How to apply law of conservation of energy in mechanics

We often comes across the problems in mechanics where we need to apply the law of conservation of energy where gravitational potential energy or gravity is involved . For solving such problems you can consider the following problem solving strategy,

1. First of all define the system which includes all the interacting bodies . Now choose a zero point for gravitational potential energy according to your convenience.
2. Select the body of interest and identify the point about which information is given in the question. Also identify the point where you want to find out asked quantity about the body of interest.
3. Check for the possibility of the presence of non-conservative forces. If there are no non-conservative forces present then write down the energy conservation equation for the system and identify the unknown quantity asked in the question.
4. Solve the equation for the unknown quantities asked in the question by substituting the given quantities in the equation obtained.

Mechanics :- Work and Energy (points to remember)

1. Work done on an object by a constant force F is
W=F.Δx
where F is the magnitude of the force, Δx is the magnitude of the displacement,and F and Δx point in the same direction.
SI unit of work : joule ( J) newton.meter
2. Work done is a simple number that is it is a scalar quantity not a vector. So there is no direction associated with it. Energy and energy transfer are also scalar quantities.
3. The kinetic energy KE of an object of mass m moving with a speed v is
defined by
T=1/2(mv²)
SI unit: joule ( J) kg.m²/s²
4. The net work done on an object is equal to the change in the object’s
kinetic energy: W_tot=T_f-T_i=ΔT
where the change in the kinetic energy is due entirely to the object’s change in speed
5. A force is conservative if the work it does moving an object between two points is the same no matter what path is taken.
6. The gravitational potential energy of a system consisting of the Earth and an object of mass m near the Earth’s surface is given by
V=mgy
where g is the acceleration due to gravity and y is the vertical position of the mass with respect to the surface of the Earth (or some other reference point).

Mass on inclined plane

Rigid Body Rotation Question 1

Question:
A rigid body of radius of gyration k, mass m and radius R rolls without slipping down a plane inclined at an angle θ  with the horizontal. (a) Find its acceleration and frictional force acting on it . (b) If the body were in the form of a disc and θ=30º , what will be the acceleration and the frictional force acting on it.

Solution:
for solution click this link

conceptual physics Centripital force

Hi all
I came across this video while browsing you tube and i like the way it explains about what is the concept of centripetal force.

Kinematics problem

Question
A car accelerates from rest at a constant rate p for some time after which it decelerates at a constant rate q to come to rest. If total time lapse is t, find (1) the maximum velocity attained and (2) the total distance traveled.

Solution

For solution visit the link

Inertial and non inertial frames

• Both rest and motion are relative terms and there is nothing like absolute rest or absolute motion.
• Position or state of motion of a body may appear different in different frame of reference i.e.,  an object at rest in one frame of reference might appear to be in motion in another frame of reference.
• Newton's first law of motion also known as law if inertia does not always holds for all frame of reference and the frames in which this law holds good are called inertial frames of reference.
• Inertial frames of reference are non-accelerating frames this means that either they are fixed or move with constant velocity.
•  Non-inertial frames are the frames in which newton's law of inertia does not holds true. 
• Rotating and acceleration frame of references falls in the category of non inertial frames.
• In this frame acceleration is caused by fictitious or pseudo forces.

For more or full notes on kinematics visit this link

Surface tension fact file

1. Those factors which increases cohessive forces between liquid molecules increases surface tension and those decreases cohesive forces decreases surface tension.
2. If impurity is completely soluble , then on mixing it in the liquid , its surface tension increases.
3. With the increase in temperature surface tension decreases.
4. Surface tension depands on the medium present on other side of the liquid surface.
5. The surface tension decreases due to electrification.
6. On mixing partial soluble impurities in a liquid , its surface tension decreases like mixing of detergent in water decreases surface tension of water.

Challenging mechanics problem : Kinematics

Question :
A body of mass m rests on horizontal table which has coefficient of static friction μ . It is desired to move the body by applying minimum possible force F. Find the magnitude and direction of the force F to which it has to be applied.
Solution :
For solution visit this link

kinematics: position and displacement

Here in this post i have mad an attempt to explain the concept of position , distance and displacement .Hope you like it

Solve out mechanics problems (IITJEE tips and tricks for mechanics)

Hi all here in this article i am giving some tips and tricks to solve problems in mechanics. To master problem solving skill you need to practice more and more problems as said practice makes a man perfect. Look at examples in your text books identify the steps and then try solving out problems on your own.
 
Motion in a Two dimensional Plane

1. Select a coordinate system and resolve the initial velocity vector into x and y components.
2. Find out acceleration in each direction and solve in each direction according to one rectilinear motion equation.
3. If the acceleration is in vertical direction only.Follow the techniques for solving constant-velocity problems to analyze the horizontal motion. Follow the techniques for solving constant-acceleration problems to analyze the vertical motion. The x and y motions share the same time of flight t.
4. There might be question about trajortory in the Problem ,find out the motion in x and y direction with respect to time from previous point.And then find the value of t from one equation and then put that value in another equation to find out the equation of trajactory

Motion in a Three dimensional Plane
1. Select a coordinate system and resolve the initial velocity vector into x , y and z components
2. Find out acceleration in each direction and solve in each direction according to one rectilinear motion equation.

Uniform Circular Motion
1. Draw a simple, neat diagram of the system.
2. Firstly consider the origin of the forces acting on the each object.To do this find out the field forces acting on the each object.Wherever contact in available account the contact force carefully
3. Find out the force acting on the body.The resultant force should provide the required centrepatal required for Circular motion
4. Centrepatal force=mv2/R will give the velotiy accordingly
For more visit http://physicscatalyst.com/articles.php

Comparison between linear motion of a particle and rotational motion of a rigid body





Concept of force

• Concept of force is central to all of physics whether it is classical physics,nuclear physics,quantum physics or any other form of physics
• So what is force? when we push or pull anybody we are said to exert force on the body
• Push or pull applied on a body does not exactly define the force in general.We can define force as an influence causing a body at rest or moving with constant velocity to undergo an accleration
• There are many ways in which one body can exert force on another body
  Few examples are given below
  (a)Stretched springs exerts force on the bodies attached to its ends
  (b)Compressed air in a container exerts force on the walls of the container
  (c) Force can be used to deform a flexible object
  There are lots of examples you could find looking around yourself
• Force of gravitational attraction exerted by earth is a kind of force that acts on every physical body on the earth and is called the weight of the body
• Mechanical and gravitation forces are not the only forces present infact all the forces in Universe are based on four fundamental forces
  (i) Strong and weak forces: These are forces at very short distance (10^-05 m) and are responsible for interaction between neutrons and proton in atomic nucleus
  (ii) Electromagnetic forces: EM force acts between electric charges
  (iii) Gravitational force -it acts between the masses
• In mechanics we will only study about the mechanical and gravitational forces
• Force is a vector quatity and it needs both the magnitude as well as direction for its complete description
• SI unit of force is Newton (N) and CGS unit is dyne where
  1 dyne= 10⁰⁵ N

Kinetic Energy

For full notes on Work , Energy and Power visit Physicscatalyst.com

• Kinetic energy is the energy possesed by the body by virtue of its motion
• Body moving with greater velocity would posses greater K.E in comparison of the body moving with slower velocity
• Consider a body of mass m moving under the influenece of constant force F.From newton's second law of motion
  F=ma
  Where a is the acceleration of the body
• If due to this acceleration a,velocity of the body increases from v₁ to v₂ during the displacement d then from equation of motion with constant acceleration we have
  v₂² -v₁²=2ad or
  a=v₂² -v₁²/2d Using this acceleration in Newton's second law of motion
  we have
  F=m(v₂² -v₁²)/2d
  or
  Fd=m(v₂² -v₁²)/2
  or
  Fd=mv₂²/2 -mv₁²/2           (7)
  We know that Fd is the workdone by the force F in moving body through distance d
• In equation(7),quantity on the right hand side mv²/2 is called the kinetic energy of the body
  Thus
  K=mv²/2
• Finally we can define KE of the body as one half of the product of mass of the body and the square of its speed
• Thus we see that quantity (mv²/2) arises purely becuase of the motion of the body
• In equation 7 quantity
  K₂=mv₂²/2
  is the final KE of the body and
  K₁=mv₁²/2
  is the initial KE of the body .Thus equation 7 becomes
  W=K₂-K₁=ΔK           (9)
• Where ΔK is the change in KE.Hence from equation (9) ,we see that workdone by a force on a body is equal to the change in kinetic energy of the body
• Kinetic energy like work is a scalar quantity
• Unit of KE is same as that of work i.e Joule
• If there are number of forces acting on a body then we can find the resultant force ,which is the vector sum of all the forces and then find the workdone on the body
• Again equation (9) is a generalized result relating change in KE of the object and the net workdone on it.This equation can be summerized as
  K_f=K_i+W           (10)
  which says that kinetic energy after net workdone is equal to the KE before net work plus network done.Above statement is also known as work-kinetic energy theorem of particles
• Work energy theorem holds for both positive and negative workdone.if the workdone is positive then final KE increases by amount of the work and if workdone is negative then final KE decreases by the amount of workdone

Pseudo or Fictitious forces

Consider a particle P in stationary frame of refrance S in which no force is acting i.e., a=0. If any other frame of refrance S' is moving with acceleration a w.r.t. frame S, then an acceleration -a₀ appears to be acting on the particle P w.r.t. S'. Therefore a force -ma₀ seems to be acting on the particle P due to the accelerated motion of the frame S'. This force is known as Pseudo force or fictitious force. Therefore pseudo forces are the force which does not actually acts on the particle but seems so because of the accelerated motion of the refrance frames are. This accelerated frame of refrance S' is a non inertial frame of refrance . If due to force F_R on the inertial frame of refrance , the particle seems to suffer an acceleration a_R then by Newtonian mechanics F_R=ma_R Therefore resultant force on particle in accelerated frame is
F=F_R+F_a
ma=ma_R+(-mF₀)
or,
a=a_R-a₀
where F_a is fictitious force. Direction of fictitious force is opposite to the accelerated motion of S' frame of refrance.

Physics for IITJEE:Exercise your Brain with Miscellaneous Physics Question

Question 1:
A 2 kg body moving to the right with speed 6 m/s coliides elastcally with a stationary body of mass 4 kg.
a.Find the velocity vector of each mass relative to center of mass before and after the collsion
b.Find the velocity vector of the each mass in original frame after the collision
c.How much energy was transfered to the 4 kg body.

Take i as the unit vector in the right direction

Answer
a.
Before collision
Velocity vector of mass 2 kg=4i m/s
Velocity vector of mass 4 kg=-2i m/s

After collision
Velocity vector of mass 2 kg=-4i m/s
Velocity vector of mass 4 kg=2i m/s

b
Velocity vector of mass 2 kg=-2i m/s
Velocity vector of mass 4 kg=4i m/s

c. 32 J

Hints:
Velocity of the center of mass=m₁v₁+m₂v₂/m₁+m₂
Velocity w.r.t CM=v-v_cm

Applying law of linear momentum and energy conservation,veloclities after collosion could be find out


Question 2:
A small solid cylinder of radius r rolls down from the top of sphere of Radius R without slipping.
a. find the angle made from the vertical at which the cylinder loses contact with the surface of the sphere
b.find the angular velocity of the cylinder at the moment it loses contact with the surface of the sphere

Answer
a. cos^-1(4/7)
b.(2/r)[(R+r)g/7]^1/2

hints:

Normal reaction will become zero at the point of losing contact
so centripetal force will become equal to the weight components across the radius.
Also applying law of conservation of energy ...we can find the solution


Question 3:

An gas obeys Vander waal equation

(P+a/V²)(V-b)=nRT

Find following for such a gas

A. The dimensional formula for constant a is
a.ML⁵T^-2
b.ML⁵T^-1
c.ML^-1T^-1
d. M⁰L^-0T⁰


B. Find the workdone by this gas when gas expand from V to 2V at constant temperature T
a.nRTlog_e(2V-b/V-b) -a/V
b.nRTlog_e(2V-b/V-b) +a/V
c. nRTlog_e(2V-b/V-b)
d. none of the above

C.Bulk modulus at constant pressure is defined as
K=(-1/V)(dP/dV)_T
Find the value of k for this gas
a.(1-b/V)(p-a/V²+2a/V³)
b. (1+b/V)(p-a/V²+2a/V³)
c.(1-b/V)(p-a/V²-2a/V³)
d.None of the above

Answer:
A.ML⁵T^-2
B.nRTlog_e(2V-b/V-b) -a/V



Question 4:
An electron (mass m and charge e) is moving along the axis of the ring of radius r and carrying a charge q.Find the time period of small oscillation of the electron about the center of the ring

Answer
T=4π(πε₀mr³/eq)^1/2

Hint:
Electric field at a point on the axis of the ring at the distance x from its center is
E=qx/4πε₀(r²+x²)^3/2

Rotation Objective Test

1. A mass is whirled in a circular path with constant angular velocity and its angular momentum is L.If the string is now halve keeping the angular velocity same then angular momentum is
a. L
b. L/4
c. L/2
d. 2L



2.A mass is moving with constant velocity along a line parallel to xaxis away from origin.its angular momentum with respect to origin is
a. is zero
b. remains constant
c. goes on increasing
d. goes on decreasing


3.A cylinder rolls up the incline plane reaches some height and then roll down without slipping through out this section.The direction of the frictional force acting on the cylinder are
a. Up the incline while ascending and down the incline while descending
b.Up the incline while ascending and desending
c. down the incline while ascending and up the incline while descending
d.down the incline while ascending and desending


4.A uniform sold sphere rolls on the horizontal surface at 20 m/s.it then rolls up the incline of 30.If friction losses are negligible what will be the value of h where sphere stops on the incline
a. 28.6 m
b 30 m
c. 28 m
d. none of these




5 A cylinder of Mass M and radius R rolls down a incline plane of inclination θ.Find the linear accleration of the cylinder
a. (2/3)gsinθ
b.(2/3)gcosθ
c gsinθ
d none of these



6 An ice skater spins with arms outstretch at 1.9 rev/s.Her moment of inertia at this time is 1.33 kgm².She pulls her arms to increase her rate of spin.Her moment of inertia after she pulls her arm is .48kgm².What is her new rate of spinning
a. 5.26 rev/s
b. 5.2 rev/s
c 4.7 rev/s
d. none of thes



7. MOment of inertia of a uniform rod of lenght L and mass M about an axis passing through L/4 from one end and perpendicular to its lenght
a. 7ML²/36
b.7ML²/48
c. 11ML²/48
d.ML²/12



8. A wheel starts from rest and spins with a constant angular acceleration. As time goes on the
acceleration vector for a point on the rim:

a. increases in magnitude but retains the same angle with the tangent to the rim

b.increases in magnitude and becomes more nearly radial

c. increases in magnitude and becomes more nearly tangent to the rim

d. decreases in magnitude and becomes more nearly radial


9.
Two wheels are identical but wheel B is spinning with twice the angular speed of wheel A. The ratio of the
magnitude of the radical acceleration of a point on the rim of B to the magnitude of the radial acceleration of
a point on the rim of A is:

a. 4
b . 2
c 1/2
d 1/4



10. For a wheel spinning with constant angular acceleration on an axis through its center, the ratio of
the speed of a point on the rim to the speed of a point halfway between the center and the rim is:

a 2
b 1/2
c 4
d 1/4




11. A wheel initially has an angular velocity of 18 rad/s. It has a constant angular acceleration of 2.0 rad/s2 and is
slowing at first. What time elapses before its angular velocity is18 rad/s in the direction opposite to its initial
angular velocity?

a 3 sec
b 6 sec
c 18 sec
d none of these


12. One solid sphere X and another hollow sphere Y are of same mass and same outer radii. Their moment of inertia about their diameters are respectively I_x and I_y such that
(A) I_x= I_y
(B) I_x > I_y
(C) I_x < I_y
(D) I_x/I_y=D_x/D_y
Where D_x and D_y are their densities.





Solutions

1.b
2.b
3.b
4.a
5. a
6 a
7 b
8 b
9 a
10 a
11. c
12 c

Rotational Short Notes -II

Torque

τ=rXF

Also τ=Iα

Kinetic Energy is pure Rotating body
KE=(1/2)Iω²

Rotational Work Done

-If a force is acting on a rotating object for a tangential displacement of s = rθ (with θ being the angular displacement and r being the radius) and during which the force keeps a tangential direction and a constant magnitude of F, and with aconstant perpendicular distance r (the lever arm) to the axis of rotation, then the work done by the force is:

W=τθ

• W is positive if the torque τ and θ are of the same direction,
otherwise, it can be negative.

Power
P =dW/dt=τω


Angular Momentum

L=rXp
=rX(mv)
=m(rXv)

For a rigid body rotating about a fixed axis
L=Iω

dL/dt=τ

if τ=0 and L is constant

For rigid body having both translation motion and rotational motion

L=L₁+L₂

L₁ is the angular momnetum of Center mass about an stationary axis
L₂ is the angular momentum of the rigid body about Center of mass

Law of Conservation On Angular Momentum

If the external torque is zero on the system then Angular momentum remains contants

dL/dt=τ_ext

if τ_ext=0
then dL/dt=0


Equilibrium of a rigid body

F_net=0 and τ_ext=0



Angular Impulse:

∫τdt term is called angular impluse..It is basically the change in angular momentum


Pure rolling motion of sphere/cylinder/disc

-Relative velocity of the point of contact between the body and platform is zero
-Friction is responsible for pure rolling motion
-Friction is non disipative in nature

E = (1/2)mv_cm²+(1/2)Iω²+mgh

Rotational Short Notes -II

Torque

τ=rXF

Also τ=Iα

Kinetic Energy is pure Rotating body
KE=(1/2)Iω²

Rotational Work Done

-If a force is acting on a rotating object for a tangential displacement of s = rθ (with θ being the angular displacement and r being the radius) and during which the force keeps a tangential direction and a constant magnitude of F, and with aconstant perpendicular distance r (the lever arm) to the axis of rotation, then the work done by the force is:

W=τθ

• W is positive if the torque τ and θ are of the same direction,
otherwise, it can be negative.

Power

P =dW/dt=τω


Angular Momentum

L=rXp
=rX(mv)
=m(rXv)

For a rigid body rotating about a fixed axis
L=Iω

dL/dt=τ

if τ=0 and L is constant

For rigid body having both translation motion and rotational motion

L=L₁+L₂

L₁ is the angular momnetum of Center mass about an stationary axis
L₂ is the angular momentum of the rigid body about Center of mass

Law of Conservation On Angular Momentum

If the external torque is zero on the system then Angular momentum remains contants

dL/dt=τ_ext

if τ_ext=0
then dL/dt=0


Equilibrium of a rigid body

F_net=0 and τ_ext=0



Angular Impulse:

∫τdt term is called angular impluse..It is basically the change in angular momentum


Pure rolling motion of sphere/cylinder/disc

-Relative velocity of the point of contact between the body and platform is zero
-Friction is responsible for pure rolling motion
-Friction is non disipative in nature

E = (1/2)mv_cm²+(1/2)Iω²+mgh

IIT PHYSICS:Rotational Short notes -I

Angular Displacement

-When a rigid body rotates about a fixed axis, the angular displacement is the angle Δθ swept out by a line passing through any point on the body and intersecting the axis of rotation perpendicularly
-Can be positive (counterclockwise) or negative (clockwise).
-Analogous to a component of the displacement vector.
-SI unit: radian (rad). Other
units: degree, revolution.

Angular Velocity

Average angular velocity, is defined by
$ = (angular displacement)/(elapsed time) = Δθ/Δt .

Instantanous Angular Velocity ω=dθ/dt


Some points
-Angular velocity can be positive or negative.
-It is a vector quantity and direction is perpendicular to the plane of rotation
-Angular velcity of a particle is diffrent about diffrent points
-Angular velocity of all the particles of a rigid body is same about a point

Angular Acceleration:

Average angular acceleration, is defined by
= (change in angular velocity)/(elapsed time) = Δω/Δt

Instantanous Angular Acceleration
α=dω/dt

Kinematics of rotational Motion
ω=ω₀ + αt
θ=ω₀t+1/2αt²
ω.ω=ω₀.ω₀ + 2 α.θ;

Also
α=dω/dt=ωdω/dθ


Vector Nature of Angular Variables
-The direction of an angular variable vector is along the axis.
- positive direction defined by the right hand rule.
- Usually we will stay with a fixed axis and thus can work in the scalar form.
-angular displacement cannot be added like vectors
-angular velocity and acceleration are vectors


Relation Between Linear and angular variables

v=ωXr

Where r is vector joining the location of the particle and point about which angular velocity is being computed

a=αXr


Moment of Inertia

Rotational Inertia (Moment of Inertia) about a Fixed Axis

For a group of particles,

I = ∑mr²

For a continuous body,

I = ∫r²dm

For a body of uniform density

I = ρ∫r²dV

Parallel Axis Therom
I_xx=I_cc+ Md²

Where I_cc is the moment of inertia about the center of mass

Perpendicular Axis Therom
I_xx+I_yy=I_zz

It is valid for plane laminas only