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Contemporary physics

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Figure 4.1. Two particles interact with each other. According to Newton’s third law, we must have F12 = - F21.

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Слайд 1Course of lectures «Contemporary Physics: Part1»
Lecture №5
Linear Momentum and Collisions.
Rotation

of a Rigid Object about a Fixed Axis.

Course of lectures «Contemporary Physics: Part1»Lecture №5Linear Momentum and Collisions.Rotation of a Rigid Object about a Fixed

Слайд 2Figure 4.1. Two particles interact with each other. According to

Newton’s third law, we must have F12 = - F21.

Figure 4.1. Two particles interact with each other. According to Newton’s third law, we must have F12

Слайд 3The linear momentum of a particle or an object that

can be modeled as a particle of mass m moving

with a velocity v is defined to be the product of the mass and velocity:

(4.1)

(4.2)

Linear momentum is a vector quantity because it equals the product of a scalar quantity m and a vector quantity v. Its direction is along v, it has dimensions ML/T, and its SI unit is kg · m/s.

The linear momentum of a particle or an object that can be modeled as a particle of

Слайд 4As you can see from its definition, the concept of

momentum provides a quantitative distinction between heavy and light particles

moving at the same velocity. For example, the momentum of a bowling ball moving at 10 m/s is much greater than that of a tennis ball moving at the same speed. Newton called the product mv quantity of motion; this is perhaps a more graphic description than our present-day word momentum, which comes from the Latin word for movement.

If a particle is moving in an arbitrary direction, p must have three components

As you can see from its definition, the concept of momentum provides a quantitative distinction between heavy

Слайд 5Using Newton’s second law of motion, we can relate the

linear momentum of a particle to the resultant force acting

on the particle. We start with Newton’s second law and substitute the definition of acceleration:

(4.3)

As m=const:

The time rate of change of the linear momentum of a particle is equal to the net force acting on the particle.

Using Newton’s second law of motion, we can relate the linear momentum of a particle to the

Слайд 6Using the definition of momentum, Equation 4.1 can be written
where

p1i and p2i are the initial values and p1f and

p2f the final values of the momenta for the two particles for the time interval during which the particles interact.

(4.5)

(4.4)

Using the definition of momentum, Equation 4.1 can be writtenwhere p1i and p2i are the initial values

Слайд 7This result, known as the law of conservation of linear

momentum, can be extended to any number of particles in

an isolated system. It is considered one of the most important laws of mechanics. We can state it as follows:

Whenever two or more particles in an isolated system interact, the total momentum of the system remains constant.

This law tells us that the total momentum of an isolated system at all times equals its initial momentum.
Notice that we have made no statement concerning the nature of the forces acting on the particles of the system. The only requirement is that the forces must be internal to the system.

(4.6)

This result, known as the law of conservation of linear momentum, can be extended to any number

Слайд 8The momentum of a particle changes if a net force

acts on the particle.
According to Newton’s second law
(4.7)
(4.8)

The momentum of a particle changes if a net force acts on the particle.According to Newton’s second

Слайд 9(4.9)
To evaluate the integral, we need to know how the

force varies with time. The quantity on the right side

of this equation is called the impulse of the force F acting on a particle over the time interval ∆t=tf - ti. Impulse is a vector defined by

Equation 4.8 is an important statement known as the impulse–momentum theorem:

The impulse of the force F acting on a particle equals the change in the momentum of the particle.

(4.9)To evaluate the integral, we need to know how the force varies with time. The quantity on

Слайд 10The direction of the impulse vector is the same as

the direction of the change in momentum. Impulse has the

dimensions of momentum—that is, ML/T. Note that impulse is not a property of a particle; rather, it is a measure of the degree to which an external force changes the momentum of the particle. Therefore, when we say that an impulse is given to a particle, we mean that momentum is transferred from an external agent to that particle.

Figure 4.2 (a) A force acting on a particle may vary in time. The impulse imparted to the particle by the force is the area under the force-versus-time curve. (b) In the time interval (t, the time-averaged force (horizontal dashed line) gives the same impulse to a particle as does the time-varying force described in part (a).

(a)

(b)

The direction of the impulse vector is the same as the direction of the change in momentum.

Слайд 11Because the force imparting an impulse can generally vary in

time, it is convenient to define a time-averaged force
where ∆t=tf

- ti.

(4.10)

(4.12)

The calculation becomes especially simple if the force acting on the particle is constant. In this case, and Equation 4.11 becomes

In many physical situations, we shall use what is called the impulse approximation, in which we assume that one of the forces exerted on a particle acts for a short time but is much greater than any other force present.

(4.11)

Because the force imparting an impulse can generally vary in time, it is convenient to define a

Слайд 12We use the term collision to represent an event during

which two particles come close to each other and interact

by means of forces. The time interval during which the velocities of the particles change from initial to final values is assumed to be short. The interaction forces are assumed to be much greater than any external forces present, so we can use the impulse approximation.

Figure 4.3 (a) The collision between two objects as the result of direct contact. (b) The “collision” between two charged particles.

We use the term collision to represent an event during which two particles come close to each

Слайд 13 The total momentum of an isolated system just before a

collision equals the total momentum of the system just after

the collision.
The total kinetic energy of the system of particles may or may not be conserved, depending on the type of collision. In fact, whether or not kinetic energy is conserved is used to classify collisions as either elastic or inelastic.
An elastic collision between two objects is one in which the total kinetic energy (as well as total momentum) of the system is the same before and after the collision. Collisions between certain objects in the macroscopic world, such as billiard balls, are only approximately elastic because some deformation and loss of kinetic energy take place. For example, you can hear a billiard ball collision, so you know that some of the energy is being transferred away from the system by sound. An elastic collision must be perfectly silent! Truly elastic collisions occur between atomic and subatomic particles.
The total momentum of an isolated system just before a collision equals the total momentum of the

Слайд 14An inelastic collision is one in which the total kinetic

energy of the system is not the same before and

after the collision (even though the momentum of the system is conserved).
Inelastic collisions are of two types.
When the colliding objects stick together after the collision, as happens when a meteorite collides with the Earth, the collision is called perfectly inelastic.
When the colliding objects do not stick together, but some kinetic energy is lost, as in the case of a rubber ball colliding with a hard surface, the collision is called inelastic (with no modifying adverb).
In most collisions, the kinetic energy of the system is not conserved because some of the energy is converted to internal energy and some of it is transferred away by means of sound. Elastic and perfectly inelastic collisions are limiting cases; most collisions fall somewhere between them.
The important distinction between these two types of collisions is that momentum of the system is conserved in all collisions, but kinetic energy of the system is conserved only in elastic collisions.

An inelastic collision is one in which the total kinetic energy of the system is not the

Слайд 15Figure 4.4 Schematic representation of a perfectly inelastic head-on collision

between two particles: (a) before collision and (b) after collision.
(a)
(b)
(4.13)
(4.14)

Figure 4.4 Schematic representation of a perfectly inelastic head-on collision between two particles: (a) before collision and

Слайд 16Figure 4.5 Schematic representation of an elastic head-on collision between

two particles: (a) before collision and (b) after collision.
(4.15)
(4.16)

Figure 4.5 Schematic representation of an elastic head-on collision between two particles: (a) before collision and (b)

Слайд 17Next, let us separate the terms containing m1 and m2

in Equation 4.15 to obtain
(4.17)
(4.18)
(4.19)
To obtain our final result, we

divide Equation 4.17 by Equation 4.18 and obtain
Next, let us separate the terms containing m1 and m2 in Equation 4.15 to obtain(4.17)(4.18)(4.19)To obtain our

Слайд 18Suppose that the masses and initial velocities of both particles

are known.
(4.20)
(4.21)
Let us consider some special cases. If m1 =

m2, then Equations 4.20 and 4.21 show us that v1f = v2i and v2f = v1i .

That is, the particles exchange velocities if they have equal masses. This is approximately what one observes in head-on billiard ball collisions - the cue ball stops, and the struck ball moves away from the collision with the same velocity that the cue ball had.

Suppose that the masses and initial velocities of both particles are known.(4.20)(4.21)Let us consider some special cases.

Слайд 19If particle 2 is initially at rest, then v2i =

0, and Equations 4.20 and 4.21 become
(4.22)
(4.23)
If m1 is much

greater than m2 and v2i = 0, we see from Equations 4.22 and 4.23 that v1f ≈ v1i and v2f ≈ 2v1i .

If m2 is much greater than m1 and particle 2 is initially at rest, then v1f ≈ -v1i and v2f ≈ 0.

If particle 2 is initially at rest, then v2i = 0, and Equations 4.20 and 4.21 become(4.22)(4.23)If

Слайд 20The momentum of a system of two particles is conserved

when the system is isolated. For any collision of two

particles, this result implies that the momentum in each of the directions x, y, and z is conserved.

For such two-dimensional collisions, we obtain two component equations for conservation of momentum:

The momentum of a system of two particles is conserved when the system is isolated. For any

Слайд 21Figure 4.6 An elastic glancing collision between two particles.
glancing collision
(4.25)
(4.24)

Figure 4.6 An elastic glancing collision between two particles.glancing collision(4.25)(4.24)

Слайд 22(4.26)
If the collision is elastic, we can also use Equation

4.16 (conservation of kinetic energy) with v2i = 0 to

give

If the collision is inelastic, kinetic energy is not conserved and Equation 4.26 does not apply.

(4.26)If the collision is elastic, we can also use Equation 4.16 (conservation of kinetic energy) with v2i

Слайд 23Figure 4.7 Two particles of unequal mass are connected by

a light, rigid rod. (a) The system rotates clockwise when

a force is applied between the less massive particle and the center of mass. (b) The system rotates counterclockwise when a force is applied between the more massive particle and the center of mass. (c) The system moves in the direction of the force without rotating when a force is applied at the center of mass.
Figure 4.7 Two particles of unequal mass are connected by a light, rigid rod. (a) The system

Слайд 24Figure 4.8 The center of mass of two particles of

unequal mass on the x axis is located at xCM,

a point between the particles, closer to the one having the larger mass.

(4.27)

(4.28)

(4.29)

Figure 4.8 The center of mass of two particles of unequal mass on the x axis is

Слайд 25(4.30)
Figure 4.9 An extended object can be considered to be

a distribution of small elements of mass ∆mi. The center

of mass is located at the vector position rCM, which has coordinates xCM, yCM, and zCM.

The center of mass of any symmetric object lies on an axis of symmetry and on any plane of symmetry.

a continuous mass distribution

(4.30)Figure 4.9 An extended object can be considered to be a distribution of small elements of mass

Слайд 26Assuming M remains constant for a system of particles, that

is, no articles enter or leave the system, we obtain

the following expression for the velocity of the center of mass of the system:

where vi is the velocity of the ith particle. Rearranging Equation 4.34 gives

(4.34)

(4.35)

Therefore, we conclude that the total linear momentum of the system equals the total mass multiplied by the velocity of the center of mass. In other words, the total linear momentum of the system is equal to that of a single particle of mass M moving with a velocity vCM.

Assuming M remains constant for a system of particles, that is, no articles enter or leave the

Слайд 27If we now differentiate Equation 4.34 with respect to time,

we obtain the acceleration of the center of mass of

the system:

Rearranging this expression and using Newton’s second law, we obtain

where Fi is the net force on particle i.

(4.36)

(4.37)

If we now differentiate Equation 4.34 with respect to time, we obtain the acceleration of the center

Слайд 28That is, the net external force on a system of

particles equals the total mass of the system multiplied by

the acceleration of the center of mass. If we compare this with Newton’s second law for a single particle, we see that the particle model that we have used for several chapters can be described in terms of the center of mass:

The center of mass of a system of particles of combined mass M moves like an equivalent particle of mass M would move under the influence of the net external force on the system.

(4.38)

That is, the net external force on a system of particles equals the total mass of the

Слайд 29The angular position of the rigid object is the angle

θ between this reference line on the object and the

fixed reference line in space, which is often chosen as the x axis.

Figure 5.1 A compact disc rotating about a fixed axis through O perpendicular to the plane of the figure.

Rotation of a Rigid Object About a Fixed Axis

(5.1)

The angular position of the rigid object is the angle θ between this reference line on the

Слайд 30Rotation of a Rigid Object About a Fixed Axis
Figure 5.2

A particle on a rotating rigid object moves from A

to B along the arc of a circle. In the time interval ∆t = tf - ti , the radius vector moves through an angular displacement ∆θ = θf - θi.

The average angular speed

The instantaneous angular speed

(5.2)

(5.3)

Rotation of a Rigid Object About a Fixed AxisFigure 5.2 A particle on a rotating rigid object

Слайд 31The average angular acceleration
The instantaneous angular acceleration
When a rigid object

is rotating about a fixed axis, every particle on the

object rotates through the same angle in a given time interval and has the same angular speed and the same angular acceleration.

(5.4)

(5.5)

The average angular accelerationThe instantaneous angular accelerationWhen a rigid object is rotating about a fixed axis, every

Слайд 32Direction for angular speed and angular acceleration
Figure 5.3 The right-hand

rule for determining the direction of the angular velocity vector.

Direction for angular speed and angular accelerationFigure 5.3 The right-hand rule for determining the direction of the

Слайд 33Rotational Kinematics: Rotational Motion with Constant Angular Acceleration
(5.6)
is the angular

speed of the rigid object at time t = 0.
Equation

5.6 allows us to find the angular speed ωf of the object at any later time t.
Rotational Kinematics: Rotational Motion with Constant Angular Acceleration(5.6)is the angular speed of the rigid object at time

Слайд 34(5.7)
is the angular position of the rigid object at time

t = 0.
Equation 5.7 allows us to find the angular

position θf of the object at any later time t.
(5.7)is the angular position of the rigid object at time t = 0.Equation 5.7 allows us to

Слайд 35If we eliminate t from Equations 5.6 and 5.7, we

obtain
This equation allows us to find the angular speed ωf

of the rigid object for any value of its angular position θf .

If we eliminate α between Equations 5.6 and 5.7, we obtain

(5.8)

(5.9)

If we eliminate t from Equations 5.6 and 5.7, we obtainThis equation allows us to find the

Слайд 36Table 5.1

Table 5.1

Слайд 37Angular and Linear Quantities
Figure 5.4 As a rigid object rotates

about the fixed axis through O, the point P has

a tangential velocity v that is always tangent to the circular path of radius r.

(5.10)

That is, the tangential speed of a point on a rotating rigid object equals the perpendicular distance of that point from the axis of rotation multiplied by the angular speed.

Angular and Linear QuantitiesFigure 5.4 As a rigid object rotates about the fixed axis through O, the

Слайд 38We can relate the angular acceleration of the rotating rigid

object to the tangential acceleration of the point P by

taking the time derivative of v:

That is, the tangential component of the linear acceleration of a point on a rotating rigid object equals the point’s distance from the axis of rotation multiplied by the angular acceleration.

(5.11)

We can relate the angular acceleration of the rotating rigid object to the tangential acceleration of the

Слайд 39Figure 5.5 As a rigid object rotates about a fixed

axis through O, the point P experiences a tangential component

of linear acceleration at and a radial component of linear acceleration ar. The total linear acceleration of this point is a = at + ar.

A point moving in a circular path undergoes a radial acceleration ar of magnitude v2/r directed toward the center of rotation .
Because v = rω for a point P on a rotating object, we can express the centripetal acceleration at that point in terms of angular speed as

(5.12)

(5.13)

Figure 5.5 As a rigid object rotates about a fixed axis through O, the point P experiences

Слайд 40Rotational Kinetic Energy
Figure 10.7 A rigid object rotating about the

z axis with angular speed ω.
(5.14)

Rotational Kinetic EnergyFigure 10.7 A rigid object rotating about the z axis with angular speed ω.(5.14)

Слайд 41We simplify this expression by defining the quantity in parentheses

as the moment of inertia I:
From the definition of moment

of inertia, we see that it has dimensions of ML2 (kg ·m2 in SI units). With this notation, Equation 5.14 becomes

(5.15)

Where KR is Rotational kinetic energy.

(5.16)

We simplify this expression by defining the quantity in parentheses as the moment of inertia I:From the

Слайд 42Calculation of Moments of Inertia
(5.17)

Calculation of Moments of Inertia(5.17)

Слайд 43Table 5.2

Table 5.2

Слайд 45Torque
Figure 5.8 The force F has a greater rotating tendency

about O as F increases and as the moment arm

d increases. The component Fsinφ tends to rotate the wrench about O.

When a force is exerted on a rigid object pivoted about an axis, the object tends to rotate about that axis. The tendency of a force to rotate an object about some axis is measured by a vector quantity called torque τ (Greek tau).

(5.18)

TorqueFigure 5.8 The force F has a greater rotating tendency about O as F increases and as

Слайд 46Figure 5.9 The force F1 tends to rotate the object

counterclockwise about O, and F2 tends to rotate it clockwise.
(5.19)

Figure 5.9 The force F1 tends to rotate the object counterclockwise about O, and F2 tends to

Слайд 47Relationship Between Torque and Angular Acceleration
Figure 5.10 A particle rotating

in a circle under the influence of a tangential force

Ft. A force Fr in the radial direction also must be present to maintain the circular motion.
Relationship Between Torque and Angular AccelerationFigure 5.10 A particle rotating in a circle under the influence of

Слайд 48The torque acting on the particle is proportional to its

angular acceleration, and the proportionality constant is the moment of

inertia.

(5.20)

The torque acting on the particle is proportional to its angular acceleration, and the proportionality constant is

Слайд 49Figure 5.11 A rigid object rotating about an axis through

Figure 5.11 A rigid object rotating about an axis through O.

Слайд 50Although each mass element of the rigid object may have

a different linear acceleration at , they all have the

same angular acceleration α.

So, again we see that the net torque about the rotation axis is proportional to the angular acceleration of the object, with the proportionality factor being I, a quantity that depends upon the axis of rotation and upon the size and shape of the object.

(5.21)

Although each mass element of the rigid object may have a different linear acceleration at , they

Слайд 51Work, Power, and Energy in Rotational Motion
Figure 5.12 A rigid

object rotates about an axis through O under the action

of an external force F applied at P.

The work done by F on the object as it rotates through an infinitesimal distance

where Fsinφ is the tangential component of F, or, in other words, the component of the force along the displacement.

(5.22)

Work, Power, and Energy in Rotational MotionFigure 5.12 A rigid object rotates about an axis through O

Слайд 52The rate at which work is being done by F

as the object rotates about the fixed axis through the

angle dθ in a time interval dt is

(5.23)

The rate at which work is being done by F as the object rotates about the fixed

Слайд 54That is, the work–kinetic energy theorem for rotational motion states

that
the net work done by external forces in rotating a

symmetric rigid object about a fixed axis equals the change in the object’s rotational energy.
That is, the work–kinetic energy theorem for rotational motion states thatthe net work done by external forces

Слайд 55Table 5.3

Table 5.3

Слайд 56Rolling Motion of a Rigid Object
Figure 5.13 For pure rolling

motion, as the cylinder rotates through an angle θ, its

center moves a linear distance s = Rθ.

(5.25)

(5.26)

Rolling Motion of a Rigid ObjectFigure 5.13 For pure rolling motion, as the cylinder rotates through an

Слайд 57Figure 5.14 All points on a rolling object move in

a direction perpendicular to an axis through the instantaneous point

of contact P. In other words, all points rotate about P. The center of mass of the object moves with a velocity vCM, and the point P’ moves with a velocity 2 vCM .
Figure 5.14 All points on a rolling object move in a direction perpendicular to an axis through

Слайд 58Figure 5.15 The motion of a rolling object can be

modeled as a combination of pure translation and pure rotation.

Figure 5.15 The motion of a rolling object can be modeled as a combination of pure translation

Слайд 59Find v1f and v2f.

Quiz
Figure 4.5 Schematic representation of an

elastic head-on collision between two particles: (a) before collision and

(b) after collision.
Find v1f and v2f. QuizFigure 4.5 Schematic representation of an elastic head-on collision between two particles: (a)

Слайд 60Quick Quiz 1 A block of mass m is projected

across a horizontal surface with an initial speed v. It

slides until it stops due to the friction force between the block and the surface. The same block is now projected across the horizontal surface with an initial speed 2v. When the block has come to rest, how does the distance from the projection point compare to that in the first case? (a) It is the same. (b) It is twice as large. (c) It is four times as large. (d) The relationship cannot be determined.

Quick Quiz 2 A car and a large truck traveling at the same speed make a head-on collision and stick together. Which vehicle experiences the larger change in the magnitude of momentum? (a) the car (b) the truck (c) The change in the magnitude of momentum is the same for both. (d) impossible to determine.

Quick Quiz 1 A block of mass m is projected across a horizontal surface with an initial

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