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PH 221-1D Spring 2013
ROTATION
Lectures 24-26
Chapter 10 (Halliday/Resnick/Walker, Fundamentals of Physics 9th edition)
1
Chapter 10Rotation
In this chapter we will study the rotational motion of rigid bodies about a fixed axis. To describe this type of motion we will introduce the following new concepts:
-Angular displacement -Average and instantaneous angular velocity (symbol: ω )-Average and instantaneous angular acceleration (symbol: α )-Rotational inertia also known as moment of inertia (symbol I ) -Torque (symbol τ )
We will also calculate the kinetic energy associated with rotation, write Newton’s second law for rotational motion, and introduce the work-kinetic energy for rotational motion
2
The Rotational Variables
In this chapter we will study the rotational motion of rigid bodies about fixed axes. A rigid body is defined as one that can rotate with all its parts locked together and without any change of its shape. A fixed axis means that the object rotates about an axis that does not move. We can describe the motion of a rigid body rotating about a fixed axis by specifying just one parameter. Consider the rigid body of the figure.
We take the the z-axis to be the fixed axis of rotation. We define a reference line which is fixed in the rigid body and is perpendicular to the rotational axis. A top view is shown in the lower picture. The angular position of the reference line at any time t is defined by the angle θ(t) that the reference lines makes with the position at t = 0. The angle θ(t) also defines the position of all the points on the rigid body because all the points are locked as they rotate. The angle θ is related to the arc length s traveled by a point at a distance r from the axis via
the equation: Note: The angle θ is measured in radianssr
3
1
2 1 2
2 1
In the picture we show the reference line at a time and at a later time . Between and the body undergoes an angular displacement
An
.
gular Displace
All the points of th
ment
erigid
tt t t
body have the same angular displacement because they
rotate locked together.
1
2 1
2 1
2We define as average angular velocity for the time interval , the ratio:
We define as the instantaneous ang
The SI unit for angular velocity is
An
rad
gular Velocity
ians/secondavg t t t
t t
0
ular velocity the limit of as 0
lim This is the definition of the first deriva
Algerbraic sign of angular f
tive with
If a rigid body rotre atquen es cy: counterclockwise (CC
W
t
tt
tt
) has a positive sign. If on the other hand the rotation is clockwise (CW) has a negative sign
t1
t2
ddt
4
If the angular velocity of a rotating rigid object changes with time we can describe the time rate of change of by defining the
Angula
angula
r Accelerati
r acelera
on
tion
1 2
1 1 2 2
In the figure we show the reference line at a time and at a later time . The angular velocity of the rotating body is equal to at and at . We define as average angular acceleration fo
t tt t
1 2
0
2 1
2
2
1
The SI unit for angular velocity is radia
r the time interval , the ratio:
We define as the instantaneous angular acceleration the limit of
ns/sec
as
lim
ond
0
av
t
g
t t
t
t t
t
t
This is the definition of the first derivative with tt
ω1
ω2
t1
t2
ddt
5
For rotations of rigid bodies about a fixed axis we can describe accurately the angular velocityby asigning an algebraic sigh. Positive for counterclockwi
Angul
se ro
ar Velocity Ve
tation and ne
ctor
gative forclockwise rotation
We can actually use the vector notation to describe rotational motion which is more complicated. The angular velocity vector is defined as follows:The of is along the rotationdirecti axis.
hn
To
e of is defined by the right hand rule (RHL)Curl the right hand so that the fingers point in the direction
of the rotation. The thumb of the right
sense R
hand giight hand rule
ves the sense :
f
o
6
When the angular acceleration is constant we can derive simple e
Rotation with Constant Angular Accelerati
xpressionsthat give us the angular velocity and the angular p
on
osit ion as function of time
We could derive these equations in the same way we did in chapter 2. Instead we will simply write the solutions by exploiting the analogy between translationaland rotational m
Rotatio
otion u
nal Mot
sing the follo
ion
wing correspondance between
Tran
the two motions
slational Motion xv
02
02
2 2 22
(eqs.1)
(eqs.2)
2
(
2
2 2
o o
o oo o
o
av v at
atx x v t
v v x
tt
a x
t
eqs.3) 7
Consider a point P on a rigid body rotating about a fixed axis. At 0 the reference line which connects the origin
Relating the Linear and A
O with point P is on the
ngular Var
x-axis (po
iable
int
s
A)
D
t
uring the time interval point P moves along arc APand covers a distance . At the same time the reference line OP rotates by an angle .
ts
Aθs
O
The arc length s and the angle are connected by the equation:
where is the dista
Relation b
nce OP. T
etween angular velocity a
he speed of point P
nd speed
d rds dr v r
dt dt dts r
v r
circumference 2 2 2The period of revolution is given by: speed
r rT Tv r
2T
1Tf
2 f 8
rO
The acceleration of point P is a vector that has twocomponents. A "radial" componet along the radius and pointing towards point O. We have enountered this component i
The Accele
n chapter
ration
4 where we called it "centripetal" acceleration. Its magnitude is:
22
rva rr
The second component is along the tangent to the circular path of P and is thus known as the "tangential" component. Its magnitude is:
The magnitude of the acceleration vector is
t
d rdv da r rdt dt dt
2 2: t ra a a
ta r
9
10
Example. Motion with constant angular acceleration.
Rotation with constant angular acceleration.
11
12
The equations of rotational kinematics.
13
The equations of kinematics.
14
Angular variables and tangential variables.
15
Angular variables and tangential variables.
16
Centripetal acceleration and tangential acceleration.
Ori
iv
mi
1 2 3
Consider the rotating rigid body shown in the figure. We divide the body into parts of masses , , ,..., ,...The part (or "element") at P
Kinetic Energy of Rotatio
has an index and mass
n
Th
i
i
m m m mi m
2 2 21 1 2 2 3 3
e kinetic energy of rotation is the sum if the kinetic 1 1 1energies of the parts ...2 2 2
K m v m v m v
22
2 2 2 2
1 1 The speed of the -th element 2 2
1 1 The term
rotational i
is known as 2 2
or about the axis nertia moment of inerti of rotation.a The axis
i i i i i ii i
i i i ii i
K m v i v r K m r
K m r I I m r
of
rotation be specified because the value of for a rigid body depends on its mass, its shape as well as on the position of the rotation axis. The rotationalinertia of an object describe
mu
s w
st
ho
I
the mass is distributed about the rotation axis2
i ii
I m r 212
K I2 I r dm 17
In the table below we list the rotational inertias for some rigid bodies2 I r dm
18
An automobile of mass 1400kg has wheels 0.75m in diameter weighing 27kg each. Taking into account the rotational kinetic energy of the wheels about their axes, what is the total kinetic energy of the automobile traveling at 80km/h? What percent of kinetic energy belongs to the rotational motion of the wheels about their axes? Pretend that each wheel has a mass distribution equivalent to that of a uniform disk.
19
2The rotational inertia This expression is useful for a rigid body that
has a discreet disstribution of mass. Fo
Calculating the Rotational In
r a continuous distribution of mass the
a
s
erti
i ii
I m r
2
um
becomes an integral I r dm
We saw earlier that depends on the position of the rotation axisFor a new axis we must recalculate the integral for . A simpler method takes advantage
Parall
of the
el-Axis Theorem
parallel-a he xis t
II
Consider the rigid body of mass M shown in the figure. We assume that we know the rotational inertia about a rotation axis that passes through the center of mass O and is perpendicul
ore
o
m
ar t t
comI
he page. The rotational inertia about an axis parallel to the axis through O that passesthrough point P, a distance h from O is given by the equation:
I
2comI I Mh 20
A We take the origin O to coincide with the center of mass of the rigid body shown in the figure. We assume that we know the ro
Proof of the Parallel-Axis Theo
tational inertia for a
rem
n axis thacomI t is perpendicular to the page and passes through O.
We wish to calculate the rotational ineria about a new axis perpendicular to the page and passes through point P with coodrinates , . Consider
an element of mass at point A with coordinates ,
Ia b
dm x y
2 2
2 22
2 2 2 2
. The distance
between points A and P is:
Rotational Inertia about P:
2 2 The second
and third integrals are zero. The first i
r
r x a y b
I r dm x a y b dm
I x y dm a xdm b ydm a b dm
2 2 2 .
2 2
ntegral is The term
Thus the fourth integral is equal to
comI a b h
h dm Mh
2
comI I Mh 21
According to spectroscopic measurements, the moment of inertia of an oxygen molecule about an axis through the center of mass and perpendicular to the line joining the atoms is 1.95x10-46kg·m2. The mass of an oxygen atom is 2.66x10-26g. What is the distance between atoms?
22
In fig.a we show a body which can rotate about an axis through
point O under the action of a force applied at point P a distance
from O. In fig.b we resolve into two compone
Torq
ts, a
ue
r dial
F
r F
and tangential. The radial component cannot cause any rotationbecause it acts along a line that passes through O. The tangentialcomponent sin on the other hand causes the rotation of the
ob
r
t
F
F F
ject about O. The ability of to rotate the body depends on themagnitude and also on the distance between points P and O.Thus we define as sitorque The distance is known a
nt
trF rF r F
FF r
r
s the and it is the
perpendicular distance between point O and the vector The algebraic sign of the torque is asigned as follows:
If a force tends to rotate an object in
moment arm
the counterc
F
F
lockwise
direction the sign is positive. If a force tends to rotate an object in the clockwise direction the sign is negative.
F
r F 23
For translational motion Newton's second law connects the force acting on a particle with the resulting accelerationThere is a similar r
Newton's Second
elationship betw
Law for Ro
een the to
tation
rque of a forceapplied on a rigid object and the resulting angular acceleration
This equation is known as Newton's second law for rotation. We will explore this law by studying a simple body which consists of a point mass at the end
of a massless rod of length . A force F is
m
r
applied on the particle and rotates
the system about an axis at the origin. As we did earlier, we resolve F into a tangential and a radial component. The tangential component is responsible for the
2
rotation. We first apply Newton's second law for . (eqs.1)The torque acting on the particle is: (eqs.2) We eliminate
between equations 1 and 2:
t t t
t t
t
F F maF r F
ma r m r r mr I
(compare with: )F maI 24
We have derived Newton's second law for rotation for a special case. A rigid body which consists of a pointmass at the end of a massl
Newton's Second Law for Rotation
ess rod of length . We willnow
m r derive the same equation for a general case.
O
12
3
i
ri
Consider the rod-like object shown in the figure which can rotate about an axisthrough point O undet the action of a net torque . We divide the body into parts or "elements" and label them. The e
net
1 2 3
1 2 3
1 1 2 2
3 3
lements have masses , , ,..., and they are located at distances , , ,..., from O. We apply Newton's secondlaw for rotation to each element: (eqs.1), (eqs.2),
(eqs
n
n
m m m mr r r r
I II
21 2 3 1 2 3
1 2 3
.3), etc. If we add all these equations we get:
... ... . Here is the rotational inertiaof the -th element. The sum ... is the net torque applied.
n n i i i
n net
I I I I I m ri
1 2 3The sum ... is the rotational inertia of the body. Thus we end up with the equation:
nI I I I I
net I 25
A block of mass m hangs from a string wrapped around a frictionless pulley of mass M and radius R. If the block descends from rest under the influence of gravity, what is the angular acceleration of the pulley?
26
In chapter 7 we saw that if a force does work on an object, this results in a change of its kinetic energy
. In a similar
Work and Rotational Kin
way, when a torque does
etic Energ
work on ro
y
a
W
K W W tating rigid body, it changes its rotational kinetic
energy by the same amount
Consider the simple rigid body shown in the figure which consists of a mass
at the end of a massless rod of length . The force does work The radial component does zero work becau
t
r
m
r F dW F rd dF
2 2 2 2 2 2
se it is at right angles to the motion.
The work . By virtue of the work-kinetic energy theorem we
1 1 1 1have a change in kinetic energy 2 2 2 2
f
i
t
f i f i
W F rd d
K W mv mv mr
W K
mr
f
i
W d
2 21 12 2f iK I I
W K
27
Conservation of Energy in Rotational MotionA meter stick is initially standing vertically on the floor. If it falls over, with what angular velocity will it hit the floor? Assume that the end in contact with the floor doesn’t slip.
28
Power has been defined as the rate at which work is done by a force and in the case of rotational motion by a torqueWe saw that a torque produces work as it rotates an object
Power
by an angldW d
e .
(Compare with )
ddW d dP d P Fvdt dt dt
Below we summarize the results of the work-rotational kinetic energy theorem
f
i
W d
For constant torque f iW
2 21 12 2f iW K I I Work-Rotational Kinetic Energy Theorem
P 29
Rotational Motion
Analogies betw
een translational and rotational Motio
Translational Motion
n
xv
0
2 22 2
20
2
2
2
22
o o o o
o oo o
av v at
atx x v t
v v a x x
t
tt
2 2
2
2
mvK
mF ma
FP Fv
IK
II
P 30