IIT JEE , AIEEE Forum - Physics , Mechanics - Important points and formulae..CONTRIBUTE

tarun_bits

Hi friends ..its` time for revesion..

so let us take the important topics of physics one by one and discuss about the important points to remember and the general formulae and mistakes we do..

please contribute and discuss the things which will help us do score high..

RATES TO EVERY CONTRIBUTION....c`mon people,,,,,

here comes first from me...

tarun_bits

WORK ENERGY AND POWER

1)In the below diagram...work done by the applied force is same wether the force is applied horizontally or it is applied along the plane if the friction is absent.. if friction has been there than the work would have been different.....
see it yourself in ques>>12 of hcv

2) Work Energy Theorem

The work done by ALL the forces is equal to the change in kinetic energy,,,,wether the forces are conservative, non-conservative or external..
W(c) + W(NC) + W(ext) = K(final) - K(initial)

3)In some questions like ques 17 of hcv ..if nothing is mentioned about friction..than also u hav to consider it and explore all the cases.

4)one important point I witnessed is that you should use g as 9.8 and not 10 when it is not given.It may cause some good variations otherwise especially in objective questions.

5)the work done is ?FRAME DEPENDENT??
For example>>>you cannot say that the work done by static frictional force is zero?.if you observe it from ground frame for a two block system..it actually does work?.its true that the point of contact does not move relative to any body.

6) Energy is a scalar quantity and so no direction is required ..see ques number 21 in hcv.

7)In the same ques 21 and many other ques u will witness that the mass of the particle is not given??u urself think now what does it signify??

8)In some questions like ques 31 of hcv?u should develop the required constraint relation and than analyse each body separately??

Than see all the forces acting and the work they do..and you can apply the Work ?Energy theorem to it?.

But you should be Very careful in applying Law of conservation of mechanical energy?

9)The Law of conservation of mechanical Energy..

The 2 conditions you sould strictly follow while applying the law of conservation of mechanical energy are as follows>>>

(A) The internal forces should be conservative..If the forces which are non-conservative like Friction are present than you can`t apply it?.
(B) The external forces should be absent or the work done by them sould be zero?
Do rate if you feel a single word useful//
@@@Have a chilling time@@@

tarun_bits

WORK= FORCE * DISPLACEMENT

1)

W= F.S= Fscos theta,,,,,

,,,,,,,,,theta is the angle between the line of action of force and the displacement..

2)

for a variable force

W=( integral of) F.ds

3)

W=(integral: of ) F.v dt

where F and v are all instantaneos

brainbank

hi tarun,

let me discuss the imp.formulas of circular motion:

=d

/dt

=d

/dt

v=r

centripetal force=mv²/r

tarun_bits

the centrifugal force is not a new force but a pseudo force

deedee

centripatl force is nt pesudo force

itz d force actin towrds d centre

magnitude=mass*radial acc

bt centrifugal force is a pseudo force

wich arise due 2 acc of d frame

itz nt real force

bt ahs magnitude saem as centripetal force

centrifugal force acts away frm d centre

hope it helps!!

tarun_bits

cheer up guys and gals...more..going good

brainbank

P.E.OF A SPRING IS 1/2KX²

biki

i wud like to discuss about frames...
i wrote it many days bfore.. but i think it to be very important..
plz wait ... i m searching

KEERU

the path of one projectile relative to another projectile in flight is a straight line

biki

lets discuss frames

once someone asked this ques. (fig.1)

a block of mass m is kept on the surface of a rough car moving with acceleration "a"
in +icap direction.
what all forces are acting on 'm' whan it is viewed in ground frame as well as car frame

in ground (inertial) frame....(no pseudo force)
the body is pushed back due to motion of car.
Forces are:-
1) mg ( -j^)
2)reaction(N) by floor (+j^)
and N = mg

3)frictional force(f) = umg(+i^)
So the frictional force is supposed to provide the necessary force ma(+i^) on the body which makes it move as seen from ground
i,e, f = ma

in car (non inertial frame)
the block is still pushed back and remain at rest for an observer in the car at the pushed position
1) mg ( -j^)
2)reaction(N) by floor (+j^)
and N = mg

3)frictional force(f) = umg(+i^)
4)pseudo force = ma(-i^)
Since the body is at rest as seen from the car,
so f = ma

In both cases , f = ma.
But from ground .... its a single force which actually is f but its magnitude is equal to ma
and from car.... they are two different forces in opposite direction making the body stay at rest.

_________________________________________________________________

same concept is applied for a ball hung from the ceiling of a car... (fig.2)

in ground (inertial) frame....(no pseudo force)
the ball is pushed back due to motion of car.
Forces are:-
1) mg ( -j^)
2)tension component Tsin

(+i^) and tension component Tcos

(+j^)
Now the tension component Tcos

balances the weight of the body

Tcos

= mg

So,tension component Tsin

provides the necessary force whic makes the body move forward with accl^n a (as seen from ground)

i,e, Tsin

= ma

in car (non inertial frame)
the block is still pushed back and remain at rest for an observer in the car at the pushed position
1) mg ( -j^)
2)tension component Tsin

(+i^) and tension component Tcos

(+j^)
3)pseudo force = ma(-i^)
Since the ball is at rest as seen from the car,
so Tcos

= mg

and Tsin

= ma

In both cases , Tsin = ma
But from ground .... its a single force which actually is Tsin but its magnitude is equal to ma
and from car.... they are two different forces in opposite direction making the body stay at rest.

This proves ........ pseudo forces come only in non-inertial frames.

We may consider pseudo forces in inertial frame (here, ground frame) and arrive at the same result....... BUT THE CONCEPT REMAINS WRONG

biki

discussing a little theory on Work Power Energy

Consider the figure ...

In this fig. we will find out what should be the velocity of the bob at the lowest position so that the string becomes slack ( i,e, T = 0 ) at the horizontal diameter..

Now the weight acts perpendicular to the tension at this position and so has no component along the string....

So taking v =velocity at that point and T = tension and l = length of string....

we have T = mv²/l..

Now let u = velocity of the bob at lowest position so that the bob acquires velocity = v at the horizontal diameter....

So ... 1/2.mu² = 1/2mv² + mgl

or v² = u²- 2gl

Using this in equ^n of T.... we have

T = m(u² - 2gl)/l

If string becomes slack.... T = 0

i,e, m(u²- 2gl)/l = 0

or u² - 2gl = 0

or u =

2gl

So we see that if velocity at lowest position should be equal to

2gl such that the string becomes slack at the horizontal diameter...

FIGURE 1

And see HCV-1 page 128 , example 8 where you will find that the velocity at lowest position such that the bob complets a full circle is 5gl .

So we conclude that....

if velocity is less than 2gl, then string becomes slack below the horizontal diameter.

if velocity = 2gl, then string becomes slack at horizontal diameter..

if velocity is greater than 2gl but less than 5gl, then string becomes slack in the region between the horizontal diameter and the topmost position....

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