Tuesday, October 03, 2017

AHSEC/CBSE - Class 11 Notes (Subject - Physics): Work, Energy and Power

Chap – 6
Work energy and power
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Work is said to be done if on application of force on a body, the body is displaced in the direction of force. Mathematically work is the product of force and displacement of a body provided, both force and displacement of the same direction.
i.e.


Work done by a constant force: Let a force F applied on a body such that the direction of force makes an anglewith displacement S, the force can be resolved into two componentsalongandperpendicular toand shown in figure below.

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According to mathematical definitions we can write –
Work may also be defined as the dot product of a force and displacement of a body.
Special Case:
  1. If


  1. If


  1. If
Nature of work:
  1. Positive work: Work is said to be if force and displacement are along the same direction. In this case, then. E.g. If a body falls freely under the effect of gravitational force than work will be.
  2. Negative work: Work is said to beif force and displacement are in opposite direction. In this case, then. E.g. when a body is thrown up against gravitational force then the direction of force and displacement is opposite and work will be.
  3. Zero work: Work is said to be zero if application of force the body does not move. E.g. a man pushing the wall which does not move or work can be zero if force and displacement are 1 to each other
i.e.
Work done by a variable force: A force which changes with displacement of force a object is called variable force. A plot of variable force with displacement is shown in the figure below:

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For a very small displacement, forceis almost constant.
By using the formula work done by a constant force, we can write work done for a small displacement
Area of
In a similar way if we decide the entire displacement betweenandand final work done for each such displacement then the total work done can be obtained by adding all these individual work done.
i.e.
= area under the graph betweenand
In integration form this can be written as –
Q. A force which depends upon displacement of the object is given by. Calculate work done by these force if object move fromto.
Ans.
Q. A forceact on a body and the body move from positionand. Calculate the work done in this force.
Ans.


Energy: energy is the capacity of doing work.
Mechanical Energy: Mechanical energy is basically of two types –
  1. Potential energy: The energy possessed by an object by virtue of its position or configuration is called potential energy. The potential energy of an object at an height ‘h’ above the ground is given by.
  2. Kinetic energy: The energy possessed by an object by virtue of its motion is called kinetic energy. The K.E. of an object with a velocity ‘V’ is given by.
Work energy theorem: According to this theorem the work done on a body is equal to the change in Kinetic energy of a body. i.e.
Proof: For constant force.

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The final velocity of an object which move through a distance ‘S’ is given by, where= initial velocity and= acceleration.
Multiplying both side with
Equation (i)
Work = change in K.E. of a body.


Proof: The Kinetic energy of an object moving with velocity ‘v’ is given
Differentiating both sidetimes we get –


Let at, then, and
at, then, and
Integrating equation (ii) in the above limit we get –




Laws of conservation of energy: According to laws of conservation of energy it can neither be created nor be destroyed it can only converted into one form to another form.
Q. Proof that total mechanical energy is conserved at every point of the path of a freely falling object. D:\Logo\New Physics Final\Untitled-16 copy.jpg











Let an object of massbe held at pointat a heightabove the ground.
Total energy,
Let the particle be allowed to fall down. Let a point, the velocity of the object be –
then
Letbe the velocity with which the object strikes the ground then
From equation (i) (iii) and (v) we find that the total energy of A, B and C is same, indicating them mechanical energy is conserved at every point of the path of a freely falling object.
Collision: A kind of interaction between the particle in which energy of a particle may change is called collision.
Types of Collision:
  1. Elastic collision: A kind of collision in which momentum and kinetic energy thus not change is called elastic collision. E.g. when the north pole of an magnet is bought close to the north pole of another suspended magnets then the suspended magnet is displaced from its position. This types of collision is called elastic collision.
  2. Inelastic collision: A kind of collision in which momentum is conserved but kinetic energy is not conserved is called inelastic collision. E.g. (i) when a stone smashes a window then collision between a stone and the window is inelastic. (ii) A ball drove from a certain height will not rebound to the same height if it has inelastic collision with the striking surface.
  3. Perfectly inelastic collision: A collision is said to be perfectly inelastic if the two bodies after collision stick together and move as one body. In these collisions linear momentum of the system remains conserved but kinetic energy is not conserved. E.g. when a moving bullet hits the stationary wooden block then it is embedded into the wooden block and both move as one body.

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Elastic collision in one Dimension:

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A body of massmoving with initial velocitycollides with another body of massmoving with initial velocity. After collision let their velocity changes toandrespectively. Then using the laws of conservation of momentum –
Which is the expression for final velocity of massafter collision?
Special Cases:
  1. If both the particles are of equal masses i.e. , then equation
Therefore, if two bodies of equal masses collide elastically then show.
Case II: If the second body at rest
But Case I: If
Therefore, when a very large object collides with lighter object at rest then there is practically no change in velocity of the heavier object but the lighter object moves with double the speed of heavier object.
Sub Case 4:
If
If a small body collides with a very large body at rest then the lighter body bounces back with same speed while the heave body remains at rest.
Q. Show that in a perfectly inelastic collision kinetic energy is lost and hence calculate the laws of energy.

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The figure shows a perfectly inelastic collision with massmoving with initial bodycollides with another body of massat rest. After collision the two bodies stick together and move with a common velocity.
Applying the laws of conservation of linear momentum –


∴ There is a loss of K.E. in a perfectly inelastic collision.
Therefore loss of energy
Which is the expression for loss of K.E. in an one dimensional perfectly inelastic collision.
Collision in two dimensions: D:\Logo\New Physics Final\Untitled-19 copy.jpg








Consider two body A and B of massesandrespectively. Let body initially at rest before the collision. Let body A moving in a straight line with velocityalong x-axis collides with the body B. Such that their centre are not in a line as shown in the figure. Such a collision is known as off-centre collision or collision at a distance. The perpendicular distance between the timeand the centre of body B is known as impact parameter. The impact parameter measure the closure of the collision of impact parameter is O, the collision is head on collision.
Let after collision body A moves with velocityand making an angle withwith x-axis. Let body B moves with velocitymaking an angle with x-axis. The angleis known as the deflecting angle or scathe angle and the angleis known as the angle of recoiled.
Since in all type of collision momentum is conserved. Therefore, the momentum is conserved along x-axis and     y-axis.
If the collision is perfectly inelastic then the kinetic energy is also conserve. Therefore K.E. before collision is equal to K.E. after collision.
Assume thatandare known now the motion after collision in values four unknown quantities.
To evaluate the values of these four quantities we need four questions. However we have only three equations. To overcome these problems, the easiest way of developing fourth equation is to measure the angle of deflectionand the angle of recoilexperimentally.


Chap – 7
Rotational Motion and System of Particles


System of particles: A system of particle consisting large number of particle which interact among themselves via internal forces is called system of particles.
The internal forces do not cause motion in a body. Motion is only caused by internally applied force.
Centre of mass: A point within or outside the body at which whole mass of the body is suppose to be concentrated is called centre of mass.
Expression for position vector of the centre of mass of the two particle system:

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Let us consider a system of two particles of massesandhaving position vectorandrespectively. andare the internal forces acting between the particles
Ifandbe the external forces acting on the particle (1) and (2) respectively, then the equation of the motion of the particles of, of massis –
Assuming that the whole mass of the system is at centre of mass, which has a position vector, we can write the equation of motion of this point us –
Comparing (4) and (5) we get –
Which is the expression for position vector of the centre of mass of a two particle system?
Special Case:
  1. If both the particle are of same mass then –
The position vector of centre of mass of two particles of equal masses is equal to the mean of the position vector of the two particles.


  1. If the centre of mass lies in the origin of the coordinate system then
From the above equation we find that ifis position thenwill be negative and vice-versa.
If,
Centre of mass will be closer to the heavier particle.


  1. If there arenumber of particles in the system, then the position vector of the centre of mass can be written as –
Motion of the centre of mass: The position vector of the centre of mass ofparticles system is given by –
From equation (2) we find that centre of mass is moving with accelerationsuch that all external forces are acting on the centre of mass resulting in this acceleration.
Momentum Conservation of centre of mass: The force acting on a system of particle can be written as –
Where
If no external force acts on the system of particles, then the momentum of the centre of mass of a system of particles will be constant which is the law of conservation of momentum.
Torque: The product of linear force and perpendicular distance of the line of action force from the axis of rotation is called torque.
The rotational effect produced by a linear force is called torque.

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Let us consider a particle of massat point. Letbe the position vector of the particle. Ifbe the linear force acting on the particle andbe the angle betweenand. Then by the definition of torque,
Special Case:
  1. If
A door can be easily rotate by applying a force perpendicular to the plane of the door.


  1. If
a door cannot due rotate by puling it away from.



  1. Torque will be large ifis large. This is the reason why wrenches are provided with long handles.
Angular momentum: D:\Logo\New Physics Final\Untitled-22 copy.jpg







It is define as the product of linear momentum the perpendicular distance of the line of action of force from the axis of rotation is called angular momentum.
I.e.
Moment of inertia: The moment of inertia of a system of particles may be defined as the sum of the product of masses of the particles constituting in the system and square of the perpendicular distances of the warders particles from the axis of rotation.

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The figure shows a system ofparticles of masses. Letare the perpendicular distances of the particle from the axis of rotation.
By the definition of moments of Inertia can be written as –
Radius of gyration: It is the perpendicular distance of a point from the axis of rotation at which whole of mass the body is supposed to be concentrated, and then the moment of inertia would be same as with actual distribution of masses. D:\Logo\New Physics Final\Untitled-24 copy.jpg








Let us consider a system ofparticles having massesand the perpendicular distance of these particle from the axis of rotation are, then
Ifbe the radius of gyration,
From (1) and (2) we get
If all the particles are of equal masses then
Radius of gyration may also be defined as the root mean square distance of the particles.
Theorems of moment of Inertia:
  1. Parallel axis theorem: The moment of inertia of a plane lamina or body about any axis AB is equal to the sum of the moment of inertia of the lamina about the parallel axis passing through the mass and the product of total mass of the lamina with square of the distance between two axis.

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i.e.
Proof: Let us consider a massat pointat a distance offrom centre of mass then the moment of inertia of the particle about
Moment of inertia of the whole lamina of about the axis
Perpendicular axis theorem: According to these theorems the moment of inertia of the lamina about an axisperpendicular to the plane of the lamina is equal to the sum of the moment of inertia of two mutually perpendicular axis lying on the plane of the lamina.
i.e. D:\Logo\New Physics Final\Untitled-26 copy.jpg







Let us consider a particle of massat a distancefromaxis from the figure
The moment of inertia of massatabout-axis is equal to
The total moments of inertia bout-axis
[Whereandare the moment of inertia of the body about y-axis and x-axis respectively.


Relation between Torque and Moment of Inertia: The torque acting on a system of particle given by –
Whereare the perpendicular distance of the particle from the axis of rotation andare the various forces acting of the particle constituting on the system –
The above equation is the Newton’s 2nd Law of motion in terms of rotational motion.
Relation between angular momentum and moment of Inertia: Angular momentum of a system of particle is given by –
Whereare the perpendiculars distance of the particle from the axis of rotation andis the linear momentum of the system of particle.
Which is the rotation ship between angular momentum and moment of Inertia?
Relationship between torque and angular momentum: The angular momentum of a system of particle is given by,
Differentiating both side with respect to time,
Laws of conservation of angular momentum: According to this law, the angular momentum of a system of particles remains constant if the net torque acting on the particle is 0.
Proof: The relation between angular momentum of a system of particle and torque is given by,


If the torque acting on the system of particle is zerothen,
Which is the law of conservation of angular momentum?
the law of conservation of momentum can also be written as,


Application of the law of conservation of angular momentum:

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When the earth is closed to the sum in its elliptical orbit then the angular velocity of the earth increases. It is because when earth closer to the sun the moment of Inertia will be less. So to keep angular momentum constant the angular velocity increases. It is in accordance with the laws of conservation of angular momentum in which, = Constant. Similarly, when earth is away from the sun, then its moment of Inertia increases so to keep angular momentum constant angular velocity decreases.


Rotational Kinetic Energy: D:\Logo\New Physics Final\Untitled-28 copy.jpg









Let us consider a system ofparticles having masses. Letbe the perpendicular distances from the axis of rotation andbe the corresponding linear velocity. Then the Kinetic energy of the system will be –


Kinetic energy of a rolling body:

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Let us consider a body of massrolling on the floor with linear velocityand angular velocity. The body will have two types of motion (i) Rotational motion, (ii) Translational motion, so the kinetic energy of the rolling body can be written as –
Which is the expression for total kinetic energy of a rolling body?


Equilibrium of a rigid body: A rigid body is said to be in equilibrium if it is in translational as well as rotational equilibrium state. For translational and rotational equilibrium the net force and net torque acting on the system must be equal to zero.
Two condition of equilibrium:
  1. 1st condition of equilibrium (Translational equilibrium): A body is said to be in translational equilibrium if it remains stationary or if it remains move with constant velocity. Therefore for translational equilibrium
or
According to Newton 2nd law
Also,
For translational equilibrium the net force acting on the system must be zero (0).
  1. 2nd Condition of Equilibrium (Rotational Equilibrium): A body is said to be in rotational equilibrium of it does not rotates (i.e. w = 0) or if it rotate with constant angular velocity (i.e. w = 0). The torque acting on the system is given by –
For rotational equilibrium, or
Equation
If the torque acting on the system is 0 then the system is said to be in rotational equilibrium
[The above object is in rotational D:\Logo\New Physics Final\Untitled-30 copy.jpg
equilibrium because it is not
rotational]

D:\Logo\New Physics Final\Untitled-30-30 copy.jpg

[Here, the body is in
translational equilibrium]
Principle of moments:

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According to the principle of moment if the anti-clock wise torque is equal to clock wise torque then the net torque acting on the system becomes zero and the body remains in rotational equilibrium.
From the figure
If
According to principle of moments, for rotational equilibrium,
Moment of Inertia of a circular ring:
  1. Moment of Inertia of a ring about an axis through its centre and perpendicular to its plane. D:\Logo\New Physics Final\Untitled-32 copy.jpg


  1. of ring about any diameter
D:\Logo\New Physics Final\Untitled-32-32 copy.jpg


Moment of Inertia of ring: D:\Logo\New Physics Final\Untitled-33 copy.jpg
  1. About an axis passing through the centre of ring andto its plane.
  1. About the diameter the ring letbe the moment of inertia of the ring about each diameter→(2)
According to theorem ofaxis
D:\Logo\New Physics Final\Untitled-33-33 copy.jpg


  1. About an axis tangential also the ring and it’s to the diameter of ring. Atbe the moment of inertia of the ring about the tangent AB parallel to diameter.
Applying the theorem of axis
D:\Logo\New Physics Final\Untitled-33-33-33 copy.jpg
  1. About the tangent II to the axis passing through the centre of the ring andto its plane. Letbe the moment of inertia of the ring about the tangent t. AB II to an axis passing through the centre of ring andto its plane. D:\Logo\New Physics Final\Untitled-34 copy.jpg
Applying theorem of II axis we get
From the about this equation we conclude that.
(i)of a ring is maximum about the tangent passing through the centre of the ring andto its plane.
(ii) of ring is minimum about the diameter of the ring. Thusof a body depends upon axis of rotation.


M.I. of a circular disc:
(1) About axis passing through the centre of disc andto its planeD:\Logo\New Physics Final\Untitled-35 copy.jpg


(2) About the diameter of the disc letbe theof the disc about each diameter,of the disc. According to the arum ofaxis we have –
D:\Logo\New Physics Final\Untitled-35-35 copy.jpg

(3) About an axis tangential to the disc and II to diameter of the disc.
Applying theorem of II axisD:\Logo\New Physics Final\Untitled-35-35-35 copy.jpg


(3) About an axis tangential to the disc andbe the plane of the disc.
Applying the theorem of II axisD:\Logo\New Physics Final\Untitled-35-35-35-35 copy.jpg


M.I. of a uniform rod:
  1. About an axis passing through its centre andto its plane.

D:\Logo\New Physics Final\Untitled-36 copy.jpg





  1. About an axis passing through its end andto its length.

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According to theorem of II axis we get


M.I. of a solid cylinder
(i) About its an axis D:\Logo\New Physics Final\Untitled-37 copy.jpg
M.I. of a hollow sphere:
(ii) About its diameter
M.I of a solid sphere:
(iii) About its diameter


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