-23)

Analogies between linear and rotational motion
Linear Motion
Rotational Motion
x (distance)
 (rotation angle)
v (velocity)
 (angular velocity)
a (acceleration)
 (angular acceleration)
m (mass)
I (moment of inertia)
F (force)
 (torque)
p = (linear momentum)
p = mv
L (angular momentum)
L=I
(9-24)
Angular momentum L
L
By analogy to the definition of the
linear momentum p  mv
we define the vector of angular
momentum L as follows:
L  I
units: kg.m2/s
Since I > 0  L is parallel to 
d L d (I )
d

I
 I  
dt
dt
dt
dp
 F for linear motion
dt

dL

dt
(9-25)
Conservation of angular momentum
L
dL

dt
dL
If   0 
0
dt
 L is a constant vector
If the torque  of all external forces on
a rigid body is zero then the angular
momentum L does not change (it is
conserved
L  0
(9-26)
Rolling
When a wheel rolls on flat ground it
executes two types of motion
a. All points on the rim rotate about the center of mass with
angular velocity 
b. The center of mass and all other points on the wheel move
with velocity vcm
Note 1: The total velocity of any point on the wheel is the
vector some of the velocities due to these two motions
Note 2:  and vcm are connected
(9-27)
Rolling
vcm
vcm
L1
L1 is the distance traveled in one revolution of the wheel
Thus: L1  vcmT
L1 is given by the equation:
L1 =2 R = circumference of the rim of the wheel
 vcmT  2 R
 vcm
vcm   R
2

R  R
T
(9-28)
Kinetic energy of a rolling object
vcm
vcm
P
I P 2
K
I P is the wheel's moment of inertia about point P
2
I P  I cm  mR 2 (parallel axis theorem) 
2
2
2
2
I

I

mv
m
(
R

)
K  ( I cm  mR 2 )
 cm

 cm
 cm
2
2
2
2
2
The first term in the expression for K is the rotational energy
2
about the center of mass. The second term is the kinetic energy
due to the translation of the center of mass.
(9-29)
Find the acceleration of the center of mass of the rolling
object (moment of inertia about the center of mass Icm ,
mass m, radius R) as it rolls down an inclined plane of
angle 
mg sin 
a
m  I / R2
(9-30)
(9-31)
Center of mass motion: Fxnet  mg sin  - f  ma
(eqs.1)
Rotation about the center of mass:   fR also   I 
Ia
Ia
fR  I 
 f  2 (eqs.2)
R
R
Ia
Substitute f from eqs.1 in eqs.2  mg sin   ma
2
R
mg sin 
g sin 
 a

Note: a does not depend on m
2
2
m  I / R 1  I / mR
(9-32)
Rolling object 1 = cylinder
Rolling object 2 = hoop
mg sin 
a
m  I / R2
Cylinder
mR 2
I1 
2
mg sin 
a1 
m  I1 / R 2
mg sin 
a1 
m  m/2
2 g sin 
a1 
 (0.67) g sin 
3
Hoop
I 2  mR 2
mg sin 
a2 
m  I2 / R2
mg sin 
a2 
mm
g sin 
a2 
 (0.5) g sin 
2