Lecture 7:Rotational motion
Physics 1, NTC
Angular Motion, General Notes
•When a rigid object rotates about a fixed axis in a given time interval,
every portion on the object rotates through the same angle in a given
time interval and has the same angular speed and the same angular
acceleration.
• So q, w, a all characterize the motion of the entire rigid object as well as the
individual particles in the object.
Section 10.1
Angular Position
•As the particle moves, the only
coordinate that changes is q
•As the particle moves through q, it
moves though an arc length s.
•The arc length and r are related:
•s = q r
Section 10.1
Conversions
•Comparing degrees and radians
360
1 rad 
 57.3
2
•Converting from degrees to radians
q  rad  

180
q  degrees 
Section 10.1
Directions
•Strictly speaking, the speed and
acceleration (w, a are the
magnitudes of the velocity and
acceleration vectors.
•The directions are actually given by
the right-hand rule.
Section 10.1
Relationship Between Angular and Linear
Quantities
•Every point on the rotating object has the same angular motion.
•Every point on the rotating object does not have the same linear
motion.
•Displacements
• s=θr
•Speeds
• v=ωr
•Accelerations
ds
dq
v
r
 rw
dt
dt
• a=αr
Section 10.3
Acceleration Comparison
•The tangential acceleration is the
derivative of the tangential velocity.
dv
dw
at 
r
 ra
dt
dt
•Centripetal acceleration
v2
aC 
 rw2
r
Section 10.3
Rotational Kinetic Energy
•The total rotational kinetic energy of the rigid object is the sum of the
energies of all its particles.
1
K R   K i   mi ri 2w 2
i
i 2
1
1 2
2 2
K R    mi ri w  Iw
2 i
2

•I is called the moment of inertia.
•Rotational kinetic energy is not a new type of energy, the form is different because it is applied to a rotating
object.
•The units of rotational kinetic energy are Joules (J).
Section 10.4
Moment of Inertia
I   ri mi
2
•Defined by
dimensions =
ML2
or unit of
i
kg.m2.
•Mass = inherent property, but the moment of inertia depends on the
choice of rotational axis.
•Moment of inertia is a measure of the resistance of an object to
changes in its rotational motion, similar to mass being a measure of an
object’s resistance to changes in its translational motion.
• The moment of inertia depends on the mass and how the mass is distributed
around the rotational axis.
Section 10.5
Moment of Inertia, cont
•For a continuous rigid object, imagine the object to be divided into
many small elements, each having a mass of Δmi.
I
lim
mi 0
r
i
i
2
mi   r dm
2
I   r r dV
2
•If r is constant, the integral can be evaluated with known geometry,
otherwise its variation with position must be known.
Section 10.5
Moments of Inertia of Various Rigid Objects
Section 10.5
Moment of Inertia of a Uniform Rigid Rod
•The shaded area has a mass
• dm = l dx
•Then the moment of inertia is
M
I y   r dm  
x
dx
L / 2
L
1
2
I
ML
12
2
L/2
2
Section 10.5
Moment of Inertia of a Uniform Solid Cylinder
•Divide the cylinder into concentric shells with radius
r, thickness dr and length L.
•dm = r dV = 2rLr) dr
•Then for I
Iz   r 2dm 
2
r
  2  r L r dr 
1
Iz  MR 2
2
Section 10.5
9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis
Example 9 The Moment of Inertia Depends on Where
the Axis Is.
Two particles each have mass and are fixed at the
ends of a thin rigid rod. The length of the rod is L.
Find the moment of inertia when this object
rotates relative to an axis that is
perpendicular to the rod at
(a) one end and (b) the center.
(a)
 
I   mr 2  m1r12  m2 r22  m0  mL 
m1  m2  m
I  mL
2
2
2
r1  0 r2  L
9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis
 
(b) I   mr 2  m1r12  m2 r22  mL 22  mL 22
m1  m2  m
I  mL
1
2
2
r1  L 2 r2  L 2
Parallel-Axis Theorem
•In the previous examples, the axis of rotation coincided with the axis
of symmetry of the object.
•For an arbitrary axis, the parallel-axis theorem often simplifies
calculations.
•The theorem states I = ICM + MD 2
• I is about any axis parallel to the axis through the center of mass of the object.
• ICM is about the axis through the center of mass.
• D is the distance from the center of mass axis to the arbitrary axis.
Section 10.5
Moment of Inertia for a Rod Rotating Around One End
– Parallel Axis Theorem Example
•The moment of inertia of the rod
about its center is
ICM
1

ML2
12
•D is ½ L
•Therefore,
I  ICM  MD 2
2
1
1 2
L
2
I
ML  M    ML
12
3
2
Section 10.5
Torque
•Torque, t, is the tendency of a force to rotate an object about some
axis.
• Torque is a vector, but we will deal with its magnitude here:
• t = r F sin f = F d
• F is the force
• f is the angle the force makes with the horizontal
• d is the moment arm (or lever arm) of the force
• There is no unique value of the torque on an object.
• Its value depends on the choice of a rotational axis.
Section 10.6
Torque is a vector!
•The moment arm, d, is the perpendicular
distance from the axis of rotation to a line
drawn along the direction of the force.
• d = r sin Φ
•The horizontal component of the force (F
cos f) has no tendency to produce a
rotation.
•Torque will have direction.
• If the turning tendency of the force is
counterclockwise, the torque will be
positive.
• If the turning tendency is clockwise,
the torque will be negative.
Section 10.6
Net Torque
•The force F1 will tend to cause a
counterclockwise rotation about O.
•The force F2 will tend to cause a
clockwise rotation about O.
•St  t1  t2  F1d1 – F2d2
Section 10.6
Torque and Angular Acceleration, Extended
•Consider the object consists of an infinite number of mass
elements dm of infinitesimal size.
•Each mass element rotates in a circle about the origin, O.
•Each mass element has a tangential acceleration.
•From Newton’s Second Law
• dFt = (dm) at
•The torque associated with the force and using the angular
acceleration gives
• d ext = r dFt = atr dm = r 2 dm
t
• This becomes
a
St  Ia
Section 10.7
Torque and Angular Acceleration, Extended cont.
•rigid body under a net torque.
•The result also applies when the forces have radial components.
• The line of action of the radial component must pass through the axis of
rotation.
• These components will produce zero torque about the axis.
Section 10.7
Falling Smokestack Example
•When a tall smokestack falls over,
it often breaks somewhere along its
length before it hits the ground.
•Each higher portion of the
smokestack has a larger tangential
acceleration than the points below
it.
•The shear force due to the
tangential acceleration is greater
than the smokestack can withstand.
•The smokestack breaks.
Section 10.7
9.3 Center of Gravity
Conceptual Example 7 Overloading a Cargo Plane
This accident occurred because the plane was overloaded toward
the rear. How did a shift in the center of gravity of the plane cause
the accident?
Summary of Useful Equations
Section 10.8
9.2 Rigid Objects in Equilibrium
Example 3 A Diving Board
A woman whose weight is 530 N is
poised at the right end of a diving board
with length 3.90 m. The board has
negligible weight and is supported by
a fulcrum 1.40 m away from the left
end.
Find the forces that the bolt and the
fulcrum exert on the board.
9.2 Rigid Objects in Equilibrium
t  F 
2 2
 W W  0
W W
F2 
2
F2

530 N 3.90 m 

 1480 N
1.40 m
9.2 Rigid Objects in Equilibrium
F
y
 F1  F2  W  0
 F1  1480 N  530 N  0
F1  950 N
9.2 Rigid Objects in Equilibrium
Example 5 Bodybuilding
The arm is horizontal and weighs 31.0 N. The deltoid muscle can supply
1840 N of force. What is the weight of the heaviest dumbell he can hold?
9.2 Rigid Objects in Equilibrium
t  W 
a a
 Wd  d  M M  0
 M  0.150 msin 13.0
9.2 Rigid Objects in Equilibrium
 Wa  a  M M
Wd 
d
 31.0 N 0.280 m   1840 N 0.150 m sin 13.0

 86.1 N
0.620 m
9.3 Center of Gravity
DEFINITION OF CENTER OF GRAVITY
The center of gravity of a rigid
body is the point at which
its weight can be considered
to act when the torque due
to the weight is being calculated.
9.3 Center of Gravity
When an object has a symmetrical shape and its weight is distributed
uniformly, the center of gravity lies at its geometrical center.
9.3 Center of Gravity
W1 x1  W2 x2  
xcg 
W1  W2  
9.3 Center of Gravity
Example 6 The Center of Gravity of an Arm
The horizontal arm is composed
of three parts: the upper arm (17 N),
the lower arm (11 N), and the hand
(4.2 N).
Find the center of gravity of the
arm relative to the shoulder joint.
9.3 Center of Gravity
W1 x1  W2 x2  
xcg 
W1  W2  
xcg

17 N 0.13 m   11 N 0.38 m   4.2 N 0.61 m 

 0.28 m
17 N  11 N  4.2 N
Energy in an Atwood Machine, Example
•The system containing the two blocks, the pulley,
and the Earth is an isolated system in terms of energy
with no non-conservative forces acting.
•The mechanical energy of the system is conserved.
•The blocks undergo changes in translational kinetic
energy and gravitational potential energy.
•The pulley undergoes a change in rotational kinetic
energy.
•Find 𝒂, 𝑻𝟏 , 𝑻𝟐 ?
Section 10.8
Rolling Object
•The red curve shows the path moved by a point on the rim of the
object.
•
This path is called a cycloid.
•The green line shows the path of the center of mass of the object.
•In pure rolling motion, an object rolls without slipping.
•In such a case, there is a simple relationship between its rotational
and translational motions.
Section 10.9
The tangential speed of a
point on the outer edge of
the tire is equal to the speed
of the car over the ground.
v  rw
a  ra
Pure Rolling Motion, Object’s Center of Mass
•The translational speed of the center of
mass is
ds
dq
v CM 
R
 Rw
dt
dt
•The linear acceleration of the center of
mass is
aCM
dvCM
dw

R
 Ra
dt
dt
Section 10.9
Rolling Motion Cont.
•Rolling motion can be modeled as a combination of pure translational motion
and pure rotational motion.
•The contact point between the surface and the cylinder has a translational
speed of zero (c).
Section 10.9
Total Kinetic Energy of a Rolling Object
•The total kinetic energy of a rolling object is the sum of the
translational energy of its center of mass and the rotational kinetic
energy about its center of mass.
K = ½ ICM w2 + ½ MvCM2
• The ½ ICMw2 represents the rotational kinetic energy of the cylinder about its center of
mass.
• The ½ Mv2 represents the translational kinetic energy of the cylinder about its center of
mass.
Section 10.9
Total Kinetic Energy, Example
•Accelerated rolling motion is possible only if
friction is present between the sphere and the
incline.
• The friction produces the net torque
required for rotation.
• No loss of mechanical energy occurs
because the contact point is at rest
relative to the surface at any instant.
• In reality, rolling friction causes
mechanical energy to transform to
internal energy.
• Rolling friction is due to deformations of
the surface and the rolling object.
Section 10.9
Find 𝑎 and 𝑣, 𝜔 at the
bottom of the incline?
9.5 Rotational Work and Energy
Example 13 Rolling Cylinders
A thin-walled hollow cylinder (mass = mh, radius = rh) and
a solid cylinder (mass = ms, radius = rs) start from rest at
the top of an incline.
Determine which cylinder
has the greatest translational
speed upon reaching the
bottom.
9.5 Rotational Work and Energy
E  12 mv 2  12 Iw 2  mgh
ENERGY CONSERVATION
1
2
1
2
mv2f  12 Iw 2f  mghf  12 mvi2  12 Iwi2  mghi
mv2f  12 Iw 2f  mghi
wf  vf r
9.5 Rotational Work and Energy
1
2
mv2f  12 I v 2f r 2  mghi
2mgho
vf 
m  I r2
The cylinder with the smaller moment
of inertia will have a greater final translational
speed.
Lecture 8: Angular Momentum
Vector Product (cross product)
•The quantity AB sin q is
equal to the area of the
parallelogram formed by
𝐴 𝑎𝑛𝑑 𝐵
•The direction of 𝐶 = 𝐴 × 𝐵
is perpendicular to the plane
formed by 𝐴, 𝐵
.
•The best way to determine
this direction is to use the
right-hand rule.
Section 11.1
Properties of the Vector Product
•The vector product is not commutative. The order in
which the vectors are multiplied is important.
A  B  B  A
• If A // B (parallel q = 0o or 180o), then
• Therefore
A A  0
A B  0
A  B  AB
•If A is perpendicular to B, then
•The vector product obeys the distributive law.
A x (B + C) = A x B + A x C
Section 11.1
Final Properties of the Vector Product
•The derivative of the cross product with respect to some variable such
as t is


d
dA
dB
A B 
B  A 
dt
dt
dt
where it is important to preserve the multiplicative order of the
vectors.
Section 11.1
Vector Products of Unit Vectors
ˆi  ˆi  ˆj  ˆj  kˆ  kˆ  0
ˆi  ˆj  ˆj  ˆi  kˆ
ˆj  kˆ  kˆ  ˆj  ˆi
kˆ  ˆi  ˆi  kˆ  ˆj
Section 11.1
Signs in Cross Products
•Signs are interchangeable in cross products
•
• and
 
A  -B  A  B
 
ˆi   ˆj  ˆi  ˆj
Section 11.1
Using Determinants
•The cross product can be expressed as
ˆi
ˆj
kˆ
A  B  Ax
Ay
Az 
Bx
By
Bz
Ay
By
Az
ˆi  Ax
Bz
Bx
Az ˆ Ax
j
Bx
Bz
Ay
By
kˆ
•Expanding the determinants gives
A  B   Ay Bz  Az By  ˆi   Ax Bz  Az Bx  ˆj   Ax By  Ay Bx  kˆ
Section 11.1
Vector Product Example
•Given
•Find
•Result
A  2ˆi  3ˆj; B  ˆi  2ˆj
A B
A  B  (2ˆi  3ˆj)  ( ˆi  2ˆj)
 2ˆi  ( ˆi )  2ˆi  2ˆj  3ˆj  ( ˆi )  3ˆj  2ˆj
 0  4kˆ  3kˆ  0  7kˆ
Section 11.1
Torque Vector Example
•Given the force and location
F  (2.00 ˆi  3.00 ˆj) N
r  (4.00 ˆi  5.00 ˆj) m
•Find the torque produced
t  r  F  [(4.00ˆi  5.00ˆj)N]  [(2.00ˆi  3.00ˆj)m]
 [(4.00)(2.00)ˆi  ˆi  (4.00)(3.00)ˆi  ˆj
(5.00)(2.00)ˆj  ˆi  (5.00)(3.00)ˆi  ˆj
 2.0 kˆ N  m
Section 11.1
Angular Momentum
•Consider a particle of mass m located at the vector
position r and moving with linear momentum p .
•Find the net torque.
r   F  t  r 
dp
dt
dr
Add the term
 p  sinceit  0 
dt
d (r  p )
t  dt
•This looks very similar to the equation for the net force
in terms of the linear momentum since the torque plays
the same role in rotational motion that force plays in
translational motion.
Section 11.2
Angular Momentum, cont
•The instantaneous angular momentum
of a particle relative to the origin O is
defined as the cross product of the particle’s
instantaneous position vector and its
instantaneous linear momentum.
L  r p
Section 11.2
Torque and Angular Momentum
•The torque is related to the angular momentum.
• Similar to the way force is related to linear momentum.
t 
dL
dt
•The torque acting on a particle is equal to the time rate of change of
the particle’s angular momentum.
•This is the rotational analog of Newton’s Second Law .
• t and L must be measured about the same origin.
• This is valid for any origin fixed in an inertial frame.
Section 11.2
Angular Momentum
•The SI units of angular momentum are (kg.m2)/ s.
•Both the magnitude and direction of the angular momentum depend
on the choice of origin.
•The magnitude is L = mvr sin f
• f is the angle between r and p .
•The direction of L is perpendicular to the plane formed by r and p
Section 11.2
.
Angular Momentum of a Particle, Example
•The vector
diagram.
L = r  p is pointed out of the
•The magnitude is L = mvr sin 90o = mvr
• sin 90o is used since v is perpendicular to r.
•A particle in uniform circular motion has a
constant angular momentum about an axis
through the center of its path.
Section 11.2
Angular Momentum of a Rotating Rigid Object
•The rigid object is a non-deformable system.
•Each particle of the object rotates in the xy plane
about the z axis with an angular speed of w
•The angular momentum of an individual particle is
Li = mi ri2 w
•L and 𝝎 are directed along the z axis.
𝑳 = 𝐼𝝎
Section 11.3
9.6 Angular Momentum
DEFINITION OF ANGULAR MOMENTUM
The angular momentum L of a body rotating about a
fixed axis is the product of the body’s moment of
inertia and its angular velocity with respect to that
axis:
L  Iw
Requirement: The angular speed must
be expressed in rad/s.
SI Unit of Angular Momentum: kg·m2/s
9.6 Angular Momentum
PRINCIPLE OF CONSERVATION OFANGULAR MOMENTUM
The angular momentum of a system remains constant (is
conserved) if the net external torque acting on the system
is zero.
9.6 Angular Momentum
Conceptual Example 14 A Spinning Skater
An ice skater is spinning with both
arms and a leg outstretched. She
pulls her arms and leg inward and
her spinning motion changes
dramatically.
Use the principle of conservation
of angular momentum to explain
how and why her spinning motion
changes.
9.6 Angular Momentum
Example 15 A Satellite in an Elliptical Orbit
An artificial satellite is placed in an
elliptical orbit about the earth. Its point
of closest approach is 8.37x106m
from the center of the earth, and
its point of greatest distance is
25.1x106m from the center of
the earth.
The speed of the satellite at the
perigee is 8450 m/s. Find the speed
at the apogee.
9.6 Angular Momentum
L  Iw
angular momentum conservation
I AwA  I PwP
I  mr 2 w  v r
vA
2 vP
mr
 mrP
rA
rP
2
A
9.6 Angular Momentum
vA
2 vP
mr
 mrP
rA
rP
2
A
rA v A  rP vP


rP vP
8.37 106 m 8450 m s 
vA 

 2820 m s
6
rA
25.110 m