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Chapter 10 - Rotation
• 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 (symbol: Δθ)
• Avg. and instantaneous angular velocity (symbol: ω )
• Avg. and instantaneous angular acceleration (symbol: α )
• Rotational inertia, also known as moment of inertia
(symbol I )
• The kinetic energy of rotation as well as the rotational
inertia;
• Torque (symbol τ )
• We will have to solve problems related to the above
mentioned concepts.
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, also called the axis of rotation or the
rotation axis.
• We can describe the motion of a rigid body rotating about
a fixed axis by specifying just one parameter.
• Every point of the body moves in a circle whose center
lies on the axis of rotation, and every point moves
through the same angle during a particular time interval.
Consider the rigid body of the figure.
We take the z-axis to be the fixed axis
of rotation. We define a reference line
that is fixed in the rigid body and is
perpendicular to the rotational axis.
The angular position of this line is the
angle of the line relative to a fixed
direction, which we take as the zero
angular position.
In Fig. 11-3 , the angular position  is measured relative to
the positive direction of the x axis.
From geometry, we know that  is given by
s
(10-1)
  (in radians)
r
Here s is the length of arc (arc distance)
along a circle and between the x axis (the
zero angular position) and the reference
line; r is the radius of that circle.
This angle is measured in radians (rad) rather than in
revolutions (rev) or degrees. The radian, being the
ratio of two lengths, is a pure number and thus has
no dimension. Because the circumference of a circle
of radius r is 2r, there are 2 radians in a complete
circle: 1rev  3600  2r  2 _ rad
(10-2)
r
1 rad = 57,30 = 0,159 rev
(10-3)
Angular Displacement
• If the body of Fig. 10-3 rotates about the
rotation axis as in Fig. 10-4 , changing the
angular position of the reference line
from 1 to 2, the body undergoes an
angular displacement  given by:
Δθ = θ2 – θ1
(10-4)
• This definition of angular displacement
holds not only for the rigid body as a
whole but also for every particle within
that body.
• An angular displacement in the counterclockwise direction is positive, and one
in the clockwise direction is negative.
Angular Velocity
• Suppose (see Fig. 10-4 ) that our
rotating body is at angular position
1 at time t1 and at angular position
2 at time t2. We define the average
angular velocity of the body in the
time interval t from t1 to t2 to be
 2  1 
Unit of ω:
 avg 

(10-5)
rad/s or rev/s
t2  t1
t
Instantaneous angular velocity, ω:
 d
  lim

t 0 t
dt
(10-6)
Angular Acceleration
• If the angular velocity of a rotating body is not
constant, then the body has an angular
acceleration, α
• Let 2 and 1 be its angular velocities at times t2
and t1, respectively. The average angular
acceleration of the rotating body in the interval
from t1 to t2 is defined as:
2  1  (10-7)
 avg 
t 2  t1

t
• Instantaneous angular acceleration, α:
Unit: rad/s2 or rev/s2
 d
  lim

t 0 t
dt
(10-8)
Do Sample Problems 10-1 & 10-2, p244-246
Relating the Linear and Angular Variables :
Consider a point P on a rigid body rotating about
a fixed axis. At t  0 the reference line that connects
O
θ
s
A
the origin O with point P is on the x-axis (point A).
During the time interval t , point P moves along arc AP
and covers a distance s. At the same time, the reference
line OP rotates by an angle  .
Relation between Angular Velocity and Speed :
The arc length s and the angle  are connected by the equation
ds d  r 
d
s  r , where r is the distance OP. The speed of point P is v  
r
.
dt
dt
dt
(10-17)
v  r
(10-18)
circumference 2 r 2 r 2
The period T of revolution is given by T 



.
speed
v
r 
T
2

1
T
f
  2 f
(10-19)
(10-20)
r
O
The Acceleration :
The acceleration of point P is a vector that has two
components. The first is a "radial" component along
the radius and pointing toward point O. We have
enountered this component in Chapter 4 where we
called it "centripetal" acceleration. Its magnitude is
v2
ar    2 r
r
(10-23)
The second component is along the tangent to the circular path of P and is thus
known as the "tangential" component. Its magnitude is
dv d  r 
d
at 

r
 r
dt
dt
dt
at  r
(10-22)
The magnitude of the acceleration vector is a  at2  ar2 .
(10-8)
Rotation with Constant Angular Acceleration :
When the angular acceleration  is constant we can derive simple expressions
that give us the angular velocity  and the angular position  as a 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
translational and rotational motion using the following correspondence
between the two motions.
Translational Motion
x
v
a
v  v0  at
at 2
x  xo  v0t 
2
v 2  v02  2a  x  xo 
Rotational Motion








  0   t
    0 t 
(eq. 1)
t2
2
  2  02  2    0 
(eq. 2)
(eq. 3)
(10-6)
Kinetic Energy of Rotation :
vi
mi
O
ri
Consider the rotating rigid body shown in the figure.
We divide the body into parts of masses m1 , m2 , m3 ,..., mi ,....
The part (or "element") at P has an index i and mass mi .
The kinetic energy of rotation is the sum of the kinetic
1
1
1
2
2
energies of the parts: K  m1v1  m2 v2  m3v32  ....
2
2
1
12
2
2
K   mi vi The speed of the ith element vi   ri  K   mi  ri  .
(10-32)
(10-34)
(10-33)
i 2
i 2
(10-31)
1
1

K    mi ri 2   2  I  2 The term I   mi ri 2 is known as
2 i
2
i

rotational inertia or moment of inertia about the axis of rotation. The axis of
rotation must be specified because the value of I for a rigid body depends on
its mass, its shape, as well as on the position of the rotation axis. The rotational
inertia of an object describes how the mass is distributed about the rotation axis.
1 2
K  I
2
I   mi ri 2
i
I   r 2 dm
(10-35)
Calculating the Rotational Inertia :
The rotational inertia I   mi ri2 . This expression is useful for a rigid body that
i
has a discreet distribution of mass. For a continuous distribution of mass the sum
becomes an integral, I   r 2 dm.
In the table below we list the rotational inertias for some rigid bodies.
I 
r
2
dm
(10-10)
Parallel - Axis Theorem :
We saw earlier that I depends on the position of the rotation axis.
For a new axis we must recalculate the integral for I . A simpler
method takes advantage of the parallel-axis theorem.
Consider the rigid body of mass M shown in the figure.
We assume that we know the rotational inertia I com
about a rotation axis that passes through the center
of mass O and is perpendicular to the page.
The rotational inertia I about an axis parallel to the axis through O that passes
through point P, a distance h from O, is given by the equation
I  I com  Mh
2
(10-36)
A
Proof of the Parallel - Axis Theorem : 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 rotational inertia
I com for an axis that is perpendicular to the page and passes
through O.
We wish to calculate the rotational inertia I about a new axis perpendicular
to the page and passing through point P with coordinates  a, b  . Consider
an element of mass dm at point A with coordinates  x, y  . The distance r
between points A and P is r 
 x  a   y  b .
2
2
2
2
Rotational Inertia about P : I   r 2 dm    x  a    y  b   dm.


I    x 2  y 2  dm  2a  xdm 2b  ydm    a 2  b 2  dm. The second
and third integrals are zero. The first integral is I com . The term  a 2  b 2   h 2 .
2
Thus the fourth integral is equal to h 2  dm  Mh 2  I  I com  Mh
Do Sample Problems 10-6 to 10-8, p254-256
Torque: Draaimoment
In fig. a we show a body that can rotate about an axis through
point O under the action of a force F applied at point P a distance
r from O. In fig. b we resolve F into two components, radial and
tangential. The radial component Fr cannot cause any rotation
because it acts along a line that passes through O. The tangential
component Ft  F sin  on the other hand causes the rotation of the
object about O. The ability of F to rotate the body depends on the
magnitude Ft and also on the distance r between points P and A.
(10-40)
Thus we define as torque   rFt  rF sin   r F .
(10-41)
(10-39)
The distance r is known as the moment arm and it is the
perpendicular distance between point O and the vector F .
The algebraic sign of the torque is assigned as follows:
If a force F tends to rotate an object in the counterclockwise
direction the sign is positive. If a force F tends to rotate an
  r F
object in the clockwise direction the sign is negative.
(10-13)
Newton's Second Law for Rotation :
For translational motion, Newton's second law connects
the force acting on a particle with the resulting acceleration.
There is a similar relationship between the torque of a force
applied 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 that consists of a point mass m at the end
of a massless rod of length r. A force F is 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 rotation. We first apply Newton's second law for Ft . Ft  mat (eq. 1)
The torque  acting on the particle is   Ft r
(eq. 2). We eliminate Ft
between equations 1 and 2:   mat r  m  r  r   mr 2    I  .
(compare with: F  ma)
  I
(10-45)
3
i
ri
O
2
1
Newton's Second Law for Rotation :
We have derived Newton's second law for rotation
for a special case: a rigid body that consists of a point
mass m at the end of a massless rod of length r. We will
now derive the same equation for a general case.
Consider the rod-like object shown in the figure, which can rotate about an axis
through point O under the action of a net torque  net . We divide the body into
parts or "elements" and label them. The elements have masses m1 ,m2 , m3 ,..., mn
and they are located at distances r1 , r2 , r3 ,..., rn from O. We apply Newton's second
law for rotation to each element:  1  I1 (eq. 1),  2  I 2 (eq. 2),
 3  I 3 (eq. 3), etc. If we add all these equations we get
 1   2   3  ...   n   I1  I 2  I 3  ...  I n   . Here I i  mi ri 2 is the rotational inertia
of the ith element. The sum  1   2   3  ...   n is the net torque  net applied.
The sum I1  I 2  I 3  ...  I n is the rotational inertia I of the body.
Thus we end up with the equation
 net  I
(10-42)
(10-15)
Analogies Between Translational and Rotational Motion
Translational Motion
x 
v 
a 
v  v0  at 
at 2
x  xo  v0t 
2
v 2  v02  2a  x  xo 
mv 2
K
2
m
F  ma
F
P  Fv
Rotational Motion



  0   t
    0  0 t 






t2
2
 2  02  2    0 
I 2
K
2
I
  I

P  
(10-18)