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)
y O
r x P y A x Cartesian
coordinates: (x, y)
Polar
coordinates: (r, ) How to convert Cartesian to polar coordinates and vice versa )
From triangle OAP we have:
x
r
y
r
sin
r
x
2
y
2
, tan
y 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 of point P and the x-axis. We express the angle of time (t) between the position vector
r
as function
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
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
av
(
t
lim
t o t
t
d
dt
t
Average angular velocity Instantaneous angular velocity
Units: rad/s (9-5)
The angular velocity vector
The vector
1. Magnitude
is defined as follows:
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
d
dt
Integrate both sides from 0 to t
t
0
d
t
0
dt
t
o
0
t
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) = x o + 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 From the definition Relationship between f and and = 2 2 = T =
f
/f 1
T
: In eqs. 1 we set o 2
f
= 0
t +
t t
In general the angular velocity is
not
constant but it changes with time. For this reason we introduce the notion of
angular acceleration
describe the rate at which to changes with t
av
(
t t
t
lim
t
0
d
dt
d
2
dt
2
t
Average angular acceleration
Units: rad/s 2
Instantaneous angular acceleration
(9-9)
Motion with
constant
angular acceleration
d
dt
d dt
Integ rate both sides (9-10) from 0 to t t 0 d
t
0
dt
t
0
t
0
o
t
d
dt o
t d
Compar e wi th
dt
:
v
o at
Integrate both s ides f o 0 to t t 0 d
o
t
0
dt
o t
0
dt
t
0
o t
t
2 2
t
0
o
t
0 Compar e with:
x
x o
v t o
a t
2 2
o
o t
t
2 2
t
2
t
0
(9-11)
Non-uniform acceleration
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.
a
c (centripetal acceleration) that points towards the center C
2.
a
t (tangential acceleration) that points along the tangent
a c
v
2
r
2
r
,
a t
dv dt
dt
)
r d
dt
r
a t
r
O
(9-12)
R i y R i v i
m i x Rotational kinetic energy
of a rigid body that rotates about an axis with angular velocity . Divide the object into N elements with masses m 1 , m 2 , …, m N Each mass element m i moves on a circle of radius R i v i = R i with speed . The kinetic energy of m i is given by:
K i
m v i i
2 2
m i
2
R i
2 2
O R i y R i v i
m i x
(9-13) i
K i
m i
2
R i
2 2 The total rotational kinetic energy K is the sum of all the kinetic energies
K K
K
1 2 2
K
2
m R
1 1 2
K N
m R
1 2 2
m R
1
N
2 The sum in the brackets is known a s the "moment of inertia" or "rotational inertia" (symbol I) of the object
K
I
2 2
K
mv
2 2 in linear motion
R i
(9-14)
Moment of Inertia I
I
m R
1 1 2
m R
2 2 2
m R N N
2 Recipe for the determination of I for a given object
1.
Divide the object into N elements with masses m 1 , m 2 , …, m N
2.
The contribution of each element I i = m i R i 2
3.
Sum all the terms I = I 1 + I 2 + …. + I N
4.
Take the limit as N
I
2
R dm
(9-15)
I
2
R 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
2
x dm
2
x dx
I CM
L L
/ 2 / 2 2
x dx
x
3 3
L
L
/ 2 / 2
I CM
L
3 12 ( ) 2 12
ML
2 12 For a rotation axis through the end point A we have:
I
0
L
I
x
3 3
L
0
L
3 3 ( ) 2 3
ML
2 3 (9-16)
Parallel axis theorem
I = I
CM
+ md
2
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 I CM about a
parallel
axis that passes through the center of mass
+
md 2 (9-17)
Torque
of a force acting on a point
r
P F
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
O r
Magnitude of
r = Fr = rsin r = “arm of the moment” = Frsin (9-18)
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 m 1 , m 2 , … m N On each of these elements we apply a force
F
1 ,
F
2 , ...
F
N For simplicity we assume that these forces are perpendicular to the object. Consider one element of mass m i results in a torque i = F i r i , F i Total torque = 1 + 2 + …+ N = m i a i = m i r i The force F i = m i r i 2 = (m 1 r 1 2 +…+ m N r N 2 ) = I i Thus:
I
Compare this with:
F
= m
a
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)
R a
System bucket
F ynet T
mg mg
ma
ma
(eq s.1) System b ucke t
TR
(eqs.2) We substiture T from eqs.1 into eq s.2
(
mg
ma R
(eqs.3)
I
I a R
MR a
2 2
R
MRa
(eqs.4
) 2 Compare eqs.3 and eqs.4
MRa
2 (
a
m g m
M
2 9.8
12 2.1 m/s 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 R o
2.1
0.5
t
4.2 rad/s 2 21 rad/s (9-22)
R
(9-23)
Torque of the gravitational force
The torque of the gravitational force on a rigid body
F
g is equal to the torque of
F
g acting
mass
of the object
at the center of
Thus = mgR R = Rsin = mgRsin
Analogies between linear and rotational motion Linear Motion Rotational Motion
x (distance) v (velocity) a (acceleration) m (mass) F (force) p = (linear momentum) p = mv (rotation angle) (angular velocity) (angular acceleration) I (moment of inertia) (torque) L (angular momentum) L = I (9-24)
L Angular momentum L
By analogy to the definition of the linear momentum
p
mv
we define the vector of angular momentum
L
as follows:
L
I
Since I > 0 units: kg.m
2 /s
L
is parallel to
d L
dt dt
)
I d
dt
I d p
F
for linear motion
dt
d L
dt
(9-25)
L Conservation of angular momentum
d L
dt
If
0
d L dt
0 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 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
v
cm
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 v cm are connected (9-27)
Rolling v cm v cm L 1
L is the distance traveled in one revolution of the wheel 1 Thus:
L
1
v T cm
L is given by the equation: 1
L
1
v T cm
v cm
R
v cm
2
T R
R
(9-28)
Kinetic energy of a rolling object v cm v cm P
K I P K
I P
I cm
2 2 I is the wheel's moment of inertia about point P P
mR
2 (parallel axis theorem) (
I cm
mR
2 ) 2 2
I cm
2 2 2 ) 2
I cm
2 2 2
mv cm
2 The first term in the expression for K is the rotatio nal energy 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 I cm , mass m, radius R) as it rolls down an inclined plane of angle
a
mg m
sin 2 (9-30)
(9-31) Center of mass motion:
Fxnet
mg
fR
ma
(eqs.1)
I
fR
I
Substitu
a
Ia R mg
sin
m
2
Ia R
2 (eqs.2) te f from eqs.1 in eqs.2 1
g
sin
mg Ia R
2
ma
2 Note: a does not depend on m
(9-32) Rolling object 1 = cylinder Rolling object 2 = hoop
a
mg
sin
m
2 Cylinder Hoop
I
1
a
1
a
1
a
1
mR
2
m
2
mg
sin
I
1
I
2 /
R
2
a
2
mg m
3 sin
m
/ 2
a
2
a
2
mR
2
m mg mg m g
sin sin 2 sin
I
2
m
/
R
2