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Course Handout

1. (i) Angular Momentum:

10.1 Angular Momentum of a particle

10.2 Systems of particles

10.3 Angular Momentum and Angular Velocity

10.4 Conservation of angular momentum

10.5 The spinning top

10.6 Review of rotational dynamics

(1) Physics: Resnick, Halliday, Krane vol 1 (5th edition)Ref. Book: Fundamental of Physics: Halliday, Resnick, and Walker

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Some comments before starting the lesson:

Use only .ppt files for presentation. Please don’t copy the slides. All are taken from those two books. Importantly, everything would be uploaded after respective classes.

Problems: for self study; solutions would be posted two weeks after giving assignment

For any clarification, please mail to: amitabh.ghoshray@niituniversity.in

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What is Classical Mechanics? What do we study in it?

Recapitulation

The principle of the relativity of motion:The fundamental concept of “mechanics” is that of motion of a body with respect to other bodies. In the absence of such other bodies it is clearly impossible to speak of motion, which is always relative. Absolute motion of a body irrespective of other bodies has no meaning.

The kinematics (describe motion) developed by Galileo and mechanics by Newton form the basis of classical physics. Understanding of the motion of the planets and the use of kinetic theory to explain the certain observed properties of gases are two important areas, where classical physics had put its mark.

However, a number of experimental observation can not be understood with classical theories. Advent of modern physics e. g. quantum mechanics, special theory of relativity solved those problems. The study of modern physics starts with special theory of relativity. From relativity emerges a new mechanics wherewe have relationship between space and time, mass and energy. This helps us to understand the microscopic world within the atom.

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What is Classical Mechanics? What do we study in it?

Translational Motion: Kinematics and DynamicsForce lawUniform Circular motionLinear MomentumRotational Kinematics and DynamicsConcept of Torque

Note: Follow metric system, which is called the SI system of units. In this system, the units of length, mass, and time are: meter, kilogram, and second, respectively.

Vectors are to be represented in bold face. Unit vectors are i, j, k

Scalar quantities are usually denoted in italic font.

Recapitulation

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Classical Mechanics: study of the motion of objects

Kinematics Dynamics

Describe motionAnalyze cause of motion

Position, velocity, acceleration

Vector quantity: magnitude, direction and follow rules of vector addition and multiplication

Position vector: r = xi+ yj + zkVelocity: v= dr/dtAcceleration: a = dv/dt

Eq. in one-dimensional:vx=dx/dt

A particular object interacting with the surroundings (environments)

This Interaction is described in terms of force F.

Newton’s laws of force:

First law: law of inertia and the reference frame is called inertial frame

Second law: ∑F = ma

Third law: Fab = -FbaKinematic equation

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A body on which no force is acts. If the body is at rest it will remain at rest. If the body is moving with constant velocity, it will continue. The tendency of a body to remain at rest or in uniform motion is called inertia, and Newton’s first law is often calledthe law of inertia.

If the net force on the body is zero, then it is always possible to find a referenceframe in which the body has acceleration. This is inertial frame of reference.

Newton’s first law and frame of reference

You as passenger in a moving car held by a seat belt and a book on the seat.If brakes are applied suddenly – what will you see?

Manish standing by the side of the road sees no unusual. He sees the car, you and the book are moving with a reduced speed. No violation of Newton’s first law.

You in the car and Manish standing by the side of the road – define two different reference frames. A reference frame requires a set of co-ordinates and clock which enable the observer to measure the variables like position, velocity etc. So the observer in different reference frames will get different result.

The book slides forward. No apparent force on the book, but relative to you it appears start move. Violation of Newton’s first law.

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Uniform Circular Motion:a particle moves at constant speed in circular path

Relation between acceleration and velocity: ar= v2/r Home work

(a) A car moving along a circular path at constant speed: uniform circular motion

(b) As the car moves from “A” to “B”, its velocity vector changes from vi to vf .

(c) Determining the direction of the change in velocity ∆v, which is toward the center of the circle for small ∆r.

The acceleration vector is always perpendicular to the path and always points toward the center of the circle. An acceleration of this nature is called a centripetal (center-seeking) acceleration.∑F = ma ⇒ |∑F| = ma = mv2/r

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Collision problem

Two objects exert forces on each other for a short (but measurable) time interval. So, three situations are clearly separable;

(1) before collision (2) during collision and(3) after collision

To apply Newton’s second law in case (2) is not easy. The basic problem is: even though, we know the motion before and after the collision; but during brief collision period we do not observe or measure what is happening.

A new dynamical variable called “Linear Momentum or translational momentumis introduced. It is again a vector quantity, and is defined as: p = mv.

Newton’s second law: ∑F = dp/dt

For a single particle of constant mass ∑F = dp/dt = d(mv)/dt = m(dv/dt) = ma

how to apply Newton’s second law?

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Linear momentum is a conserved quantity, meaning that a closed system is not affected by external forces. Its total linear momentum cannot change. In classical mechanics, conservation of linear momentum is implied by Newton's laws; but it also holds in special relativity (with a modified formula) and, with appropriate definitions, a (generalized) linear momentum conservation law holds in electrodynamics, quantum mechanics, quantum field theory.

Conservation of Linear Momentum 

Consider a body (1) of mass m1 moving with initial velocity v1i and momentum p1i = m1 v1i colliding with a second body having mass m2 moving with initial velocity v2i and momentum p2i = m2 v2i. We define the motion of two bodies as our system and does not interact with the environment i.e. no external force is acting on each of them, except the colliding force for a short timeAfter the collision, m1 moves with final velocity v1f and momentum p1f = m1 v1f

body (2) with mass m2 moves with final velocity v2f and momentum p2f = m2 v2f.

At any time, total momentum P = p1 + p2 : dP/dt = dp1/dt + dp2/dt = ∑F1 + ∑F2

Before collision, ∑F1 and ∑F2 are zero; dP/dt = 0; similarly after collision, dP/dt = 0

During collision, force on (1) is F12 due to (2); force on (2) is F21 due to (1);

As, F12 = -F21 ⇒dP/dt = 0; i.e. P is constant.

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System of Particles

In simple translational motion, treatment of object as point particle having mass but no size does not have any problem. As all points in an object move in a similar fashion in a translational motion.

In case of rotation while moving, this idea creates problem. In this case, there is one point of the object whose motion under the influence of external forces can be analyzed as that of a simple particle. This point is called center of mass.

The center of mass of a rigid body consisting of two particles:

y

x

m1

m2

rcm

r2

r1

Consider a rigid body, consisting of two masses m1 and m2 connected by a massless rigid rod.

We set the coordinate system in such a way thattwo particles lie in the xy plane.

r1 and r2: position vectors of m1 and m2 at a particular time.

The center of mass is located by rcm

rcm

rcm = (r1m1 + r2m2)/(m1 + m2)

It is a fixed point on the rigid body and is determined by the distribution of mass.

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y

x

m1

m2

rcm

r2

r1

rcm

rcm = (r1m1 + r2m2)/(m1 + m2)

m1

m2

Later, the system would moved to new location and cm would similarly change its position. Let us see why the motion of cm is special.

r’cm

vcm = drcm/dt = {m1(dr1/dt) + m2(dr2/dt)} /(m1 + m2)

vcm

vcm = {m1 v1 + m2v2}/(m1 + m2)

acm = {m1 a1 + m2a2}/(m1 + m2)

So, motion of the cm system becomes simple

For many particles system rcm = (∑ rnmn)/M

M = ∑mn

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Rotational Kinematics

Pure rotation of a rigid body: if every point of the body moves in a circular path.The centers of these circles lie on a common straight line, called axis of rotation.

A rigid object is one that is non-deformable —that is, it is an object in which the separations between all pairs of particles remain constant.

A rigid object of arbitrary shape rotating about a fixed axis through o perpendicular to the plane of the figure. (z is the axis of rotation) A particle at P rotates in a circle of radius r centered at o.

It is convenient to represent the position of P with its polar coordinates (r, θ), r = the distance from the origin to P and θ is measured counterclockwise from some preferred direction - say, the positive x axis.

As the particle moves along the circle from the positive x axis (θ = 0) to P, it moves through an arc of length s, which is related to the angular position θ through the relationship θ = s/r .

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Being a ratio of two lengths, θ has no dimension, but we commonly give θ the artificial unit radian (rad), where one radian is the angle subtended by an arc length equal to the radius of the arc. θ = r/r = 1 radian. If the particle moves through the circumference of the circle (i.e. one revolution = 2πr/r =2π radian =360°.

A particle at P at time ti moves to Q at time tf

so radius vector r sweeps out ∆θ = θi – θf

This is angular displacement; the angular velocityω=dθ/dt, so is angular acceleration α = dω /dt

Note: this discussion is also valid for the motion of a single particle in a circular path. (held by a rigid mass less rod of length r)

It can be proved that rotational variables are vectors like variables associated with translational motion.

Rotational variables

Note: angles and associated quantities is expressed on either radians, degreesor revolutions. But in equation, if both rotational and translational variables are used, only radian should be used.

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Under constant angular acceleration, kinematic expression for rotational motion are of the same form as those for linear motion under constant linear acceleration. We need substitution like x →θ, v → ω, a → α.

Relation between linear and angular variables: The velocity v is always tangent to the circular path and is called tangential velocityθ = s/r, s = θr ; differentiate w. r. t. time t

ds/dt = (dθ/dt)r ; vT = ωr ; differentiate again

dvT/dt = (dω/dt)r so, tangential accn. aT= αr

We have seen radial or centripetal component aR = vT2/r = ω2r

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aT= αr aR = vT2/r = ω2r

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Rotational Dynamics

In the study of dynamics, we define “force” in terms of the accelerationit produced when acting upon a body of known mass. Same analogy: angular acceleration is produced when a force acts on a rigid body that is free to rotate about a fixed axis. Note: Angular acceleration also depends on where the force is applied to the body. Thus a given force applied at different locations of a body will in general produce different angular accelerations. Torque is the quantity in rotational dynamics that takes account of both the magnitude of the force and the direction and location at which it is applied.Torque is the tendency of a force to rotate an object about an axis. Just as a force is a push or a pull, a torque can be thought of as a twist to an object. Mathematically, torque is defined as the cross product of the lever-arm distance and force, which tends to produce rotation. τ = r × F

r

F0

τ

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Besides the magnitude of the tangential FT component, the distance r would also determine the amount of angular acceleration. The rotational quantity that includes effect of both is called torque τ and is given by τ = r F sinθ.

x

y

rP

F

θ

x

y

rP

F

θ FRFT

Line of action of Fx

y

rP

F

θF⊥

θ

r⊥τ = r × F

A force F is applied to a rigid body at point P, located a perpendicular distance r from the axis of rotation say z axis. Cross sectional view in the xy plane is shown. We assume F lies in this plane and has only x and y components.

F has radial FR and tangential FT component. FR = F cosθ has no effect on the rotation about z axis. Only FT = F sinθ has effect on rotation.

τ = rF⊥τ = r⊥ F

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Besides the magnitude of the tangential FT component, the distance r would also determine the amount of angular acceleration. The rotational quantity that includes effect of both is called torque τ and is given by τ = r F sinθ.

x

y

rP

F

θ

x

y

rP

F

θ FRFT

Line of action of Fx

y

rP

F

θF⊥

θ

r⊥τ = r × F

Torque is zero if (1) r =0, (2) F =0, and (3) θ =0 or 180°.Another way of interpreting τ will be clear from this figure. The component of force ⊥ to r is F⊥ = FT = F sinθ. The component of r, ⊥ to the line of action of F is r⊥ = r sinθ.τ = r F sinθ = rF⊥ or one can write τ = r sinθ F = r⊥ F. r⊥ is also called as moment arm

τ = rF⊥τ = r⊥ F

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The tendency of a body to remain at rest or in uniform linear motion is the inertia of the system. The Newton’s first law is called law of inertia. The frame of reference to which it applies is “inertial frame”. The translational inertia has single value as mass of a body is unique. But the rotational inertia of an object may vary if we choose different axis of rotation.

The rotational inertia

The rotational inertia of a single particle:

x

y

rP

F

θ FRFT

So net tangential force responsible for rotation =∑FT = F sinθ. Newton’s second law for tangential motionof the particle ∑FT = maT; where aT = αzr

F sinθ = m αzr or rF sinθ = m αzr2 τ= m r2αz

The rotational inertia I = m r2

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x

y

r2

P

r1

m1

m2

x

y

T2

P

T1

m1

m2

Newton’s second law for rotation

A rigid body consisting of two particles m1 and m2, each can rotate in the xy plane about z axis. The particles are connected to the axis by thin rods of negligible mass of lengths r1 and r2. They are also connected to each other by a similar rod.

T1 and T2, tensions experiencing by the particles respectively along the rod connected to the origin. Tension acting along the rod connecting two particles are T1r and T2r.

T1r T2r

For, T1r action-reaction pair Tr1 similar for T2r & Tr2 ,’ As it is mass less Tr1 + T2r =0, T1r = -T2r

The net force on m1 ∑F1 = P+ T1 + T1r The net force on m2 ∑F2 = T2 + T2r

We resolve each of ∑F into radial and tangential components.

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x

y

m1

m2

∑F1

∑F2

∑F1R ∑F2R

∑F1T

∑F2TNo radial motion and no torque for radial component, as they pass through origin and their moment arm is zero.

Net torque would be only from the tangential component of the force.

∑τz = ∑τz + ∑τz= (∑F1T)r1 + (∑F2T)r2

= (m1a1T)r1 + (m2a2T)r2

Tangential force --- tangential acceleration

= (m1αzr1)r1 + (m2αzr2)r2

= (m1r12

+ m2r22)αz

I = (m1r12

+ m2r22)

∑τz = I αz

Total rotational inertia

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We define the angular momentum l of the particle with respect to the origin O: I = r x p …(1)

1. (i) Angular momentum

Consider a particle of mass m and linear momentum p at a position r relative to the origin O of an inertial reference frame.

Magnitude: I =rpsinθ …(2)θ is the smaller angle between r and p when These vectors are tail to tail. Its direction is perpendicular to the plane formed by r and p .

The sense is given by the right-hand rule: swing the fingers of the right hand from the direction of r into the direction of p, through the smaller angle between them; the extended right thumb then points in the direction of I.

We can write the magnitude of I either as l= (r sinθ)p =p r┴ …(3a)

or l = r (p sin θ) = r p┴ …(3b)

r p

o

x

y

z

θ

r p

o

x

y

z

m

I = r x p

Unit of I kg.m2/s

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Equation 3(b) shows that only the component of p perpendicular to r contributes to the angular momentum. When the angle between r and p is 0 and 180°, there is no perpendicular component (p = p sin θ = 0); then the line of action of p passes through the origin, and r is also zero. In this case both

equation 3(a) and 3(b) show that the angular momentum l is zero.

r┴ is the component of r at right angle to the line of the action of p, and l= (r sinθ)p =p r┴ …(3a)

l = r (p sin θ) = r p┴ …(3b)

ro

θ

p

x

ro

θ

p

x

p

θr

oθ

p

x

p

θ

r sinθ = r┴

p┴ is the component of p at the right angles to r.

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Exercise 10.2

If we are given r, p, θ we can calculate the angular momentum of a particle fromI =rpsinθ. If we are given components (x,y,z) of r and (vx, vy, vz) of v; (a) find the components of l along x, y, z axes. (b) show that if the particle moves only in the xy plane, the resultant angular momentum has only a z component.

I = r x p I = r x (mv) r= xi + yj +zk v= vxi + vyj +vzk

r x v= i j k x y z vx vy vz

I = m[i(yvz – zvy) + j(zvx – xvz) + k(xvy – yvx)]

l= lxi + lyj +lzk Ix = m(yvz – zvy)

Iy = m(zvx – xvz)

Iz = m(xvy – yvx)Particle moves in the xy plane: z = vz = 0

Only z component is present

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P. 1 Class workA particle P of mass 2.13 kg has position r and velocity v as shown in fig. It is acted on by the force F. All three vectors lie in a common plane. Presume that r =2.91 m, v = 4.18 m/s, and F = 1.88 N. Compute (a) the angular momentum of the particle and (b) the torque, about the origin, acting on the particle. What are the directions of these two vectors?[sin(33°)=0.5445, sin(26°)= 0.438]

x

y

P

r

F

v

26°

33°

49°

l = (2.91 m)(2.13 kg)(4.18 m/s) sin(147°) = 14.1 kg .m2/s

I = r x (mv)

l = rmvsinθ in the z direction θ = 147°

τ = r Fsin θ k = (2.91 m)(1.88 N) sin(26°)k = 2.40 N .m

b) τ = r × F

We choose common plane as xy plane

Both are in the z direction.

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Relation between torque and angular momentum for a single particle.

Differentiate eq. (1) w. r. t. time dI/dt = d(r × p)/dt …(4)

dI/dt = dr/dt×p + r×dp/dt; dr/dt = v, and p = mv. Making these substitutions into the first product on the right, we obtaindI/dt = (v × mv) + r × dp/dt …(5)

Now v × mv = 0, because the vector product of two parallel vectors is zero.

dp/dt = ∑F dI/dt = r × ∑F = ∑τ

This states that the net torque acting on a particle is equal to the time rate of change of its angular momentum. Both the torque τ and the angular momentum I in this equation must be defined with respect to the same origin. Equation (6) is the rotational analogue of equation F = dp/dt, which states that the net force acting on a particle is equal to the time rate of change of its linear momentum.

∑τ = dI/dt …..(6)

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Equation (6) like all three dimensional vector equation, is the equivalent to three one dimensional equations – namely,

∑τx = dIx/dt, ∑τy = dIy/dt and ∑τz = dIz/dt …(6a)

Hence, the x component of the net external torque is given by the change with time of the x component of the angular momentum. Similar results hold for the y and z directions.

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Sample problem 10.1A particle of mass m is released from rest at point P falling parallel to the y axis (see figure). (a) Find the torque acting on m at any time t with respect to origin O. (b) Find the angular momentum of m at any time t, with respect to same origin. (c) show that the relation ∑τ = dI/dt yields a correct result.

O Px

r

b

θmg

mP1

Let P1 be the position at time t. Force acting on mis mg. The velocity at this point vf= gt

r × F = τ; Right hand rule: the direction is into the paper

τ = r F sinθ = r sinθ F = bmg a constant

y

I = r x (mv) = r sinθ mgt = bmgt

dI/dt = d(bmgt)/dt = bmg = τ

If origin lies on P, no torque and angular momentum.

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Course Handout

1. (i) Angular Momentum:

10.1 Angular Momentum of a particle

10.2 Systems of particles

10.3 Angular Momentum and Angular Velocity

10.4 Conservation of angular momentum

10.5 The spinning top

10.6 Review of rotational dynamics

(1) Physics: Resnick, Halliday, Krane vol 1 (5th edition)Ref. Book: Fundamental of Physics: Halliday, Resnick, and Walker