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

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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 :
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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

Torque

• Torque of a force about any point is equal to the product of the force and perpandicular distance of the line of action of force from the point.
  τ= Force x perpandicular distance from line of action from point = Frsinθ =(Fsinθ)r
  or,
  τ= (component of force perpandicular to position vector) x (position vector)
  
• Unite of torque (a) MKS system --- N-m (b) CGS system --- dyne-cm.
  
• Dimension of torque is ML²T^-2
  
• In vector form τ=F x r
  
• Torque is a vector quantity having direction perpandicular to the plane of force and position vector and its direction is given by right hand rule.
  
• If the torque acting on a body tends to rotate in anticlockwise direction then torque is positive and if it tends to rotate the body in clockwise direction then the is negative.
  
• If the body s acted upon by more then one forces then total torque on the body is the vector sum of each torque.
  
• τ= dJ/dt , where J is the angular momentum of the body.
  
• The more is the value of r , more will be the torque and it would be easier to rotate the object.
  
• Work done by the torque is ∫>τdθ = Torque x angular displacement. here limits of integration goes from θ₁ to θ₂
  

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

Problem Solving tips for IITJEE Mechanics

Vector Mathematics

You will come across vectors in physics problem very frequently.So it is must to know to solve the vector mathematics in short time.And Making sure you have done in correctly.You will find vectors in every module be it mechanics ,electrostatics,magentics

Addition or Substraction
When you need to add or substract two or more vectors, you can use following procedure

1. Select a coordinate system that is convenient. (Try to reduce the number of
components you need to find by choosing axes that line up with as many
vectors as possible.)

2. Draw a labeled sketch of the vectors described in the problem.

3. Find the x and y components of all vectors

4. Compute the resultants in x and y direction

5. If necessary, use the Pythagorean theorem to find the magnitude of the resultant vector and select a suitable trigonometric function to find the angle
that the resultant vector makes with the x axis.

Products
When u need to perform scalar or vector product.It is nice to following procedure

1. Select a coordinate system(x,y,z) that is convenient. (Try to reduce the number of
components you need to find by choosing axes that line up with as many
vectors as possible.).Use i,j,k as the unit vector across x,y,z axis.

2. Draw a labeled sketch of the vectors described in the problem.

3. Express all the vector in form in i,j,k forms

4. Perform the multipication




Motion in a Two dimensional Plane

- Select a coordinate system and resolve the initial velocity vector into x and y
components.

-Find out acceleration in each direction and solve in each direction according to one rectilinear motion equation.

-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.

- 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

- Select a coordinate system and resolve the initial velocity vector into x , y and z components

-Find out acceleration in each direction and solve in each direction according to one rectilinear motion equation.



Uniform Circular Motion

-Draw a simple, neat diagram of the system.

-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

-Find out the force acting on the body.The resultant force should provide the required centrepatal required for Circular motion

-Centrepatal force=mv²/R will give the velotiy accordingly.


Newtons Law Problem
-Draw a neat diagram of the system.

-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

-Isolate the body whose motion is to be analyzed. Draw a free-body diagram for this body. For systems containing more than one objects, draw separate free-body diagrams for each objects This way there will not any confusion about the force acting on each object.Newtons Third law will help in obtaining the action reaction pair. Do not include in the free-body diagram forces exerted by the object on its surroundings. Establish convenient coordinate axes for each objects and find the components of the forces along these axes.

- Use pseudo force if viewing from the non intertial frame of refrence

-Apply Newton's second law, F= ma, in component form.

-Solve the component equations for the unknowns. Remember that you must have as many independent equations as you have unknowns to obtain acomplete solution.

-Make sure your results are consistent with the free-body diagram.

- If the object are attached to each others by the strings,then make use of constraints theory to find out the acceleration equation between the objects


Work And Energy

- Choose your frame of refrence.KE will differ in each refrence frame while PE will remains constant

-Define your system, which may include two or more interacting particles, as well as springs or other systems in which elastic potential energy can bestored. Choose the initial and final points.

-Identify zero points for potential energy (both gravitational and spring). If there is more than one conservative force, write an expression for the potential energy associated with each force.

-Determine whether any nonconservative forces are present. Remember that if friction or air resistance is present, mechanical energy is not conserved.

-Determine whether any external forces are present. Remember that external forces are present, mechanical energy is not conserved.

-If mechanical energy is conserved, you can write the total initial energy at some point E_i=PE_i+KE_i. Then, write an expression for the total final energy at the final point that is of interest E_f=PE_f+KE_f. Because mechanical
energy is conserved, you can equate the two total energies and solve for the quantity that is unknown.

-If frictional forces are present (and thus mechanical energy is not conserved), first write expressions for the total initial and total final energies. In this case, the difference between the total final mechanical energy and the total initial mechanical energy equals the change in mechanical energy in the system
due to friction.

-If external forces are present (and thus mechanical energy is not conserved), first write expressions for the total initial and total final energies. In this case, the difference between the total final mechanical energy and the total initial mechanical energy equals the change in mechanical energy in the system
due to external force.

-If both external forces and frictional forces are present (and thus mechanical energy is not conserved), first write expressions for the total initial and total final energies. In this case, the difference between the total final mechanical energy and the total initial mechanical energy equals the change in mechanical energy in the system due to external force and friction


Momentum and Collisions

- Choose your frame of refrence

-Set up a coordinate system and define your velocities with respect to that system.It is usually convenient to have the x axis coincide with one of the initial velocities.

-In your sketch of the coordinate system, draw and label all velocity vectors and include all the given information.

-Write expressions for the x and y components of the momentum of each object before and after the collision. Remember to include the appropriate signs for the components of the velocity vectors.

-Write expressions for the total momentum in the x direction before and after
the collision and equate the two. Repeat this procedure for the total momentum
in the y direction. These steps follow from the fact that, because the momentum of the system is conserved in any collision(law on conservation of linear momentum), the total momentum along any direction must also be constant. Remember, it is the momentum of the system that is constant, not the momentan of the individual objects.

-If the collision is inelastic, kinetic energy is not conserved, and additional information is probably required. If the collision is perfectly inelastic, the final velocities of the two objects are equal. Solve the momentum equations for the unknown quantities.

-If the collision is elastic, kinetic energy is conserved, and you can equate the
total kinetic energy before the collision to the total kinetic energy after the
collision to get an additional relationship between the velocities.And you can solve the energy and momentum equation to find out the find velocities.

- Centre of mass can be useful feature in solving the momentum problems


Body in Static equilibrium

-Draw a neat diagram of the system.

-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


-Isolate the object being analyzed. Draw a free-body diagram and then show
and label all external forces acting on the object, indicating where those forces are applied.This way there will not any confusion about the force acting on each object.Newtons Third law will help in obtaining the action reaction pair. Do not include forces exerted by the object on its surroundings.(For systems that contain more than one object, draw a separate free-body diagram for each one.) Try to guess the correct direction for each force. If the direction you select leads to a negative force, do not be alarmed; this merely means that the direction of the force is the opposite of what you guessed.

-Establish a convenient coordinate system for the object and find the components of the forces along the two axes. Then apply the first condition for equilibrium. Remember to keep track of the signs of all force components.

-Choose a convenient axis for calculating the net torque on the object. Remember that the choice of origin for the torque equation is arbitrary; therefore, choose an origin that simplifies your calculation as much as possible.
Note that a force that acts along a line passing through the point chosen as
the origin gives zero contribution to the torque and thus can be ignored.

Newtons law of Motion

Basic concepts of laws of motion

(1) Newton's first law of motion:-
'A body continues to be in state of rest or uniform motion unless it is acted upon by some external force to act otherwise'

(2) Newton's second law of motion:-
'Rate of change of momentum of a body is proportional to the applied force and takes place in the direction of action of force applied'
Mathematically,
F=dp/dt
=ma
where, F=external force applied
p=mv,momentum of the body
a=acceleration
SI unit of force is Newton (N).

(3) Impulse is the product of force and time which is equal to the change in momentum
Impulse=FΔt
=Δp

(4) Newton's third law of motion:-
'To every action there is always an equal and opposite reaction'
F_AB=-F_BA

(5) According to the law of conservation of linear momentum
Initial momentum = final momentum
m₁v1+m₂v2=m₁v1^'+m₂v2^'


For equilibrium of a body
F₁+F₂+F₃=0

Some points to note
1. An accelerated frame is called non interial frame while am non accelerated frame is called interial frame
2. Newton Ist law are valid in intertial frame only
3 Apparent weight of a body in the lift
Going Upward with acceleration a
W=m(g+a)
Going Down with acceleration a
W=m(g-a)
4.Always draw free body diagram to solve the force related problems

Relative velocity

-For two objects A and B moving with the uniform velocities V_A and V_B.
Relative velocity is defined as
V_BA=V_B-V_A
where V_BA is relative velocity of B relative to A

Similiary relative velocity of A relative to B
V_AB=V_A-V_B
Special cases.
For straight line motion,we can just add or subtract to get the relative velocity.
If the objects are moving in the same direction,relative velocity can be get by substracting other.If they are moving in opposite direction ,relative velocity will be get by adding the velocities example like train problems

For two dimensions motion,we will need to add or subtracting components along x & y direction to get the relative velocity
Suppose
v_A=v_xai + v_yaj
v_B=v_xbi + v_ybj

Relative velocity of B relative to A
=v_xbi + v_ybj -(v_xai + v_yaj)
=i(v_xb-v_xa) + j(v_yb-v_ya)

For three dimensions motion
v_A=v_xai + v_yaj +v_zaz
v_B=v_xbi + v_ybj + v_zbz

Relative velocity of B relative to A
=v_xbi + v_ybj + v_zbz -(v_xai + v_yaj +v_zaz)
=i(v_xb-v_xa) + j(v_yb-v_ya)+z(v_yb-v_ya)

Short notes in Motion in a plane

Short notes for motion in one dimension

i,j,z are the unit vector along x,y and z axis
Notes: Bold letter are used to denote vector quantity

r=xi
v=(dx/dt)i
a=dv/dt=(d²x/dt²)i
a=vdv/dr
v=u+at
s=ut+1/2at²
v²=u²+2as


r=∫vdt
v=∫adt

Short notes for motion in two dimension

r=xi +yj
v=dr/dt=(dx/dt)i + (dy/dt)j
a=dv/dt=(d²x/dt²)i+(d²x/dt²)j
a=vdv/dr
v=u+at
s=ut+1/2at²
v²=u²+2as

r=∫vdt
v=∫adt

Short notes for motion in three dimension
r=xi +yj+zk
v=dr/dt=(dx/dt)i + (dy/dt)j+(dz/dt)k
a=dv/dt=(d²x/dt²)i+(d²x/dt²)j+(d²x/dt²)k
a=vdv/dr
v=u+at
s=ut+1/2at²
v²=u²+2as


r=∫vdt
v=∫adt

Gravitation

Introduction :
Kepler's law
Acceleration due to grawing of earth
Acceleration due to grawing below and above earth surface.

(1) Introduction :

- In our dairy life we have noticed things falling freely downwords towards earth when thrown upwards or dispped from some hight.

- Fact that all bodies irrespective of their masses are accelerated towards the earth mith a constant aceeleation was first recognized by galive (1564-1642)

- The motion of celesh'al bodies such as moon. earth plametes etc. and attrachieve of moon towards earth and arth towards sun is an interasting subject of study since long time.

- Now the question's what is the force that produces such acceleration is which earth attract all bodies towards the centre and what is the law governing this force.

- Is this law is same for both earthly and weshal bodies.

- Answer to this question was given by Newton as he declared that "laws of nature are same for earthly and weshal Bodies".

- The force between any object falling freely towards earth and that between earth and moon are gowerned by the same laws.

- Johnaase kepler (1571-1631) Studied the planetary motion in detail and formulated his three laws of planetary motion, which were available Universal law of grawitation.

(2) Kepler's Law :-

Kepler's law of planetary motion are :-

(i) Law of orbits :-

Each planet revolues around the sun in an elliptical orbit with sun at one of the foci of the ellipse .

Fig (a) An ellipse traced by planet sevolary round the sun.




AO = a - Sewi major axis

BO = b - Sewi minor axis

P - hearest point between planet and sun k/as perihetion

A - farthest point between planet and sun apheiton.

(ii) Law of areas :-

"The line joining planet and the sun sweeps

equal area in equal intervals of time"

This law follows from th e observation that when planet is nearer to the sun
its velocity increases and It appears to be slower when it is farther from the sun.

(iii) Law of periods :-

"The squre of time period of any planet about the sun is propotional to the cube of the semi-major axis."

- If T is the time period of semi major axis a of elliptical orbit then.

T² x a³ (1)

- If T₁ and T₂ are time periods of any two planets and a₁ and a₂ being their semi major axis resp. then

T₁² =ka₁³

T₂² =ka₂³ (2)

This question (2) can be used to find the time period of a planet, when the time period of the other planet and the semi-major axis of orbits of two planets.

(3) Universal law of gravition :-

- Everybody in the universe attracts every other body with a force which is directly proportional to the product of their masses and invessly proportional to the square of distance between them. - Mathematically Newton's gravitation law is if F is the force acting between two bodies of masses M₁ and M₂ and the distance between them is R then majnitade of force is given as

F =	Gm1m2
	r2

In vector notation

F =	Gm1m2(-r^)
	r2

F =	-Gm1m2r^
	r2

where G- universal gravitational constant

r^ - unit vector from m₁ to m₂ and r^ = r₁^ - r₂^

Gravitational force is attractive constant is

SI - G= 6.67 x 10^-11 nm² kg^-2

CGS - G= 6.67 x 10^-8 dyn cm² g^-2

Deminsional formula of C1 is [M^-1L³T^-2]

(4) The gravitational constant :- (Measurement)

(5) Accelesation due to gravity of earth :-

- Earth attracts every object lying an its surface towards its centre with a force known as gravitational towards its centre with a force known as gravitational pull or gravity.
- Whenever force acts on any body it produces acceleration and in case of gravitation this acceleration produced under effect of gravity is known as accelesation due to gravity (g)
- Value of accelesation due to gravity is independent of mass of the body and its value near surface of earth is 9.8 ms^-2
- Expression for acceleration due to gravity
Consider mass of earth to be as M_E and its redius be R_E Suppose a body of mass M (much smaller then that fo earth) is kept at the earth surface. Force eseerted by earth on the body of mass m is

F =	-GMME
	RE2

Let the above equation is (4)
The force for the body due to earth produces acceleration due to gravity

(g) in the motion of the body. From Newton's Second law of motion

f=mg ----(5)

from (4) and (5)

g =	-GME
	RE2

which is acceleration due to gravity at earth's surface.

(6) Acceleration due to gravity below and above the earth surface :-

(i) Above earth's surface

- An object of mass m is placed at hight h above the earth's surface

F =	-GMME
	(RE+h)2

from this it can be concluded that value of g decreases as distance above surface of earth measures now,

g =	-GME
	RE(1+h/RE)2

g =	go
	(1+h/RE)2

which is equation no (7)
Where

go =	GME
	Re2

* eqn (7) tells us that for small hight h above surface of earth. value of g decreases by factor (1-2h/R_E)

- for h is smaller compared to R

g=g_o(1+h/R_E)^-2

g=g_o(1-2h/R_E) Expanding by Binomial theriom



(ii) Below the earth's surface





- If one goes inside the earth surface the value of g again decreases

- P = density of material of earth them

m=4/3R_E³P

from this accelesation due to gravity at earth’s Surface is

g =	G4/33RE3P
	RE2

g=4/3GR_EP ----(8)

g- acceleration due to gravity at depth d below earth's surface

- Body at depth d will experience force only due to portion of reduce (R_E-d) of earth's

- outer spnerical shell of thickness d will not exeperience any force

- M is mas of the portion of earth with radius (R_E-d) then

g =	-GM
	RE2

M=4/3(R_E-d)³ P

g =	-G4/3(RE-d)3 P
	(RE2-d)2

g= 4/3G (R_E-d)P (9)

Dividing epn (9) by (8)

g/go= (1-d/R_E)

or = g=go (1-d/R_E) --- (10)

from eqn (10) it is clearer that acceeleration due to gravity also decreases with depth.