Introduction to Classical Mechanics (Physics)

Classical physics involves classical mechanics (the study of the movement of fluids and particles), thermodynamics (the study of temperature and heat transfer), and electromagnetism (the study of electricity, magnets, and electromagnetic waves).

Introduction to Classical Mechanics (Physics) :it2edu

Introduction to classical mechanics

CONSERVATIVE FORCE

A force acting on a particle s conservative it the particle after going through a complete round trip and then return to its initial position with the same kinetic energy as it has initially.

Example:

I) When a ball is thrown against gravity, the ball reached a certain height coming momentarily at rest so that its kinetic energy becomes zero. Then it returns downward under gravity with the same kinetic energy with which it was thrown (because all resistance accused to be zero). Thus the force of gravity is conservative.

ii) Elastic force exerted by an ideal spring is conservative.

iii) The electrostatic forces are also conservative.

Another definition of conservative force

As per work energy theorem, the work done by the resultant forces acting on a particle is equal to the change in the kinetic energy of the particle. When a conservative force acts on a particle going through a round trip, the kinetic energy does not change. Hence the network done by the force is zero. Thus, force acting on a particle is conservative if the net work done by the force in a complete wound trip of particle is zero.

NON-CONSERVATIVE FORCE

A force acting on a particle is non-conservative it the particle, after going through v a complete round trip, return to its initial position with change kinetic energy.

Unlike conservative forces in which air resistance change it is always present in non conservative forces. They always oppose the motion, irrespective of the motion direction.

Hence, a part of kinetic energy is always spent in overcoming these forces. The viscous forces and frictional force are responsible for the decrease in kinetic energy.

Force acting on a particle is non conservative if the net work done by the force in a complete round trip is non zero.

DIFFERENCE IN CONSERVATIVE AND NON-CONSERVATIVE FORCE FORCE ON THE BASIS OF PATH TRAVERSED

Suppose a particle acted upon by conservative forces goes from A to B along path 1 and returns to A along path 2. As the force is conservative, the work done in the outgoing journey is equal and opposite to that in return journey.

W_A --> 1 --> B = -W_B-->2 --> A

It the particle moved from a to b along the 2 (b) the work done would be equal and opposite to that in moving from b to a along path 2.

W_A --> 1 --> B = W_A --> 2 --> B

Comparing both equations

W_A --> 1 --> B = W_A --> 2 --> B

This shows that work done by the conservative forces in moving the particle from A to B along path 1 is the same as that along the path 2. Here, a force is conservative if it depends only on the positions of particle and independent of path followed between A to B.

If the work done depends upon path, the force is non-conservative.

POTENTIAL ENERGY

When a body is capable to do work by virtue of its position, configuration or state of strain, it is said to possess potential energy. The potential energy of a body or a system of bodies is, a form of stored energy which cab ne fully recovered, and converted into kinetic energy usually it is reopened by U.

CONSERVATIVE FORCE AS -VE GRADIENT OF

When a body is taken from one position to another against an intrinsic conservative force (such as elastic, gravitational, electrical), then the work done in this process is stored as potential energy in the body. (Provided no work done to overcome frictional forces).

Thus, the difference in the potential energy of a body at two positions is defined as the work done in moving the body from is position to the other in the absence of dissipating forces.

Let us consider a particle acted upon by an intrinsic force conservative force F suppose a particle is moved at the x axis from a position x_₍₀ ) to a position x by mean of an external applied force F__(app ) such that (F__(app )=-F). The work done in moving the particle will appear as potential Energy.

The work done by the applied force on the particle in moving from x_₍₀ ) to x.

U(x)-U(x)=∫_(x_₍₀ ))^x▒〖F__(app ) dx〗

Since Fapp=-F

∴U(x)-U(x)=-∫_(x_₍₀ ))^x▒〖F∙dx〗

(Because the work done by the intrinsic force is –ve as the motion of the particle against force)

The position x₀ is so chosen that conservative force acting as the particle is zero.

U(x)-U(x)=-∫_(x_₍₀ ))^x▒〖F∙dx〗 equation (i)

On differentiating

U(x)=-∫_(x_₍₀ ))^x▒〖F∙dx〗=-∫_(x_₍₀ ))^x▒〖dU/dx∙dx〗=-∫_(x_₍₀ ))^x▒dU=U(x)-U(x_₍₀ ) )=U(x)

Because U(x_0 ) is assigned zero value

F=(-dU/dx)

Thus, the conservative force is given by the negative gradient of potential energy.

Another method:

The potential energy U may be defined as a function of position whose –ve gradient gives intrinsic force.

We can extend potential energy equation, U(x)= in 3-D motion. If a particle added upon by a conservative force F moves from a space point of zero potential energy to another space point (x, u, z) describe by position vector r.

U(r)=∫▒〖U(x,y,z)〗=-∫_(x_₍₀ ))^x▒〖F_x∙dx〗-∫_(y_₍₀ ))^y▒〖F_y∙dy〗-∫_(z_₍₀ ))^z▒〖F_z∙dz〗

On partially differentiate the above equation w.r.t. x,y,z

Fx=-dU/dx, Fy=-dU/dy, Fz=-dU/dz

∴F ⃗=Fxi ̂+Fyj ̂+Fzk ̂

Fx=-dU/dx i ̂-dU/dy j ̂-dU/dz k ̂

Fx=-(d/dx i ̂+d/dy j ̂+d/dz k ̂ )U

But we know that (d/dx i ̂+d/dy j ̂+d/dz k ̂ ) Is the vector differential operator ∇ ⃗ or operator (grad) ⃗

∴F=-∇ ⃗U=-grandU

This is the required expression.

POTENTIAL ENERGY

Thus, the difference in the potential energy of a body at two positions is defined as the work done in moving the body from is position to the other in the absence of dissipating forces.

The work done by the applied force on the particle in moving from x_₍₀ ) to x.

U(x)-U(x)=∫_(x_₍₀ ))^x▒〖F__(app ) dx〗

Since Fapp=-F

∴U(x)-U(x)=-∫_(x_₍₀ ))^x▒〖F∙dx〗

(Because the work done by the intrinsic force is –ve as the motion of the particle against force)

The position x₀ is so chosen that conservative force acting as the particle is zero.

U(x)-U(x)=-∫_(x_₍₀ ))^x▒〖F∙dx〗 equation (i)

On differentiating

U(x)=-∫_(x_₍₀ ))^x▒〖F∙dx〗=-∫_(x_₍₀ ))^x▒〖dU/dx∙dx〗=-∫_(x_₍₀ ))^x▒dU=U(x)-U(x_₍₀ ) )=U(x)

Because U(x_0 ) is assigned zero value

F=(-dU/dx)

Thus, the conservative force is given by the negative gradient of potential energy.

Another method:

The potential energy U may be defined as a function of position whose –ve gradient gives intrinsic force.

U(r)=∫▒〖U(x,y,z)〗=-∫_(x_₍₀ ))^x▒〖F_x∙dx〗-∫_(y_₍₀ ))^y▒〖F_y∙dy〗-∫_(z_₍₀ ))^z▒〖F_z∙dz〗

On partially differentiate the above equation w.r.t. x,y,z

Fx=-dU/dx, Fy=-dU/dy, Fz=-dU/dz

∴F ⃗=Fxi ̂+Fyj ̂+Fzk ̂

Fx=-dU/dx i ̂-dU/dy j ̂-dU/dz k ̂

Fx=-(d/dx i ̂+d/dy j ̂+d/dz k ̂ )U

But we know that (d/dx i ̂+d/dy j ̂+d/dz k ̂ ) Is the vector differential operator ∇ ⃗ or operator (grad) ⃗

∴F=-∇ ⃗U=-grandU

This is the required expression.

EQUIPOTENTIAL SURFACE

These are the surfaces in which at every point on the surface has the same potential.

Properties of equipotential surface:

They never cross each other.
They give the direction of the electrical field.
Electrical field is normal to the equipotential surfaces.
Work done in moving a test charge over an equipotential surface is zero.

An equipotential region of a scalar potential in 3-D space in generally is an equipotential surface. The gradient of the scalar potential is everywhere perpendicular to the equipotential surface and zero inside a 3-D equipotential region.

SCALAR POTENTIAL

It describes the situation where the difference in the potential energies of an object in two different potations depends only on the position, not on the path taken by the object in travelling from one point to another point.

The scalar potential is an example of scalar field. For given vector field F, the scalar potential ‘P’ is defined as

F= -∇∙P=-(dP/dx,dP/dy,dP/dz) (Where ∇P is gradient of P)

THE DEL OPERATOR

The del operator, written as grad is a vector differential operator an plays an important role in physics. It is defined as-

∇ ⃗=i ⃗ d/dx+j ⃗ d/dy+k ⃗ d/dz

RADIAN OF SCALAR FIELD

The gradient is a vector operation which operates on a scalar function to produce a vector whose magnitude is the maximum rate of change of the function at the point of the gradient and is described along the direction of max rate of change.

∇f=[i ⃗ d/dx+j ⃗ d/dy+k ⃗ d/dz]f

Let V(x,y,z) be a single valued scalar function with continuous first order derivative. Consider two region having co-ordinate (x,y,z) and (x+dx, y+dy, z+dz)

So, the position vector of point P is-

r ⃗=xi ̂+yj ̂+zk ̂

And r ⃗+dr ⃗=(x+dx) i ̂+(y+dy) j ̂+(z+dz) z ̂

dr ⃗=dxi ̂+dyj ̂+dzk ̂

This gives the displacement vector in going from P to Q now, suppose V at pt. P+Q respectively. Thus, the total change in V in going from P to Q is dv.

So, dV=dV/dx dy+dV/dy dy+dV/dz dz

Or it may write as:

dV=(i ̂ dV/dx+j ̂ dV/dy+k ̂ dV/dz)×(dxi ̂+dyj ̂+dzk ̂ )

dV=(i ̂ dV/dx+j ̂ dV/dy+k ̂ dV/dz)∙dr ⃗

dV=∇ ⃗V∙dr ⃗

CURL OF FIELD

Curl of a vector quantity indicates that the much of the vector quantity curls or twist around.

Let A is a differentiable vector field. Suppose A_x, A_y+A_z Are the components of vector A. Then curl of A is defined as:

curl A=|■(I ̂&J ̂&K ̂@d/dx&d/dx&d/dx@A_X&A_Y&A_Z )|

Curl A is equal to the cross product of operator grad

∴curl A ⃗=∇ ⃗×A ⃗

∇ ⃗×A=I ̂((DA_Z)/Dy-(DA_Y)/Dz)+j ̂((DA_X)/Dz-(DA_Z)/Dx)+k ̂((DA_Y)/Dx-(DA_X)/Dy)

A force is said to be conservative if its curls is zero.

SHOW THAT Curl F ⃗=0 FOR A CONSERVATIVE FORCE F ⃗

Proof: A conservative force is the –ve gradient of potential energy.

∴F ⃗=-grad U=-∇ ⃗∙U

curl F ⃗=∇ ⃗×F ⃗=∇ ⃗×∇ ⃗U [Ignoring –ve sign]

curl F ⃗=∇ ⃗×(dU/dx i+dU/dy j+dU/dz k)

curl F ⃗=|■(I ̂&J ̂&K ̂@d/dx&d/dx&d/dx@dU/dx&dU/dy&dU/dz)|

curl F ⃗=I ̂((D^2 U)/(Dy∙dz)-(D^2 U)/(Dz∙dy))+j ̂((D^2 U)/(Dx∙dz)-(D^2 U)/(Dz∙dx))+k ̂((D^2 U)/(Dx∙dy)-(D^2 U)/(Dy∙dx))

U is a perfect differential so that, (D^2 U)/(dy∙dz)=(D^2 U)/(Dz∙dy), (D^2 U)/(Dx∙dz)=(D^2 U)/(Dz∙dx),(D^2 U)/(Dx∙dy)=(D^2 U)/(Dy∙dx)

curl F ⃗=I ̂((D^2 U)/(Dy∙dz)-(D^2 U)/(Dy∙dz))+j ̂((D^2 U)/(Dx∙dz)-(D^2 U)/(Dx∙dz))+k ̂((D^2 U)/(Dx∙dy)-(D^2 U)/(Dx∙dy))

curl F ⃗=I ̂(0)+j ̂(0)+k ̂(0)

curl F ⃗=∇ ⃗×F ⃗=0

FRAME OF REFERENCE

A co-ordinate system relative to which the motion of an object is described is known as a frame of reference. In other words, it is a system of co-ordinates w.r.t. To which the position of the particle or an event in two or three dimensional space may be described.

There are two types of reference frames:

INITIAL FRAME

The frame at rest or moving with a constant velocity is known as initial frames. In these frames, Newton’s 1^st and 2^nd law holds good. A frame of reference at rest or moving with constant velocity (uniform motion) w.r.t. To an inertial frame is also an inertial frame.

NON-INERTIAL FRAME

These are the frames in which Newton’s la does not hold well. All the reference frames accelerated relative to an inertial frame are non-inertial frame.

CENTRIPETAL AND CENTRIFUGAL FORCES

The component of force acting on a body in curvilinear motion, which is directed towards the center of curvature or axis of rotating, is called Centripetal force.

Centrifugal force is defined as, “the apparent force, which is equal and opposite to the centripetal force acting on a rotating body away frame the center of rotation, caused by the inertia of the body.

Centripetal force is an actual force where as centrifugal force is an apparent force.

Centripetal force Fc=m∙v^2/r

units = Newton

FACTIOUS OR PSEUDO FORCES

These are the apparent forces that act on all mares whose motion is described using a non inertial frame of reference, such as rotating frame. It is also known as pseudo force. Such forces arise, when a frame of reference is accelerating compared to a non accelerating frame.

For example, if a person standing at bus top, watching an accelerating car, he infers that a force is exerted a car and hence car is accelerating.

But if the person inside the accelerating can is looking at the person standing at the bus stop, he finds that the person is accelerating with respect to the car, though no force is acting on it. Here the concept of pseudo force is required to convert the non-inertial frame of reference to an equivalent inertial frame.

Suppose that S is an inertial frame and another frame S’ moving with an acceleration a₀ relative to S. The acceleration of a particle P on which no force is acting, will be zero in frame S, but in frame S’ the observer will find that on acceleration –a_₀ is acting on it. Thus in frame S’ the observed force on the particle is –ma_₀. Such force which does not actually act on the particle, but appears blue to the acceleration of the frame is called fictions or pseudo force.

The fictionus force on the particle P is F_0=–〖Ma〗_0 [The accelerate frame s’ is non inertial]

Real force acting on the particle P is F_i=–〖Ma〗_I

So, the observer in the accelerated frame will measure the resultant force which is the sum of real and fictionus force.

CENTRIPETAL ACCELERATION

Acceleration is a change in velocity; wither in its magnitude or in direction or both. In uniform circular motion, the direction of velocity changes constantly, so there is always an associated acceleration, even the speed is constant.

Consider an object moving in a circular an object moving in a circular path at constant speed. The direction of the instantaneous velocity is shown at two points along the path. The direction of the acceleration is shown with the vector diagram.

Centripetal acceleration is defined as the rate of change tangential velocity so, a ⃗_c=〖Lim〗┬(∆t→0)⁡〖(∆ν ⃗)/∆t〗

The direction of centripetal acceleration is always inward along the radius vector of circular motion.

The triangle formed by the velocity vectors and the triangle formed by the radii r and ∆s are similar. Both triangles ∆ABC and ∆PQR are isohels triangle with two equal sides. By sing the property of two similar triangles, we have v1=v2=v3

So, ∆ν/v=∆s/r

Now, acceleration is ∆ν/v So, we have to find expression for ∆v

And ∆v=v/r∙∆s

∆v/∆t=v/r∙∆s/∆t [Dividing by ∆t both sides]

We know that ∆v/∆t=a_c And ∆s/∆t=v, lines or tangent speed

So, a_c=v/r∙v

Centripetal acceleration, a_c=v^2/r

This is the accelerations of an object in a circle of radius at speed v. So centripetal acceleration is greater at high speeds and in sharp curves –(smaller radii). As we notices while driving in car.

CORIOLIS FORCE

The change in pressure measured across a given distance is called pressure gradient. The pressure gradient results in a net force that is directed frame high to low pressure and this force is called pressure gradient force.

The air has been set in motion by the pressure gradient. It undergoes an apparent deflection from its path as seen by an observer on the earth this apparent deflection is called Coriolis force and it’s a result of earth rotation.

As the air moves from high to low pressure in northern hemisphere, it is deflected to the right by Coriolis force. In the southern hemisphere, on moving from high to low pressure is deflated to the left by the Coriolis force.

Coriolis force is zero at the equator. The amount of defilation that air makes is directly related to both the speed and its latitude.

CORIOLIS ACCELERATION

It is due to Coriolis force which is basically due to the rotation of any object rotating about its own axis due to which any object gets deflated from its original position in the rotating frame.

It is given by Coriolis acceleration=2×v×ω

Where, v(velocity) normal to ω (angular velocity)

We can realize this acceleration seeing the weather phenomena because wind circulated clockwise about regions of

High pressure in N-hemisphere and it circulates in anticlockwise in S-hemisphere and due to which India also benefits from monsoon winds.

Moving air undergoes an apparent deflection from its path as see by the observer on the earth. This apparent defilation it’s the result of Coriolis acceleration. Since the amount of deflection of air makes is directly related to both the latitude air speed. Thus slowly blowing winds will be deflected only a small amount, while strangest winds deflects more. Similarly, the winds blowing closer to the pales will be deflated more than wind sat the same speed closer to equator. The Coriolis force only acts on large objects like air manes moving considerable distance. For example, small objects ship at sea our two small to experience significant defoliation in the direction due to Coriolis acceleration. Hence, the Coriolis acceleration is particularly significant with regards to winds, ocean onset and tidal streams.

FOUCAULT PENDULUM

The Foucault pendulum is simple device named after French physicist learns Foucault used to shoe or to demonstrate the earth’s rotation. This pendulum was introduced in 1851 and was the first experiment to give simple direct evidence of the earth restoration.

Constructions: In this pendulum a relatively large mars suspended from a long line mounted so that its perpendicular place of swing it’s not confined to a particular direction and rotates in relations to the earth’s surface. Originally this pendulum consisted of 28 kg iron ball suspended from inside the dame of a building by a steel wire of 220 feet length, and set in motion by drawing the ball to one side carefully releasing it to start its swinging.

Working: When the Foucault pendulum swings back and forth in a place the earth rotates beneath it, so that relative motion exists between them. At the north pole latitude 90⁰, the relative motion as viewed from above in the plane of the pendulums suspensions is a counter clockwise rotation of the earth once approximately every 24 hrs (23:56:04). Correspondingly the plane of the pendulum as viewed from the above appears to rotate in a clockwise directions once a day a Foucault pendulum always rotates clockwise in the Northern hemisphere with a rate that became slower as the pendulum location approaches the equator. The rate of rotation depends on the latitude at the equator 0⁰, latitude; a Foucault pendulum does not rotate. In the southern hemisphere, rotation is counter clockwise.

The rotations of Foucault pendulum can be stated mathematically as equal to the rate of rotation of earth X sum of the number of degrees of latitude. Because the earth rotates once a sidereal day or 360⁰ approx every 24 hrs, its rate of rotation may be expressed as 15⁰ per hrs.

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