Piri Reis University 2011-2012/ Physics -I
Physics for Scientists &
Engineers, with Modern
Physics, 4th edition
Giancoli
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Piri Reis University 2011-2012 Fall Semester
Physics -I
Chapter 10
Rotational Motion
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Piri Reis University 2011-2012
Lecture X
I. Angular Quantities
II. Constant Angular Acceleration
III. Rolling Motion (Without Slipping)
IV. Torque
V. Rotational Dynamics; Torque and Rotational Inertia
VI. Solving Problems in Rotational Dynamics
VII. Rotational Kinetic Energy
VIII. Angular Momentum and Its Conservation
IX. Vector Nature of Angular Quantities
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I. Angular Quantities
In purely rotational motion, all points on the
object move in circles around the axis of
rotation (“O”). The radius of the circle is r. All
points on a straight line drawn through the axis
move through the same angle in the same
time. The angle θ in radians is defined:
where l is the arc length.
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I. Angular Quantities
Angular displacement:
The average angular velocity is defined as the
total angular displacement divided by time:
The instantaneous angular velocity:
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I. Angular Quantities
The angular acceleration is the rate at which the angular velocity
changes with time:
The instantaneous acceleration:
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I. Angular Quantities
Every point on a rotating body has an angular velocity ω and a linear
velocity v.
They are related:
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I. Angular Quantities
Therefore, objects farther from
the axis of rotation will move
faster.
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I. Angular Quantities
If the angular velocity of a rotating
object changes, it has a tangential
acceleration:
Even if the angular velocity is constant,
each point on the object has a
centripetal acceleration:
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I. Angular Quantities
Here is the correspondence between linear and rotational quantities:
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I. Angular Quantities
The frequency is the number of complete revolutions per second:
Frequencies are measured in hertz.
The period is the time one revolution takes:
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II. Constant Angular Acceleration
The equations of motion for constant angular acceleration are the same as
those for linear motion, with the substitution of the angular quantities for
the linear ones.
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III. Rolling Motion (Without Slipping)
In (a), a wheel is rolling without slipping. The
point P, touching the ground, is instantaneously
at rest, and the center moves with velocity v.
In (b) the same wheel is seen from a reference
frame where C is at rest. Now point P is moving
with velocity –v.
The linear speed of the wheel is related to its
angular speed:
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IV. Torque
To make an object start rotating, a force is needed; the position and direction of
the force matter as well.
The perpendicular distance from the axis of rotation to the line along which the
force acts is called the lever arm.
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IV. Torque
A longer lever arm is very
helpful in rotating objects.
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IV. Torque
Here, the lever arm for FA is the distance from the knob to the hinge; the
lever arm for FD is zero; and the lever arm for FC is as shown.
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IV.Torque
The torque is defined as:
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V. Rotational Dynamics; Torque and
Rotational Inertia
Knowing that
, we see that
This is for a single point mass; what
about an extended object?
As the angular acceleration is the
same for the whole object, we can
write:
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V. Rotational Dynamics; Torque and
Rotational Inertia
The quantity
object.
is called the rotational inertia of an
The distribution of mass matters here – these two objects have the same
mass, but the one on the left has a greater rotational inertia, as so much
of its mass is far from the axis of rotation.
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V. Rotational Dynamics; Torque and
Rotational Inertia
The rotational inertia of an object
depends not only on its mass
distribution but also the location of
the axis of rotation – compare (f)
and (g), for example.
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Rotational Dynamics: I
F  m a  m ( R  )   (torque)  F  R  (m R 2 ) 
(a)  I  m R
2
(b)  I   m i Ri
2
(moment of inertia
(moment of inertia
for single particle)
for a group of particles)
m2
m1
m3
Rotational Motion
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Rotational Dynamics: I
I  I cm (moment of intertia about the c.m.)
d
I  I cm  Md 2 (parallel - axis theorem)
Rotational Motion
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Parallel-axis Theorem
d
I  I cm  Md
2
1
2
2
 MR 0  MR 0
2
3
2
 MR 0
2
Rotational Motion
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Parallel-axis Theorem
2
1
l
1
l
2
2
Ig  If  M    M l  M  M l
4 3
 2  12
2
Rotational Motion
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VI. Solving Problems in Rotational Dynamics
1. Draw a diagram.
2. Decide what the system comprises.
3. Draw a free-body diagram for each object under consideration,
including all the forces acting on it and where they act.
4. Find the axis of rotation; calculate the torques around it.
5. Apply Newton’s second law for rotation. If the rotational inertia is not
provided, you need to find it before proceeding with this step.
6. Apply Newton’s second law for translation and other laws and
principles as needed.
7. Solve.
8. Check your answer for units and correct order of magnitude.
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VII. Rotational Kinetic Energy
The kinetic energy of a rotating object is given by
By substituting the rotational quantities, we find that the rotational kinetic
energy can be written:
A object that has both translational and rotational motion also has both
translational and rotational kinetic energy:
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VII. Rotational Kinetic Energy
When using conservation of energy, both rotational and translational
kinetic energy must be taken into account. All these objects have the same
potential energy at the top, but the time it takes them to get down the
incline depends on how much rotational inertia they have.
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VII. Rotational Kinetic Energy
The torque does work as it moves the wheel through an angle θ:
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VIII. Angular Momentum and Its Conservation
In analogy with linear momentum, we can define angular momentum L:
We can then write the total torque as being the rate of change of angular
momentum.
If the net torque on an object is zero, the total angular momentum is
constant.
Therefore, systems that can change their rotational inertia through internal
forces will also change their rate of rotation:
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IX. Vector Nature of Angular Quantities
The angular velocity vector points along the axis of rotation; its direction
is found using a right hand rule:
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IX. Vector Nature of Angular Quantities
Angular acceleration and angular
momentum vectors also point
along the axis of rotation.
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Summary of Chapter 10
• Angles are measured in radians; a whole circle is 2π radians.
• Angular velocity is the rate of change of angular position.
• Angular acceleration is the rate of change of angular velocity.
• The angular velocity and acceleration can be related to the linear velocity and acceleration.
• The frequency is the number of full revolutions per second; the period is the inverse of the
frequency.
•The equations for rotational motion with constant angular acceleration have the same form as
those for linear motion with constant acceleration.
• Torque is the product of force and lever arm.
• The rotational inertia depends not only on the mass of an object but also on the way its mass is
distributed around the axis of rotation.
• The angular acceleration is proportional to the torque and inversely proportional to the rotational
inertia.
•An object that is rotating has rotational kinetic energy. If it is translating as well, the translational
kinetic energy must be added to the rotational to find the total kinetic energy.
• Angular momentum is
• If the net torque on an object is zero, its angular momentum does not change.
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Example 1
An automobile wheel has a radius of 42 cm. If a car drives 10 km,
through what angle has the wheel rotated?
a) In revolutions
b) In radians
c) In degrees
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Solution of Ex.1
Note distance car moves = distance outside of wheel moves
a) Find N:
Basic formula
s  2rN
 rq
Known: s = 10 000 m, r = 0.42 m
N
s
2r
= 3 789
b) Find q in radians
Known: N
q  2 (radians/r evolution) N
c) Find q in degrees
Known: N
q = 2.38 x 104 rad.
q  360 (degrees/r evolution) N
q = 1.36 x 106 deg
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Example 2
A race car engine can turn at a maximum rate of 12 000 rpm. (revolutions
per minute).
a) What is the angular velocity in radians per second.
b) If helipcopter blades were attached to the crankshaft while it turns with
this angular velocity, what is the maximum radius of a blade such that the
speed of the blade tips stays below the speed of sound.
DATA: The speed of sound is 343 m/s
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Solution of Ex.2
a) Convert rpm to radians per second
 rev. 
12 000

 min   2  rad  = 1256 radians/s


rev 
 sec 

60 

 min 
b) Known: v = 343 m/s,  = 1256 rad./s
Find r
Basic formula
v  r
r
v

= 0.27 m
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Example 3.
A pottery wheel is accelerated uniformly from rest to a rate of 10 rpm
in 30 seconds.
a) What was the angular acceleration? (in rad/s2)
b)How many revolutions did the wheel undergo during that time?
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Solution of Ex.3
First, find the final angular velocity in radians/s.
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 rev. 
 rad 
 rad 
 f  10
 2 

  1.047 

 min  60sec/ min 
 rev 
 sec 
a) Find angular acceleration
Basic formula
 f  i
 t  
 f  i
= 0.0349 rad./s2
t
b) Find number of revolutions: Known i=0, f =1.047, and t = 30
First find Dq in radians
Dq 
Basic formula
Dq 
i   f
2
t
f
t  15.7 rad.
2
Dq (rad.)
N
= 2.5 rev.
2 (rad./rev. )
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Solution of Ex.3
b) Find number of revolutions:
Known i=0, f =1.047, and t = 30,
First find Dq in radians
Basic formula
Dq 
i   f
2
Dq 
t
f
t  15.7 rad.
2
Dq (rad.)
= 2.5 rev.
N
2 (rad./rev. )
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Example 4.
A coin of radius 1.5 cm is initially rolling with a rotational speed of 3.0
radians per second, and comes to a rest after experiencing a slowing
down of  = 0.05 rad/s2.
a) Over what angle (in radians) did the coin rotate?
b) What linear distance did the coin move?
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Solution of Ex.4
a) Find Dq, Given i= 3.0 rad/s, f = 0,  = -0.05 rad/s2
Basic formula
2
2
f
i
    2Dq
Dq 

2
i
 2
= 90 radians = 90/2 revolutions
b) Find s, the distance the coin rolled
Given: r = 1.5 cm and Dq = 90 rad
Basic formula
s  rDq
s  rDq , (Dq is in rad.s)
= 135 cm
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Example 5.
A space-station is constructed like a barbell with two 1000-kg
compartments separated by 50 meters that spin in a circle (r=25 m).
The compartments spins once every 10 seconds.
a) What is the acceleration at the extreme end of the compartment?
Give answer in terms of “g”s.
b) If the two compartments are held together by a cable, what is the
tension in the cable?
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Solution of Ex.5
a) Find acceleration a
Given: T = 10 s, r = 25 m
Basic formula
First, find  in rad/s
 rad   rev 
  2 
 N
   2  0.1
 rev   s 
Basic formula
2
v
2
a
 r
r
Then, find acceleration
a   r = 9.87 m/s2 = 1.006 g
2
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Solution of Ex.5
b) Find the tension
Given m = 1000 kg, a = 1.006 g
Basic formula
F  ma
T  ma = 9870 N
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Example 6.
A race car speeds around a circular track.
a) If the coefficient of friction with the tires is 1.1, what is the
maximum centripetal acceleration (in “g”s) that the race car can
experience?
a) Find the maximum centripetal acceleration
Known: m  1.1
Remember, only consider forces towards center
Basic formula
f  mn
F  ma
f  mmg
ma  mmg
a  mg
Maximum a = 1.1 g
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Solution of Ex.6(b)
b) What is the minimum circumference of the track that would permit
the race car to travel at 300 km/hr?
b) Find the minumum circumference
Known: v = 300 km/hr = 83.33 m/s, a = 1.1 g
Basic formula
2
First, find radius
v2
r
a
v
2
a
 r
r
Then, find circumference
L  2r
= 4.043 m
In the real world: tracks are banked
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Example 7.
A yo-yo is spun in a circle as shown. If the
length of the string is L = 35 cm and the
circular path is repeated 1.5 times per
second, at what angle q (with respect to the
vertical) does the string bend?
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Solution of Ex.7


F  ma
Basic formula
Apply F=ma for both
the horizontal and
vertical components.
may  0
 Fy  T cosq  mg
T cosq  mg
Basic formula
2
a  r
max  m r  m L sin q
 F2 x  T sin q
m L  T
2
2
r = Lsinq
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Solution of Ex.7
We want to find q, given =2·1.5 & L=0.35
max  m r  m L sin q
 F2 x  T sin q
m L  T
2
2
may  0
 Fy  T cosq  mg
T cosq  mg
2 eq.s & 2 unknowns (T and q)
mg
g
cos q 
 2
T
 L
q = 71 degrees
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HOMEWORK
Giancoli, Chapter 10
4, 12, 19, 22, 26, 27, 30, 36, 41, 46
References
o “Physics For Scientists &Engineers with Modern Physics” Giancoli 4th edition,
Pearson International Edition
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