kinematics formulas...

Blazing goIITian

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9 Jun 2012 16:32:00 IST
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9 Jun 2012 16:32:00 IST
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kinematics formulas...
Engineering Entrance , JEE Main , JEE Advanced , Physics , Mechanics , academic , Kinematics

Kinematics

Note: -

 

Bold letter are used to denote vector quantity

i

,j,z are the unit vector along x,y and z axis

Quick review of Kinematics formulas

S.No. Type of

Motion

Formula

1 Motion in one

dimension

r

v

a

v

=xi=(dx/dt)i=dv/dt=(d2x/dt2)i and a=vdv/dr=u+at

s

=ut+1/2at2

v

2=u2+2as

In integral form

r=∫vdt

v

=∫adt

2

dimension

r

v

a

Motion in two=xi +yj=dr/dt=(dx/dt)i + (dy/dt)j=dv/dt=(d2x/dt2)i+(d2x/dt2)j and a=vdv/dr

Constant accelerated equation same as above

3

dimension

r

v

a

a

Motion in three=xi +yj+zk=dr/dt=(dx/dt)i + (dy/dt)j+(dz/dt)k=dv/dt=(d2x/dt2)i+(d2x/dt2)j+(d2x/dt2)k=vdv/dr

Constant accelerated equation same as above

4

y=(v

v

velocity makes with the positive x axis.

Projectile Motion x=(v0cosθ0)t0sinθ0)t-gt2/2x= v0cosθ0 and vy= v0sinθ0t-gt , where θ0 is the angle initial

5

motion

Uniform circular

a=v

of is always along radius of the circle towards the centre and

a=4π

of time period T

2/R , where a is centripetal acceleration whose direction2R/T2 acceleration in uniform circular motion in terms

Concept of relative velocity

For two objects A and B moving with the uniform velocities

VA and VB.

Relative velocity is defined as

V

BA=VB-VA

where

Similarly relative velocity of A relative to B

VBA is relative velocity of B relative to A

V

AB=VA-VB

Special cases: -

S.No. Case Description

1 For straight line

motion

If the objects are moving in the same direction, relative velocity can be

get by subtracting other. If they are moving in opposite

direction ,relative velocity will be get by adding the velocities example

like train problems

2 For two dimensions

motion

if

v

v

a=vxai + vyajb=vxbi + vybj

Relative velocity of B relative to A

=v

=

xbi + vybj -(vxai + vyaj)i(vxb-vxa) + j(vyb-vya)

3 For three

dimensions motion

v

v

a=vxai + vyaj +vzazb=vxbi + vybj + vzbz

Relative velocity of B relative to A

=v

=

xbi + vybj + vzbz -(vxai + vyaj +vzaz)i(vxb-vxa) + j(vyb-vya)+z(vyb-vya)

Free fall acceleration

S.No. Point

1

accelerated motion.

Freely falling motion of any body under the effect of gravity is an example of uniformly

2

equations of motion by acceleration due to gravity

Kinematics equation of motion under gravity can be obtained by replacing acceleration 'a' in'g'.

3

v = v

Thus kinematics equations of motion under gravity are0 + gt , x = v0t + . ( gt2 ) and v2 = (v0)2 + 2gx

4

downwards and negative when the body is projected up against gravity.

Value of g is equal to 9.8 m.s-2.The value of g is taken positive when the body falls vertically

Laws of motion

S.No. Term Description

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,

p=mv , momentum of the body

a

=acceleration

3 Impulse

change in momentum

Impulse =

4 Newton's third law

of motion

Impulse is the product of force and time which is equal to theF:t =:p

'To every action there is always an equal and opposite reaction'

F

AB=-FBA

5 Law of conservation

of linear momentum

Initial momentum = final momentum

m

1v1+m2v2=m1v1'+m2v2'

For equilibrium of a body

F

1+F2+F3=0

Some points to note

S.No. Point

1

inertial frame

An accelerated frame is called non inertial frame while an non accelerated frame is called

2

Newton first law are valid in inertial frame only

3

Going Upward with acceleration a

W=m(g + a)

Going Down with acceleration a

W=m(g-a)

Apparent weight of a body in the lift

4

Always draw free body diagram to solve the force related problems

Friction and Frame of reference

S.No. Term Description

1 Friction

motion between them. When bodies slip, frictional force is called static

frictional force and when the bodies do not slip, it is called kinetic

frictional force.

Frictional force acts between the bodies whenever there is a relative

2 Kinetic

Frictional force

When bodies slip over each other

f=μ

Where N is the normal contact force between the surface and μ

coefficient of kinetic Friction. Direction of frictional force is such that

relative slipping is opposed by the friction

KNk is the

3 Static Frictional

force

Frictional force can also act even if there is no relative motion. Such

force is called static Frictional force. Maximum Static friction that a body

can exert on other body in contact with it is called limiting Friction.

f

N is the normal contact force between the surface

And μ

f

maxsN Wheres is the coefficient of static Frictionmax is the maximum possible force of static Friction. Note that μs > μk

and Angle of friction tanλ=μ

s

4 Inertial Frame

Of reference

Inertial frame of references is those attached to objects which are at rest

or moving at constant Velocity. Newton’s law are valid in inertial frame of

reference. Example person standing in a train moving at constant

velocity.

5 Non Inertial

Frame Of

reference

Inertial frame of references is attached to accelerated objects for

example: A person standing in a train moving with increasing speed.

Newton’s law are not valid.

To apply Newton’s law ,pseudo force has to be introduced in the equation

whose value will be F=-ma

Work, Energy and Power

S.No. Term Description

1

displacement vector. For constant Force

W=

Work 1. Work done by the force is defined as dot product of force andF.s

where F is the force vector and s is displacement Vector

2.

dW=

For variable ForceF.ds or W=∫F.ds

It is a scalar quantity

2

Conservative Forces

1.

is called

2.

then it is called

3.

Non Conservative Forces are frictional forces

Conservative And NonIf the work done by the force in a closed path is zero, then itconservative ForceIf the work done by the force in a closed path is not zero,non conservative ForceGravitational ,electrical force are Conservative Forces and

3

as

K.E=(1/2)mv

Kinetic Energy 1. It is the energy possessed by the body in motion. It is defined2

2.

the kinetic energy of the system W=K

Net work done by the external force is equal to the change inf-Ki

4

system. It is due to conservative force. It is defined as

dU=-

Potential Energy 1. It is the kind of energy possessed due to configuration of theF.dr

U

f-Ui=-∫F.dr

Where F is the conservative force

F

=-(∂U/∂x)i-(∂U/∂y)j-(∂U/∂z)k

For gravtitional Force

2.

where h is the height between the two points

Change in Potential Energy =mgh

3.

E=K.E+P.E

Mechanical Energy is defined as

5

Energy

Law Of conservation of

In absence of external forces, internal forces being conservative,

total energy of the system remains constant.

K.E

1+P.E1=K.E2+P.E2

6

Watt. 1W=1Js

quantity.

Power Power is rate of doing work i.e., P=work/Time. Unit of power is-1. In terms of force P= F.v and it is a scalar

Momentum and Collision

S.No. Term Description

1 Linear Momentum

velocity

The linear momentum p of an object of mass m moving withv is defined as p=mv

Impulse of a constant force delivered to an object is equal to the

change in momentum

of the object

F:t = :p = mv

f - mvi

Momentum of system of particles is the vector sum of individual

momentum of the particle

p

totalviMi

2 Conservation of

momentum

When no net external force acts on an isolated system, the total

momentum of the system is constant. This principle is called

conservation of momentum.

if Σ

Fext=0 then ΣviMi=constant

3 Collision

but kinetic energy is not.

Inelastic collision - the momentum of the system is conserved,

Perfectly inelastic collision

- the colliding objects stick together.

Elastic collision

the system are conserved.

- both the momentum and the kinetic energy of

4 Inelastic collision

While colliding if two bodies stick together then speed of the

composite body is

1 2

1 1 2 2

m m

m u m u

v

+

+

=

Kinetic energy of the system after collision is less then that before

collison

5 Elastic collision in

one dimension

Final velocities of bodies after collision are

2

1 2

2

1

1 2

1 2

1

m m

2m

u

m m

m m

v

?

?

? ??

?

+

+ ? ?

?

?

? ??

?

+

=

u? ?

2

1 2

2 1

1

1 2

1

2

m m

m m

u

m m

2m

v

?

?

? ??

?

+

+ ? ?

?

?

? ??

?

+

=

u? ?

also

1 2 2 1

u u = v v

Special cases of Elastic Collision

S.No. Case Description

1

m1=m2 v1=u2 and v2=u1

2

When one of the bodies is at rest say u2=0

1

1 2

1 2

1

m m

m m

v

?

?

? ??

?

+

=

u? ?and

1

1 2

1

2

m m

2m

v

?

?

? ??

?

+

=

u? ?

3

When m1=m2 and u2=0 i.e., m2 is at rest v 0 1 = and

2 1

v = u

4

m

When body in motion has negligible mass i.e.1<<m2 1 1 v = −u and v 0 2 =

5

m

When body at rest has negligible mass i.e.1>>m2 1 1 v = u and

2 2

v =2u

Mechanics of system of particles

S.No. Term Description

1 Centre of mass

concentrated, for consideration of its translational motion.

It is that point where entire mass of the system is imagined to be

2

centre of mass

position vector of

R

element i

cmriMi/ΣMi where ri is the coordinate of element i and Mi is mass of

3 In coordinate

system

x

cm=ΣxiMi/ΣMi

y

cm=ΣyiMi/ΣMi

z

cm=ΣziMi/ΣMi

4

Velocity of CM vCMviMi/ΣMi

The total momentum of a system of particles is equal the total mass

times the velocity of the centre of mass

5

particles we find F

Force When Newton’s second law of motion is applied to the system oftot=MaCM with aCM=d2RCM/dt2

Thus centre of mass of the system moves as if all the mass of the

system were concentrated at the centre of mass and external force

were applied to that point.

6

conservation in

COM motion

P=

particles is equal to the product of the total mass of the system and

the velocity of its centre of mass.

MomentumMvCM which means that total linear momentum of system of

Rigid body dynamics

S.No. Term Description

1 Angular Displacement

displacement is the angle :θ swept out by a line passing

through any point on the body and intersecting the axis of

rotation perpendicularly

-It can be positive (counter clockwise) or negative (clockwise).

-Analogous to a component of the displacement vector.

-SI unit: radian (rad). Other

units: degree, revolution.

-When a rigid body rotates about a fixed axis, the angular

2 Angular velocity

Instantaneous Angular Velocity

-Angular velocity can be positive or negative.

-Average angular velocity, is equal to :θ/:t .ω=dθ/dt

 

plane of rotation

-Angular velocity of a particle is different about different points

-Angular velocity of all the particles of a rigid body is same

about a point.

-It is a vector quantity and direction is perpendicular to the

3 Angular Acceleration

Instantaneous Angular Acceleration

α=d

Average angular acceleration= :ω/:tω/dt

4 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

 

5 Kinematics of rotational

Motion

ω

=ω0 + αt

θ

=ω0t+1/2αt2

ω

Also

α=dω/dt=ω(dω/dθ)

.ω=ω0.ω0 + 2 α.θ;

6 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

7 Moment of Inertia

For a group of particles,

I =

=αXrRotational Inertia (Moment of Inertia) about a Fixed AxisΣmr2

For a continuous body,

I =

For a body of uniform density

I = ρ∫r

r2dm2dV

8 Parallel Axis Therom

of mass

Ixx=Icc+ Md2 Where Icc is the moment of inertia about the centre

9 Perpendicular Axis

Therom

 

I

xx+Iyy=Izz It is valid for plane laminas only.

10 Torque

τ

=rXF also τ=Iα where α is angular acceleration of the body.

11 Rotational Kinetic

Energy

 

KE=(1/2)Iω

2 where ω is angular acceleration of the body

12 Rotational Work Done

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 a

constant 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.

-If a force is acting on a rotating object for a tangential

13 Power

 

P =dW/dt=τω

 

14 Angular Momentum L

=rXp

=

=m(

For a rigid body rotating about a fixed axis

L=Iω and d

rX(mv)rXv)L/dt=τ

if τ=0 and L is constant

For rigid body having both translational motion and rotational

motion

 

L

=L1+L2

L

axis

1 is the angular momentum of Centre mass about an stationary

L

mass.

2 is the angular momentum of the rigid body about Centre of

15 Law of Conservation On

Angular Momentum

 

If the external torque is zero on the system then Angular

momentum remains contants

d

L/dt=τext

if

then d

τext=0L/dt=0

 

body

F

16 Equilibrium of a rigidnet=0 and τext=0

17 Angular Impulse

angular momentum

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

18 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

-If friction is non dissipative in nature

E = (1/2)mv

2

cm+(1/2)Iω2+mgh

Gravitation

 

S.No. Term Description

 

1 Newton’s Law of

gravitation

1 2

2

r

Gm m

F

=

where G is the universal gravitational constant

G=6.67

×10-11Nm2Kg-2

2 Acceleration due

to gravity

 

g=GM/R

earth

2 where M is the mass of the earth and R is the radius of the

3 Gravitational

potential energy

 

PE of mass m at point h above surface of earth is

 

(

R h)

GmM

PE

 

+

= −

 

4 Gravitational

potential

(R h)

GM

V

 

+

= −

 

Law of

orbits

 

Each planet revolves round the sun in an elliptical orbit

with sun at one of the foci of elliptical orbit.

 

Law of

areas

 

The straight line joining the sun and the planet sweeps

equal area in equal interval of time.

 

5 Kepler’s Law of

planetary motion

Law of

periods

 

The squares of the periods of the planet are proportional

to the cubes of their mean distance from sun i.e.,

T

2VR3

6 Escape velocity

projected in order that it may escape earth’s gravitational pull. Its

magnitude is v

Escape velocity is the minimum velocity with which a body must bee=√(2MG/R) and in terms of g ve=√(2gR)

Orbital

Velocity

 

The velocity which is imparted to an artificial satellite few

hundred Km above the earth’s surface so that it may

start orbiting the earth v

0=√(gR)

7 Satellites

Periodic

Time

 

T=2π√[(R+h)

3/gR2]

With

altitude

?

?

??

= ? −

?

R

h

g g

h

2

1

 

With

depth

?

?

??

= ? −

?

R

d

g g

d 1

8 Variation of g

With

latitude

 

φ

φ

2

g = g 0.037cos

Elasticity

 

S.No. Term Description

 

1 Elasticity

force is withdrawn

The ability of a body to regain its original shape and size when deforming

2 Stress

Stress=F/A where F is applied force and A is area over which it acts.

3 Strain

Longitudinal strain = :l/l volume strain = :V/V and

change in shape of the body.

It is the ratio of the change in size or shape to the original size or shape.shear strain is due to

4 Hook’s Law

deformations stress is proportional to strain".

Hook's law is the fundamental law of elasticity and is stated as “for small

 

or, stress/strain = constant

This constant is known as modulus of elasticity of a given material

Thus, stress proportional to strain

Young's Modulus of Elasticity

Y=Fl/A:l

Bulk Modulus of Elasticity

K=-V:P/:V

5 Elastic

Modulus

 

Modulus of Rigidity

η=F/Aθ

6

Ratio

Poisson's

The ratio of lateral strain to the longitudinal strain is called Poisson’s ratio

which is constant for material of that body. σ=l:D/D:l

 

7

energy

Strain

Energy stored per unit volume in a strained wire is E=.(stress)x(strain)

 

Hydrostatics

 

S.No. Term Description

 

1 Fluid pressure

unit is Pascal 1Pa=1Nm

It is force exerted normally on a unit area of surface of fluid P=F/A. Its-2.

2 Pascal’s Law

Pressure in a fluid in equilibrium is same everywhere.

3 Density

Density of a substance is defined as the mass per unit volume.

4 Atmospheric

pressure

 

Weight of all the air above the earth causes atmospheric pressure which

exerts pressure on the surface of earth. Atmospheric pressure at sea

level is P

0=1.01x105Pa

5 Hydrostatic

pressure

 

At depth h below the surface of the fluid is P=ρgh where ρ is the density

of the fluid and g is acceleration due to gravity.

 

6 Gauge pressure

pressure and pressure due to all the fluid above that point.

P=P0+ ρgh , pressure at any point in fluid is sum of atmospheric

7 Archimedes

principle

 

When a solid body is fully or partly immersed in a fluid it experience a

buoyant force equal to the weight of fluid displaced by it.

 

8 Upthrust

It is the weight of the displaced liquid.

9 Boyle’s law

PV=constant

10 Charle’s law

V/T=constant

Hydrodynamics

 

S.No. Term Description

 

1 Streamline

flow

 

In such a flow of liquid in a tube each particle follows the path of its

preceding particle.

 

2 Turbulent

flow

 

It is irregular flow which does not obey above condition.

 

3 Bernoulli’s

principle

p u gh cons tan t

2

1

+ ρ 2 + ρ =

4 Continuity of

flow

variable cross section and v

crossing these areas.

1 1 2 2 A v = A v where A1 and A2 are the area of cross section of tube of1 and v2 are the velocity of flow of liquids

5 Viscosity

dv/dx is

Viscous force between two layers of fluid of area A and velocity gradient

dx

dv

F

= −ηA where η is the coefficient of viscosity.

6 Stokes’ law

viscosity η is

Viscous force on a spherical body of radius r falling through a liquid ofF = 6πηrv where v is the velocity of the sphere.

7 Poiseuilli’s

equation

 

Volume of a liquid flowing per second through a capillary tube of radius r

when its end are maintained at a pressure difference P is given by

 

l

Q

 

η

π

 

8

Pr

4

=

where l is the length of the tube.

Simple Harmonic Motion

 

S.No. Term Description

 

1 SHM

the mean position and opposes its increase. Restoring force is

In SHM the restoring force is proportional to the displacement fromF=-Kx

Where K=Force constant , x=displacement of the system from its

mean or equilibrium position

 

 


Comments (2)


Blazing goIITian

Joined: 8 May 2012 12:29:36 IST
Posts: 466
9 Jun 2012 16:35:45 IST
0 people liked this

from www.loook into.co.cc. some of the above are not clear........sorry.........

Scorching goIITian

Joined: 7 Apr 2012 14:16:14 IST
Posts: 233
13 Jun 2012 20:12:22 IST
0 people liked this

Whyyyy



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