How to Prepare Rotational Motion (Physics) for JEE Main

Cross Product
 The Cross Product (or Vector Product) of two vectors A and B is
a multiplication of vectors where the result is a vector quantity C
with a direction perpendicular to both vectors A and B, and the
magnitude equal to ABsin :
sinABC  Magnitude of the cross (vector)
product of two vectors A and BBAC


 A is the magnitude of the first vector, B is the magnitude of the
second vector and  is the angle between the two vectors.
 The direction of the cross product is perpendicular to the plane
formed by the two vectors in the product. This leaves two possible
choices which are resolved by using the Right Hand Rule.

Cross Product
Properties of the Cross Product
 Cross Product is Anti-Commutative.
 Parallel Vectors have Cross Product of zero.
 Cross Product obeys the Distributive Law.
 Product Rule for Derivative of a Cross Product.
ABBA


CABACBA

 )(
B
dt
Ad
dt
Bd
ABA
dt
d 


 )(

Cross Product
 The above formula for the cross product is useful when the magnitudes
of the two vectors and the angle between them are known.
 If you only know the components of the two vectors:
kBABAjBABAiBABABA xyyxzxxzyzzy
ˆ)(ˆ)(ˆ)( 

sinABC 
yzzyx BABAC 
zxxzy BABAC 
xyyxz BABAC 
BAC


Components of cross
product vector

Cross Product
 Right-handed coordinate
system, in which:
kji ˆˆˆ 
ikj ˆˆˆ 
jik ˆˆˆ 
 Left-handed coordinate
system, in which:
kji ˆˆˆ 
ikj ˆˆˆ 
jik ˆˆˆ 

Torque
 Net force applied to a body gives that body an acceleration.
 What does it take to give a body angular acceleration?
 Force is required! It must be applied in a way that gives a twisting or
turning action.
The quantative measure
of the tendency of a force
to cause or change the
rotational motion of a
body is called torque.
 This body can rotate about axis
through O,  to the plane. It is acted
by three forces (in the plane of
figure). The tendency of any force to
cause the rotation depends on its
magnitude and on the perpendicular
distance (lever arm) between the
line of action of the force and point O.

Torque
 Torque is a vector quantity that measures the tendency of a force to
rotate an object about an axis. The magnitude of the torque produced
by a force is defined as
rFrFFl tansin  
 where
r = distance between the pivot point and the point of application of the
force.
F = the magnitude of the force.
 = the angle between the force and a line extending thru the pivot and
the point of application.
Ftan = F sin() = the component of the force perpendicular to the line
connecting the pivot and the point of application.
L = r sin() = moment arm or lever arm = distance from the pivot to
the line of action of the force.

Torque
Fr


Some important points about torque
 Torque has units of N·m. Despite the fact that this unit is the same as a
Joule it is customary to leave torque expressed in N·m (or foot·pounds).
 Engineers will often use the term "moment" to describe what physicists
call a "torque".
 We will adopt a convention that defines torques that tend to cause
clockwise rotation as negative and torques that tend to cause counter-
clockwise rotation as positive.
 Torques are always defined relative to a point. It is incorrect to simply
say the "torque of F". Instead you must say the "torque of F relative to
point X".
 More general definition for the torque is given by the vector (or cross
product). When a force acting at a point which has position vector r
relative to an origin O the torque exerted by the force about the origin is
defined as

Torque and Angular Acceleration for a
Rigid Body

Torque and Angular Acceleration for a Rigid
Body
 If we consider a rigid body rotating about a fixed axis as made up of a collection
of individual point particles, all of which have to obey Newton's Second Law for
a particle, then we can show that the net torque acting on the body about the
given axis of rotation will equal the moment of inertia of the body about that
axis times the angular acceleration.
tan,11tan,1 amF 
 Z-axis is the axis of rotation; the first particle has
mass m1 and distance r1 from the axis of rotation.
 The net force acting on this article has a component
F1,rad along the radial direction, a component F1,tan that
is tangent to the circle of radius r1 in which particle
moves, and component F1z along axis of rotation.
 N2L for tangential component is
zra 1tan,1  zrmrF 2
111tan,1  zz I  11 
F1,rad and F1z do not contribute to the torque about z-axis.
z
i
iz
i
ii
i
iz Irm  











  2
For all particles

Body
 This expression is the rotational form of Newton's Second Law
for rigid body motion (for a fixed axis of rotation):
  zz I N2L for a rigid body in rotational form
 Valid only for rigid bodies! If the body is not rigid like a rotating tank
of water, the angular acceleration is different for different particles.
 z must be measured in rad/s2 (we used atan=rz in derivation)
 The torque on each particle is due to the net force on that particle,
which is the vector sum of external and internal forces.
 According to N3L, the internal forces that any pair of particles in the
rigid body exert on each other are equal and opposite. If these forces
act along the line joining the two particles, their lever arms with
respect to any axis are also equal. So the torques for each pair are
equal and opposite and add to ZERO.

Body
ONLY external
torques affect the
rigid body’s rotation!

N2L in Rotational Form
IDENTIFY the relevant concepts:
 The equation z=Iz is useful whenever torques act on a rigid body
- that is, whenever forces act on a rigid body in such a way as to
change the state of the body’s rotation.
 In some cases you may be able to use an energy approach instead.
However, if the target variable is a force, a torque, an acceleration,
an angular acceleration, or an elapsed time, the approach using
equation z=Iz is almost always the most efficient one.
Problem-Solving Strategy Rotational Dynamics
for Rigid Bodies

SET UP the problem using the following steps:
1. Draw a sketch of the situation and select the body or bodies to be
analyzed.
2. For each body, draw a free-body diagram isolating the body and
including all the forces (and only those forces) that act on the body,
including its weight. Label unknown quantities with algebraic symbols.
A new consideration is that you must show the shape of the body
accurately, including all dimensions and angles you will need for
torque calculations.
3. Choose coordinate axes for each body and indicate a positive sense
of rotation for each rotating body. If there is a linear acceleration, it’s
usually simplest to pick a positive axis in its direction. If you know the
sense of z in advance, picking it as the positive sense of rotation
simplifies the calculations. When you represent a force in terms of its
components, cross out the original force to avoid including it twice.
Problem-Solving Strategy

EXECUTE the solution as follows:
 For each body in the problem, decide whether it undergoes translational
motion, rotational motion, or both. Depending on the behavior of the
body in question, apply F=ma, z=Iz, or both to the body. Be
careful to write separate equations of motion for each body.
 There may be geometrical relations between the motions of two or
more bodies, as with a string that unwinds from a pulley while turning it
or a wheel that rolls without slipping. Express these relations in algebraic
form, usually as relations between two linear accelerations or between a
linear acceleration and an angular acceleration.
 Check that the number of equations matches the number of unknown
quantities. Then solve the equations to find the target variable(s).

EVALUATE your answer:
 Check that the algebraic signs of your results make sense.
 As an example, suppose the problem is about a spool of thread. If
you are pulling thread off the spool, your answers should not tell
you that the spool is turning in the direction that rolls the thread
back on the spool!
 Whenever possible, check the results for special cases or extreme
values of quantities and compare them with your intuitive
expectations.
 Ask yourself: “Does this result make sense?”

Rigid-Body Rotation about a
Moving Axis

Rigid-Body Rotation about a Moving
Axis
 Let’s extend analysis of rotational motion to
cases in which the axis of rotation moves: the
motion of a 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.
 It is applicable even when the center of mass
accelerates (so that is not at rest in any
inertial frame).
 The translation of the center of mass and the
rotation about the center of mass can be
treated as separate but related problems.
 The prove of all that is beyond of the scope
of this course. We will learn concept only.

Axis
 If a round object of cross-sectional radius R rolls without slipping then
the distance along the surface that the object covers will be the same as
the arc length along the edge of the circular object that has been in
contact with the surface (i.e. s = Rq).
 Differentiating this expression with respect to time shows that the speed
of the center of mass of the object will be given by
Rvcm  Condition for rolling without slipping
 This condition must be satisfied if an object is rolling without slipping.
Rolling motion can be thought of in two different ways:
 Pure rotation about the instantaneous point of contact (P) of the object
with the surface.
 Superposition of translation of the center of mass plus rotation about
the center of mass.

Axis
 The wheel is symmetrical, so its CM is at its geometric center.
 We view the motion in inertial frame of reference in which the surface is at rest. In
order not to slip, the point of contact (where the wheel contacts the ground) is
instantaneously at rest as well. Hence the velocity of the point of contact relative to
the CM must have the same magnitude but opposite direction as the CM velocity.
 If the radius of the wheel is R and its angular speed about CM is : vcm=R.

Axis
 The velocity of a point on the wheel is the vector sum of the velocity of CM and the
velocity of the point relative to the center of mass.
 Thus, the point of the contact is instantaneously at rest, point 3 at the top of the
wheel is moving forward twice as fast as the center of mass; points 2 and 4 at the
sides have velocities at 45 degrees to the horizontal.

Axis
 The kinetic energy of an object that is rolling without slipping is given
by the sum of the rotational kinetic energy about the center of mass
plus the translational kinetic energy of the center of mass:
22
2
1
2
1
cmcm IMvK  Rigid body with both
translation and rotation
 If a rigid body changes height as it moves, you must also consider
gravitational potential energy
 The gravitational potential energy associated with any extended body of
mass M, rigid or not, is the same as if you replace the body by a particle
of mass M located at the body’s center of mass:
cmMgyU 

Axis
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:
cmext aMF


zcmz I  
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

Work and Power in Rotational Motion
 The work done by a torque on an object that undergoes an angular
displacement from q1 to q2 is given by
q
q
q
dW z
2
1
Work done by a torque
 If the torque is constant then the work done is given by
qqq  zzW )( 12 Work done by a constant torque
 Note: similarity between these expressions and the equations for work
done by a force (W=FS).

Work and Power in Rotational Motion
 The rotational analog to the Work - Energy Theorem is
2
1
2
2
2
1
2
12
1



IIdIW zztot  
 The change in rotational kinetic energy of a rigid body equals the
work done by forces exerted from outside the body.
 The rate at which work is performed is the power
zzP 

Angular Momentum of a Particle.
Definition
 The angular momentum L of a particle relative to a point O is the
cross product of the particle's position r relative to O with the linear
momentum p of the particle.
vmrprL


Angular momentum of a particle
 The value of the angular momentum
depends on the choice of the origin O,
since it involves the position vector
relative to the origin
 The units of angular momentum: kg·m2/s
Mass m is moving in XY plane
Right-hand rule
“Lever
arm”

Angular Momentum of a Particle
 When a net force F acts on a particle, its velocity and linear momentum
change. Thus, angular momentum may also change.



 Framr
dt
Ld For a particle acted on by net force F
)()(
)(
amrvmv
dt
vmd
rvm
dt
rd
dt
vmrd
dt
Ld 


















 Rate of change of angular momentum L of a particle equals the
torque of the net force acting on it.
Vector product of
vector by itself = 0

Angular Momentum of a Rigid Body
 Rigid body rotating about Z-axis with angular speed 
 Consider a thin slice of the body lying in XY plane
 2
)( iiiiii rmrrmL 
 Each particle in the slice moves in a circle
centered in the origin O, and its velocity vi
at each instant  to its position vector ri
 Thus, =90°, and particle of the mass mi at
distance ri from O has speed vi=ri
 The direction of angular momentum Li is by
right-hand rule and the magnitude:
 The total angular momentum of the slice is the
sum of Li of particles:
   IrmLL iii   2

 For points not lying in XY plane, the position vectors have components in Z-direction
as well as in X- and Y-directions. This gives the angular momentum of each particle
a component perpendicular to Z-axis. But if Z-axis is the axis of symmetry, -
components for particles on opposite sides of this axis add up to ZERO.
 Thus, when a rigid body rotates
about an axis of symmetry, its
angular momentum vector L lies
along the symmetry axis, and
its magnitude is L=I
 The angular velocity vector lies
also along the rotation axis.
Hence for a rigid body rotating
around axis of symmetry, L and
 have the same direction


IL 
for a rigid body rotating
around a symmetry axis

Angular Momentum and Torque
 For any system of particles (including both rigid and non-rigid bodies),
the rate of change of the total angular momentum equals the sum of
the torques of all forces acting on all the particles
 Torques of the internal forces add to zero if these forces act along the
line from one particle to another, so the sum of torques includes only the
torques of external forces:
dt
Ld



for any system of particles
 If the system of particles is a rigid body rotating about its axis of
symmetry (Z-axis), then Lz=Iz and I is constant. If this axis is fixed in
space, then the vectors L and  change only in magnitude, not in direction
z
zz
I
dt
Id
dt
Ld


 

 • If body is not rigid, I may
change, and L changes even if
 is constant

 If the axis of rotation is not a
symmetry axis, L does not in
general lie along the rotation axis.
 Even if  is constant, the direction
of L changes and a net torque is
required to maintain rotation
• If the body is an unbalanced wheel of
your car, this torque is provided by
friction in the bearings, which causes the
bearings to wear out
• Balancing a wheel means distributing the
mass so that the rotation axis is an axis
of symmetry, then L points along the
rotation axis, and no net torque is
required to keep the wheel turning

Conservation of Angular
Momentum

Momentum
 When the net external force torque acting on a system is
zero, the total angular momentum of the system is constant
(or conserved)
Angular momentum conservation
 This principle is universal conservation law, valid at all scales from
atomic and nuclear systems to the motions of galaxies
 Circus acrobat, diver, ice skater use this principle:
 Suppose acrobat has just left a swing with arms and legs extended and
rotating counterclockwise about her center of mass. When she pulls her
arms and legs in, her moment of inertia Icm with respect to her center of
mass changes from a large value I1 to much smaller value I2. The only
external force is her weight, which has no torque (goes through center of
mass). So angular momentum remains constant, and angular speed
changes:
constL
dt
Ld
z 


,0
zz II 2211  

Physics of Falling Cats
How does a cat land on its legs when dropped?
… Moment of inertia is important ...
 To understand how a cat can land on it's feet, you must first know
some concepts of rotational motion, since the cat rotates as it
falls.
 Reminder: The moment of inertia of an object is determined by
the distance it's mass is distributed from the rotational axis.
 Relating this to the cat, if the cat stretches out it's legs and tail, it
increases it's moment of inertia; conversely, it can decrease it's
moment of inertia by curling up.
 Remember how it was proved by extending your
professor’s arms while spinning around on a swivel chair?
 Just as a more massive object requires more force to move, an
object with a greater moment of inertia requires more torque to
spin. Therefore by manipulating it's moment of inertia, by
extending and retracting its legs and rotating its tail, the cat can
change the speed at which it rotates, giving it control over which
part of it's body comes in contact with the ground.

Physics of Falling Cats
... and the conservation of angular momentum ...
 If a cat is dropped they almost always tend to land on their feet
because they use the conservation of angular momentum to
change their orientation
 When a cat falls, as you would expect, its centre of mass follows
a parabolic path. The cat falls with a definite angular momentum
about an axis through the cat’s centre of mass.
 When the cat is in the air, no net external torque acts on it about
its centre of mass, so the angular momentum about the cat’s
centre of mass cannot change.
 By pulling in its legs, cat can considerably reduce it rotational
inertia about the same axis and thus considerably increase its
angular speed.
 Stretching out its legs increases its rotational inertia and thus
slows the cat’s angular speed.
 Conservation of angular momentum allows cat to rotate its body
and slow its rate of rotation enough so that it lands on its feet

Momentum
 Falling cat twists different parts of its body in
different directions so that it lands feet first
 At all times during this process the angular
momentum of the cat as a whole is zero
 A free-falling cat cannot alter its total angular
momentum. Nonetheless, by swinging its tail and
twisting its body to alter its moment of inertia, the
cat can manage to alter its orientation

Falling Cats: More Information
How does a cat land on its legs when dropped?
 Cats have the seemingly unique ability to orient themselves in a fall allowing them to
avoid many injuries. This ability is attributed to two significant feline characteristics:
“righting reflex” and unique skeletal structure.
 The “righting reflex” is the cat’s ability to first, know up from down, and then
the innate nature to rotate in mid air to orient the body so its feet face
downward.
 Animal experts say that this instinct is observable in kittens as young as three
to four weeks, and is fully developed by the age of seven weeks.
 A cat’s “righting reflex” is augmented by an unusually flexible backbone and the
absence of a collarbone in the skeleton. Combined, these factors allow for amazing
flexibility and upper body rotation. By turning the head and forefeet, the rest of the
body naturally follows and cat is able reorient itself.
 Like many small animals, cats are said to have a non-fatal terminal falling velocity.
That is, because of their very low body volume-to-weight ratio these animals are
able to slow their decent by spreading out (flying squirrel style). Animals with these
characteristics are fluffy and have a high drag coefficient giving them a greater
chance of surviving these falls.

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