IIT JEE , AIEEE Forum - Physics , Mechanics

ab_batra

centre of mass concept
a good shortcut
if a problem comes on conservation of momentum of 2 bodies
then
M* x(towards left)=m*x(towards right)
where x is distance moved

pranay_robot

hey its another one :

1) FRICTION ALWAYS OPPOSES RELATIVE VELOCITY OF A BODY. HERE THE WORD" RELATIVE" IS MOST IMPORTANT.

2) IF L IS THE LENGTH OF SIMPLE PENDULUM. AND M IS THE MASS OF ITS BOB.LET THE BOB GIVEN A VELOCITY V ALONG HORIZONTAL DIRECTION.

THE BOB PERFORMS OSCILLATIONS IF " V

ROOT 2GL."

THE BOB DESCRIBES VERTICAL CIRCLE OF RADIUS L IF

"

V

ROOT5GL."

THE BOB NEITHER PERFORMS OSCILLATIONS NOR DESCRIBE VERTICAL CIRCLE IF V LIES BETWEEN" ROOT2GL" AND "ROOT5GL"

pranay_robot

CENTRIFUGAL FORCE IS PSEUDO FORCE WITH RESPECT TO INERTIAL FRAME OF REFERENCE.

CENTRIFUGAL FORCE IS REAL FORCE WITH RESPECT TO NON INERTIAL FRAME OF REFERENCE.

CENTRIFUGAL FORCE IS NOT THE FORCE OF REACTION OF CENTRIPETAL FORCE.

MOMENT OF INERTIA IN ROTATIONAL MOTION IS SIMILAR TO MASS IN TRANSLATORY MOTION.

MOMENT OF INERTIA OF ROTATING RIGID BODY IS INDEPENDENT OF ITS ANGULAR VELOCITY.

MOMENT OF INERTIA OF METALLIC BODY ALSO DEPENDS UP ON ITS TEMPERATURE.

FORCE IN TRANSLATORY MOTION IS SIMILAR TO TORQUE IN RROTATIONAL MOTION.

IN CASE OF ROLLING BODIES TRANSLATORY K.E

ROTATIONAL K.E.

rajat.khanduja

Another important to keep in mind is that, static friction can vary from 0 to

_s N .

But Kinetic friction will always be constant and equal to

_k N.

Hope you find this interesting as well as important

rajat.khanduja

A good question to test your basics

if a body has an initial velocity 'u' and its acceleration changes with time as
"at" where 'a' is a constant. then,

a) v = u + at²b) v = at
c) v = u + at² /2

If you think that the answer should be (a) and that this is an idiotic ques. then you maybe right.

But you are not.

The answer is (c).

Why?????
Because the equations of motion are valid only in cases of constant acceleration

But here 'a' changes with time.

Give it another try now (after the hint)

Well,for those who still didn't get it

use integration.

integrate v =

at dt

You will get,

v = c + at²/2

As initial velocity 'u', then c = u

PS : RATE ME IF YOU FIND THIS HELPFUL. NUDGE ME IF YOU NEED HELP.

ridhima

com on more
more contributions awaited
good job done by everyone

cheers

ridhima

HERE IS ONE FROM ME ALSO

DURING CIRCULAR MOTION

read this carefully

WHEN THE BOB IS HELD VERTICALLY AND IT IS WHIRLED IN THE HORIZONTAL PLANE

AN OBIVIOUS QUES ARRISES THAT EVEN THOUGH GRAVITATIONAL FORCE IS ACTING ON THE BOB WHY DOESNT IT ACCELERATE TOWARDS THE EARTH

THINK

AS WE START WHIRLING THE BOB - OBSERVE THAT THE VERTICAL COMPONENT OF THE TENSION FORCE SUPPORTS THE WEIGHT

U MUST BE WONDERING THAT THE THE STRING IS TOTALLY HORIZONTAL.

SEE THE BOB FALLS TO THE EARTH BUT DOES NOT ESCAPE TO THE EARTH AS THE TANGENTIAL VELOCITY OF THE BOB TAKES IT AWAY FROM THE EARTH .

SO WE CANNOT SAY THAT IT DOSENT FALL

IT FALLS BUT IS AT THE SAME TIME TAKEN AWAY ALSO . THAT IS THE REASON A CIRCULAR MOTION IS GENERATED

deedee

motion in 1D

important formulaes 4 avg velocity :

if a body covers first half distance wid velocity v1

and d next ahlf veloity v2

then d avg velocity

= (2*v1*v2)/v1+v2

if d body travels wid uniform vel v1............for time t1

n wid uniform vel v2 for tym t2

then avg velocity is

=(v1*t1 + v2*t2)/t1 + t2

if a body covers first half of d distance at d speed v1

next half at d speed v2

n d last one third at d speed v3

then avg speed is

=(3* v1*v2*v3)/v1*v2 +v2*v3 +v1*v3

hope u'll find it useful..... :)

kishan12

U all r doing a great job...........and ya ridhima if u want to know the common misconceptions then try the MTG series of interative physics.they have a whole different section cosisting of these misconceptions......... but anyways there is nothing better then one himself or herself find one........

deedee

hav some more

d distance travelled by a freely fallin body durin in 1st second is always g/2 or 4.9.

irrespective of the height h

for a freely fallin body where initial vel is zero

(1) velocity is directly proportional to time

(2)velocity is directly proportional to d root of distance fallen

(3)distance fallen is directly proportional to d square of time

rajat.khanduja

Look at the figure below ( download it if not clear). The blocks A, B & C of mass 'm' each have accelerations a₁ , a₂ & a₃ . F₁ & F₂ are external forces of magnitudes 2mg and mg respectively. Then,

a) a₁ = a₂ = a₃

b) a₁ > a₂, a₂ = a₃
c) a₁ > a₃ > a₂On first look, this appears to be a question for beginners but here's an advise, don't be too, confident.

Well those who got (b) as the answer, you need to check again and so do those who got answer as (a).

So, the answer is (c).

How?????/
On first look it appears that all these systems are subjected to the same forces and hence the accelerations should be the same. Well, I'd suggest that you make your equations and check again.

Surprised. Well we must no forget that "acceleration depends on FORCE AND MASS"

Hope you find it intersting

apurviitjee2008

many problems in mechanics can be solved by using the simple conservation of energy concept
rather that going for the resolution of forces along the direction of motion
this is especially true for pulley, spring-mass problems........
kinetic energy of a body is changed by the net work done by external forces

siddharthsaxena

Thnx for the wonderful response from you for my previous posts in dis thread........

Here r some of d best shrtcuts availabale today!!!!!!!!!!!!

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