ROTATION OF RIGID BODIES
Wind turbines converts the energy of wind into rotational
energy by means of blades. The blades have the same
angular velocity. It involves a body that rotates about an
axis that is stationary in some inertial frame of reference.
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7
Almost everywhere, we know that there exists
a rotation of an object from the motions of
electrons in atoms to the motions of the entire
galaxies. That is why it is imperative to
develop a model so that we can understand
and analyze the motion of a rotating body. In
this chapter, a rotating object is being studied.
Some analogy in rotational motions like
angular velocity, angular acceleration are
being introduce and equations were develop
to describe rotational motion. Torque was
also defined which is the twisting or turning
effort of a force. Work, power, energy and
conservation of angular momentum were
discussed which are tremendously useful for
understanding rotational motion. There are
two types of rotational motion that were being
discussed in this chapter which are the
rotational kinematics and the rotational
dynamics.
ROTATIONAL KINEMATICS
Real-world bodies are considered as a rigid body, an idealized model which has a
perfectly definite and unchanging shape and size. It is a body where all the particles maintain
their relative position as it rotates. In this section, the rotational motion is described using
kinematic language. When a rigid body rotates about a fixed axis, its motion can be described
in terms of its angular position 𝜃, angular velocity 𝜔, and angular acceleration 𝛼. Then we look
at the kinetic energy of rotation, the key to using energy methods for rotational motion.
5.1 Angular Velocity and Acceleration
In describing rotational motion, the usual way to measure
the angular displacement is in terms of radian. One radian is the
angle subtended at the center of the circle by an arc with the length
equal to the radius of the circle
=
s
r
(5-1)
where 𝑟 is the radius of the circle and 𝑠 is the arc length as seen
in Figure 5.1.
Figure 5.1: Measuring angle
in radians.
Angular velocity (𝜔) is the time rate of change of the angular position. Average angular
velocity is defined as
ave =
  2 − 1
=
t
t 2 − t1
(5-2)
85
And the instantaneous angular velocity is defined as
 ave d
=
t → 0 t
dt
 = lim
(5-3)
The SI unit for angular velocity is in radian/seconds (or rad/s).
Angular acceleration 𝛼 is the time rate of change of the angular velocity. The average angular
acceleration is defined as
 ave =
2 − 1 
=
t2 − t1
t
(5-4)
while the instantaneous angular acceleration is defined as
ave d d 2
=
=
t →0 t
dt dt 2
 = lim
(5-5)
The SI unit for angular acceleration is rad/s2.
The distinction between angular velocity  Z and ordinary velocity, or linear velocity, v x is that
if an object has a velocity v x , the object as a whole is moving along the x-axis. By contrast, if
an object has an angular velocity  Z then it is rotating around the z-axis. We do not mean that
the object is moving along the z-axis.

The figure below shows the right-hand rule for the direction of the angular velocity vector  .

Reversing the direction of rotation reverses the direction of  . The sign of  Z for rotation along
the z-axis.

Figure 5.1: The right-hand rule for the direction of the angular velocity vector  . Reversing the direction

of rotation reverses the direction of  . (Source: Young & Freedman, Univ. Physics with Modern Physics 13th
Ed.)
86
Sample Problem
A fan blade rotates with angular velocity given by z (t ) =  − t 2 , where
 =0.800 rad/s3.

=5.00 rad/s and
a) Calculate the angular acceleration as a function of time.
b) Calculate the instantaneous angular acceleration
acceleration  av− z for the time interval
Solution
z =
a)
 z (t ) =
d Z
dt
 z at t = 3.00s and the angular
t = 0 to t = 3.00s .
 av− z =
 2
t
d 2
(t ) = 2t
dt
d z
= −2t = (−1.60 rad/s3)t.
dt
b)  z (3.0 s)= (-1.60 rad/s3) (3.0 s) = -4.80 rad/s2
 av− z =
z (3.0s) − z (0)
3.0s
=
−2.20rad / s − 5.00rad / s
= −2.40rad / s 2
3.0s
5.2 Rotation with Constant Angular Acceleration
For the special case of constant angular acceleration, the equations which relate
angular displacement, angular velocity, and angular acceleration and time have the same form
as the kinematics equations for constant linear acceleration. These equations can be obtained
from the familiar ones by replacing x with  , v with  omega, and a with  .
Table 5.1: Rotational Analog of Linear Motion with Constant Acceleration
Linear Motion
Rotational Motion
v = vo + at
 = o +  t
1
1
𝑥 − 𝑥𝑜 = (𝑣 + 𝑣𝑜 )𝑡
𝜃 − 𝜃𝑜 = (𝜔 + 𝜔𝑜 )𝑡
1
x − xo = vo t + at
2
𝜃 − 𝜃𝑜 = 𝜔𝑜 𝑡 + 𝛼𝑡 2
𝑣 2 − 𝑣𝑜 2 = 2𝑎(𝑥 − 𝑥𝑜 )
𝜔2 − 𝜔𝑜 2 = 2𝛼(𝜃 − 𝜃𝑜 )
2
2
1
2
87
Sample Problem
When an electric fan is turned off, its angular velocity is decreasing uniformly from 500 rev/min
to 200 rev/min in 40 s.
a) Find the angular acceleration in rev/s 2
b) The number of revolutions made by the motor in the 40s interval.
c) How many more seconds required for the fan to come to rest if the angular acceleration
remains constant using the value calculated in (a)?
Solution
a)
 = O + t
=
b)
c)
 − o
t
rev
rev

− 500
 200
min
min
=
40s
 1 min
x
 60s = −0.125 rev
s2

rev
rev  1 min 
+ 500
  200

x
1
min
min
60 sec 



 = ( + o )t =
x(40s ) = 233rev


2
2




 = O + t
t=
− o

=
rev 1 min
x
min 60s = 26.67s
rev
− 0.125 2
s
− 200
5.3 Relating Linear and Angular Kinematics
The relationship between linear and angular speeds is
 = r
(5-6)
The tangential acceleration of a point on a rotating body is
atan =
dv
d
=r
= r
dt
dt
(5-7)
The centripetal acceleration of a point on a rotating body is
arad =
v2
=  2r
r
(5-8)
88
Sample Problem
An electric turntable 0.750 m in diameter is rotating about a fixed axis with an initial angular
velocity of 0.250 rev/s and a constant angular acceleration of 0.900 rev/s 2. (a) Compute the
angular velocity of the turntable after 0.200 s. (b) Through how many revolutions has the
turntable spun in this time interval?(c) What is the tangential speed of a point on the rim of the
turntable at t=0.200 s? (d) What is the magnitude of the resultant acceleration of a point on
the rim at t= 0.200 s?
Solution
Apply constant angular acceleration equations. A point on the rim has both tangential and
radial components of acceleration. v = r.
a tan = r
a)
arad = r 2
z = 0 z +  z t = 0.250rev / s + (0.900rev / s 2 )(0.200s) = 0.430rev / s
b) av − z t = (0.340 rev / s )( 0.2 s ) = 0.068 rev.
c)
 0.750m 
v = r = 
(0.430rev / s )( 2rad / rev ) = 1.01m / s
 2 
d)
2
2
a = arad
+ atan
= ( 2 r) 2 + (r ) 2 .
a=
((0.430rev / s)(2rad / rev)) (0.375)
2
2
+
(0.900rev / s
2
)( 2rad / rev)(0.375m) .
2
a = 3.46m / s 2 .
5.4 Energy in Rotational Motion
A rotating rigid body consists of mass in motion, so it has kinetic energy. This energy
can be expressed in terms of the body’s angular velocity and a quantity called moment of
inertia. The moment of inertia of a body is a measure of the resistance of an object to changes
in its rotational motion. For a system of particles of masses
mi at
distances
ri
from an axis
passing through a point P the rotational inertia of the system about the axis is given by:
I = m1r1 + m2 r2 + ... =  mi ri
2
2
i
2
(5-9)
The SI unit of moment of inertia is kg.m2.
When a rigid body rotates about a fixed axis, the speed
v i of the i th particle is given by:
vi = ri . The kinetic energy of the particle can be expressed as
1
1
2
2
mi vi = mi ri  2
2
2
(5-10)
89
The total kinetic energy (K) of the body is the sum of the kinetic energies of all its particles:
1
2
K =  2  mi ri
i
2
(5-11)
In terms of moment of inertia (I), the rotational kinetic energy of a rigid body is:
K=
1 2
I
2
(5-12)
The SI unit of the rotational kinetic energy is Joule (J).
Note: Moment of inertia of different solids can be found in the Appendix D.
Sample Problem
1. A 0.75 kg wooden ball 0.15 m in radius is rolling at a speed of 3.05 m/s. Find its total
kinetic energy.
Solution
1
1
𝐾 = 𝑚𝑣 2 + 𝐼𝜔 2
2
2
2
2
5
5
𝐼 = 𝑚𝑅 2 = (0.75 𝑘𝑔)(0.15 𝑚)2 = 6.75 𝑥 10−3 𝑘𝑔 ∙ 𝑚2
𝑣
3.05 𝑚/𝑠
𝑟
0.15 𝑚
𝜔= =
𝐾=
= 20.33 𝑟𝑎𝑑/𝑠
1
1
1
1
𝑚𝑣 2 + 𝐼𝜔 2 = (0.75 𝑘𝑔)(3.05 𝑚⁄𝑠 )2 + (6.75𝑥10−3 )(20.33 𝑟𝑎𝑑⁄𝑠 )2
2
2
2
2
𝐾 = 4.88 𝐽
2. Four small spheres, each with mass of 0.20 kg are
arranged in a square of 0.40 m on each side and
connected by light rods. Find the moment of inertia in
kg.m2 if the system is rotated about an axis through the
center of the square, perpendicular to its plane.
Solution
I = m1r12 + m2 r22 + m3r32 + m4 r42
Using the Pythagorean Theorem,
2
2
 0.4m   0.4m 
r= 
 +
 = 0.28m
 2   2 
90
(The distance from point O to the sphere)
Since r is just the same for all masses, then


I = 4 0.2kg (0.28m ) = 0.063kg m 2
2
5.5 Parallel-Axis Theorem
The moment of inertia of any object about an axis through its center of mass is the
minimum moment of inertia for an axis in that direction in space. The moment of inertia about
any axis parallel to that axis through the center of mass is given by:
I p = I cm + Md 2
(5-13)
Where, M the object's mass and d the perpendicular distance between the two axes.
5.6 Moment of Inertia calculation
For continuous mass distribution, the moment of inertia for linear distribution is
I =  r 2dm
(5-14)
For continuous mass distribution, the moment of inertia for volume distribution is
I =  r 2 dV
If the body is uniform in density, then we may take
I =   r 2 dV

(5-15)
outside the integral:
(5-16)
ROTATIONAL DYNAMICS
A force is required to start a stationary body rotating or to bring a spinning body to a
halt. This force must be applied in a way that it gives a twisting or turning action. The twisting
or turning effort of force is called the torque.
In this section, we will learn how to find the net torque acting on a rigid body; work and power
in rotational motion; and conservation of angular momentum.
91
5.7 Torque

The Torque (  ) is the tendency of a force to cause or change
the rotational motion of the body. It is a measure of the twisting effect
of a force on a body.
The figure shows a wrench being used to loosen a tight bolt. Force

Fb
applied near the end of the handle is more effective that an equal force

Fa
applied near the bolt. Force

Fc
doesn’t do any good at all; it is
applied at the same point and has the same magnitude as

Fb
but its
parallel along the length of the handle which contribute any torque.
The
torque
sign
conventions:
Figure 5.2: A wrench
being used to loosen a
tight bolt. (Source: Young
& Freedman, Univ. Physics
with
Modern Physics
13th Ed.)
(+) →counter-clockwise
(–) →clockwise
The torque of a force about a point is the product of the force
magnitude and the lever arm of the force. From figure 5.3 ,
𝜏1 = + 𝐹1 𝑙1
𝜏2 = − 𝐹2 𝑙2
𝜏3 = 0
Three ways to calculate the torque of the force about the point
O. In figure 5.4, and are in the plane of the page and the
torque vector points out of the page toward you.
If 𝐹⃗ and 𝑟⃗ are not perpendicular with each other,
Figure 5.3: (Source: Young &
Freedman, Univ. Physics with Modern
Physics 13th Ed.)
the magnitude of torque is
 = Fl = rF sin  = Ftan r
(5-17)
where,
𝑙 – moment arm or lever arm perpendicular with
Force 𝐹.
F – magnitude of the force
 – angle between 𝑟⃗ and 𝐹⃗
In general the definition of torque is



 = r F
(5-18)
The SI unit of Torque is N∙m. The direction of the torque is
given by the right-hand rule.
Figure 5.4 (Source: Young & Freedman,
Univ. Physics with Modern Physics 13th Ed.)
92
Sample Problem



Find the torque produced by each force through point O. The magnitudes of F1 , F2 and F3 are
8N, 12N and 10N, respectively. What is the net Torque?
30o
O
75o
2
3
Solution
a) 𝜏⃑𝐹1 = 𝑟⃑1 𝑥 𝐹⃑1 = −𝑟1 𝐹1 𝑠𝑖𝑛𝜃 = −(5 𝑚)(8 𝑁)sin90° = − 40 𝑚𝑁
b) 𝜏⃑𝐹2 = 𝑟⃑2 𝑥 𝐹⃑2 = +𝑟2 𝐹2 𝑠𝑖𝑛𝜃 = +(2 𝑚)(12 𝑁)sin30° = +12 𝑚𝑁
c)
𝜏⃑𝐹3 = 𝑟⃑3 𝑥 𝐹⃑3 = +𝑟3 𝐹3 𝑠𝑖𝑛𝜃 = +(2 𝑚)(10 𝑁)sin75° = +19.3 𝑚𝑁
The net torque is  = −8.7 mN
5.8 Equilibrium
The word equilibrium means balance. In particular, static equilibrium means that a system is
stable and at rest. This means that the net force must be zero – called translational equilibrium.
However, to be complete the net torque must also be zero – called rotational equilibrium. For
complete static equilibrium, both of these two conditions must be fulfilled.
Two conditions for equilibrium:
1. The static equilibrium which is
∑ 𝐹⃗ =0
(5-19)
2. The rotational equilibrium which is
∑ 𝜏⃗=0
(5-20)
5.9 Torque and Angular Acceleration for Rigid Body
The net force acting on this particle has a component 𝐹1𝑟𝑎𝑑 along the radial direction, a
component 𝐹1𝑡𝑎𝑛 that is tangent to the circle of radius r1 in which particle moves, and component
F1z along axis of rotation as shown in the figure.
93
The net torque is
 z = I z
(5-21)
For all particles the equation is
2
 iz =   mi ri  z =   I i  z
i
i

i 
(5-22)

As a rigid body rotates around the z-axis, a net force F1
acts on one particle of the body. Only the force component
F1,tan can affect the rotation, because only F1,tan exerts a
torque about O with a z-component (along the rotation
axis).
Figure 5.5 (Source: Young &
Freedman, Univ. Physics with
Modern Physics 13th Ed.)
Sample Problem
A cable is wrapped several times around a uniform solid cylinder
that can rotate about its axis. The cylinder has diameter 0.120
m and mass 50 kg. The cable is pulled with a force of 9.0N.
Assuming that the cable unwinds without stretching or slipping,
what is its linear acceleration?
Solution
 z = I z   z =
z =
z
I
=
z
I cy
I cy =
1
MR 2
2
FR
2F
2(9.0 N )
=
=
= 6.0rad / s 2
 MR 2  MR (50kg)( 0.060m)


 2 
a x = R x = (0.060m)(6.0.rad / s 2 ) = 0.36m / s 2
94
5.10 Rigid-Body Rotation about a Moving Axis
We can extend our analysis of the dynamics of rotational motion to some cases in which
the axis of rotation moves. When that happens, the motion of the body is combined translation
and rotation.
Every possible motion of a rigid body can be represented as a combination of translational
motion of the center of mass and rotation about an axis through the center of mass.
In this case, the kinetic energy is associated with motion of the center of mass and with rotation
about an axis through the center of mass.
1 2 1
K = Mvcm
+ I cm 2
2
2
(5-23)
An important case of combined translation and rotation is rolling without slipping, such as the
motion of the wheel shown in figure. The condition for rolling without slipping is
𝑣𝑐𝑚 = 𝑅𝜔
where, 𝑣𝑐𝑚 𝑅
𝜔 -
(5-24)
velocity of center of mass
radius of the wheel
angular speed
Figure 5.6: The motion of a rolling wheel is the sum of the translational motion of the center of mass plus
the rotational motion of the wheel around the center of mass. (Source: Young & Freedman, Univ. Physics
with Modern Physics 13th Ed.)
Sample Problem
Flywheels which are simply large rotating disks have been suggested as a means of storing
energy for solar-powered generating systems. Estimate the kinetic energy that can be stored in
a 20,000-kg (10-ton) flywheel with a diameter of 20 m (a 6-story building). Assume it could
hold together (without flying apart due to internal stresses) at 100 rpm (revolution per minute).
95
Solution


 = 100
rev  1 min  2rad 


 = 10.5rad / s
min  60 sec  rev 
𝐼𝑑𝑖𝑠𝑐 = ½𝑀𝑅2
K=
1 2 11
1

I =  MR 2  2 = ( 20,000kg)( 20m) 2 (10.5rad / s ) 2
2
22
4

K = 2.205  108 J
5.11 Dynamics of Combined Translation and Rotation
The combined translational and rotational motion of an object can also be analyzed from
the standpoint of dynamics. In this case the object must obey both of the following forms of
Newton's Second Law:
Two following conditions should be met:
1. The axis through the center of mass must be an axis of symmetry
2. The axis must not change direction
5.12 Rolling Friction
Rolling friction occurs when a wheel, ball, or cylinder rolls freely over a surface, as in
ball and roller bearings. The main source of friction in rolling appears to be dissipation of energy
involved in deformation of the objects.
Figure 5.7: Rolling down a perfectly rigid surface and a deformable surface. The deformation in the right
figure is greatly exaggerated. (Source: Young & Freedman, Univ. Physics with Modern Physics 13th Ed.)
96
Sample Problem
A solid bowling ball rolls without slipping down the return
ramp at the side of the alley. The ramp is inclined at an
angle  to the horizontal. Treat the ball as a uniform
solid sphere, ignoring the finger holes.
a) What is the ball’s acceleration?
b) Determine the frictional force.
Solution
2
MR 2
5
I solid sphere =
 Fx = Mg sin  + ( − f ) = Macm, x
(1)
If the ball rolls without slipping:
acm− x = R z   z =
acm− x
R
(2)
The equation of motion for rotation about the axis through the center of mass is
2
2 
 z = fR = I cm z =  MR  z
5

(3)
Substitute z (2) to (3), so that (3) becomes,
fR =
2
2
a

MR 2  cm− x   f = Macm− x
5
5
 R 
Substitute f to (1) and solve for
acm− x
Mg sin  + (− f ) = Macm − x
acm − x =
and
f =
5
g sin 
7
2
5
2
M ( g sin  ) = Mg sin 
5
7
7
Therefore, the acceleration is just 5/7 as large as it would be if the
ball could slide without friction down the slope.
5.13 Work and Power in Rotational Motion
When you pedal a bicycle, you apply forces to a rotating body
and do work on it. Similar things happen in many other real-life situations, such as a rotating
motor shaft driving a power tool or a car engine propelling the vehicle.
97
Consider a tangential force applied to a rotating body that does work as shown in Figure 5.8.
The work done by a torque is,
Figure 5.8: A tangential
force applied to a rotating
body does work.
𝑑𝑊 = 𝐹𝑡𝑎𝑛 𝑑𝑠 = 𝐹𝑡𝑎𝑛 𝑅𝑑𝜃
2
W =   z d
1
(5-25)
(5-26)
If the torque remains constant while the angle changes by a finite amount, then the work done
by a constant torque is,
W =  z (2 − 1 ) =  z 
(5-27)
Total work done on a rotating rigid body is,
𝜔2
1
𝑊𝑡𝑜𝑡 = ∫ 𝐼𝜔𝑧 𝑑𝜔𝑧 = 𝐼𝜔2 2
2
𝜔1
1
− 𝐼𝜔1 2
2
(5-28)
The change in the rotational kinetic energy of a rigid body equals the work done by forces
exerted from outside the body. This equation is analogous to the work–energy theorem for a
particle.
The power associated with work done by a torque acting on a rotating body,
𝑃=
𝑑𝑊
𝑑𝜃
=𝜏
𝑑𝑡
𝑑𝑡
(5-29)
Note that 𝑑𝑊/𝑑𝑡 is just the rate of doing work and 𝑑𝜃/𝑑𝑡 is the angular velocity.
P =  zz
(5-30)
Sample Problem
What is the power output in horsepower of an electric motor turning at 4800 rev/min and
developing a torque of 4.30 N-m?
Solution
98
From the equation of power in rotational motion,
P =  zz
=
 4800rev 2rad 1min 
x

 = 2161 Watts
1rev
60s 
 min
 = (4.30N .m )
The power output of an electric motor is, P = 2161 W.
In horsepower, P = 2161W 
1hp
= 2.9hp
746W
5.14 Angular momentum
The analog of linear momentum discussed in chapter 4 of a particle is angular
⃗⃗), defined by
momentum (𝑳
   

L = r  p = r  mv
(5-31)
where,

p - linear momentum

r - position vector relative to point O
v - velocity of the particle
m - constant mass of the particle


 - angle between r and p
of an inertial frame
Consider a thin slice of the body lying in xy-plane as shown in
Figure 5.9 and rotating about z-axis with angular speed . Each
particle in the slice moves in a circle centered in the origin O, and its
velocity v i at each instant perpendicular to its position vector ri
The direction of angular momentum Li is determined by right-hand
Figure 5.9: Rotating thin slice of
the body.
rule and the magnitude is
𝐿𝑖 = 𝑚𝑖 (𝑟𝑖 𝜔)𝑟𝑖 = 𝑚𝑖 𝑟𝑖2 𝜔
(5-32)
The total angular momentum of the slice is the sum of Li of particles
L =  Li = ( mi ri 2 ) = I
(5-33)
The angular momentum for rigid body rotating around a symmetry axis


L = I
(5-34)
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Then the sum of external torque for any system of particle is given by

 dL
 =
dt
(5-35)
Sample Problem
A turbine fan of an engine has a moment of inertia of 2.5 kg.m 2
about its axis of rotation. As the turbine starting up, its angular
velocity as a function of time is 
= (40rad / s 3 )t 2 .
a) Find the fans angular momentum as a function of time, and
find its value at time t=3.0s.
b) Find the net torque acting on the fan as a function of time,
and find the torque at time t=3.0s.
Solution
a) The only component of angular momentum is along the rotation (z) axis
100kg  m 2 2
L = I =  2.5kg  m 2  40rad / s 3 t 2 = (
)t
z
z 


s3
At t=3.0s : L = (100kg  m 2 / s 3 )(3.0s) 2 = 900kg  m 3 / s
z
b)
 =
z
dL
z = (100kg  m 2 / s 3 )( 2t ) = (200kg  m 2 / s 3 )t
dt
At t=3.0s :
 = (200kg  m 2 / s 3 )(3.0s) = 600kg  m 2 / s 2 = 600N  m
z
5.15 Conservation of Angular Momentum
The principle of conservation of angular momentum states that:
“When a net external Torque acting on a system is zero, then the total angular momentum of
the system is constant (conserved).”
 = 0 then
dL
= 0 and L is constant.
dt
(5-37)
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Sample Problem
An acrobatic physics professor stands at the center of a
turntable, holding his arms extended horizontally with a
5.0-kg dumbbell in each hand. He is set about a vertical
axis, making 1 rev in 2.0s. Find the Prof’s new angular
velocity if he pulls the dumbbells into his stomach.
Solution
The moment of inertia of the system is
I = I prof + I dumbells ; I dumbells = 2(mr 2 )
Initially,
I1 = 3.0kg  m2 + 2(5.0kg)(1.0m) 2 = 13kg  m2
1z =
1rev
= 0.50rev / s
2. 0 s
The final moment of inertia
I 2 = 2.2kg  m2 + 2(5.0kg)(0.20m2 ) = 2.6kg  m2
2 z =
I1
13kg  m 2
1z =
(0.50rev / s) = 2.5rev / s
I2
2.6kg  m 2
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