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Mechanics

Hot goIITian

 Joined: 27 Jun 2007 Post: 178
18 Dec 2007 10:02:37 IST
0 People liked this
109
11718
ONLY FOR BRAINY IIT ASPIRANTS
Engineering Entrance , Medical Entrance , AIPMT , JEE Main , AIIMS , JEE Advanced , Physics , Mechanics

since most of us have finished

mechanics , especially the

+1 students who have it fresh

in their minds . let us try to

make our concepts more

better .
most of us must have faced

this problem of judging the

problem in the wrong way

let us compile all these

common errors and post it

here so that every1 can

benifit fron it

5 pointer to every

contribution cheers

Blazing goIITian

Joined: 2 Mar 2007
Posts: 482
18 Dec 2007 14:45:05 IST
10 people liked this

Take one more dude................

Q.Even a sphere rolling on a FRICTIONLESS ground slows down, why???????????

Ans - Because there is a slight deformation in the contact surface (and hence it is no more a point). this shifts the centre of mass and hence the normal force.

The torque due to this normal force opposes the rolling and sphere stops .

Hope it helps

Blazing goIITian

Joined: 16 Dec 2007
Posts: 432
18 Dec 2007 15:03:16 IST
7 people liked this

hcv question: why does a fat man fall over while bending?

A.  for a fat man, the center of mass is above his waist. so if he bends, it is horizontally displaced as compared to his feet.

the normal reaction does not shift its line of action. the man's weight, due to changed position of center of mass, acts along another line.

generally, the normal reaction and weight act along same line so he doesn't fall over. however if they act along different lines, this sets up a torque that makes him fall frontwards

Forum Expert
Joined: 19 Feb 2007
Posts: 3836
18 Dec 2007 15:05:02 IST
8 people liked this

SOME IMPORTANT RELATIONS....

1st the value of  vel at bottom

for -:

a circular ring or cyl.shell , vel at bottom = (gs sin@)1/2

a circular disc or solid cyl , vel at bottom = (1.33 gs sin@)1/2

solid sphere , vel at bottom = [ (10/7) gs sin@ ] 1/2

spherical shell , vel at  bottom = [ (6/5) gs cos@ ]1/2

2nd acc at bottom

for

circular ring or cyl shell , its = (1/2) gsin@

circular disc or solid cyl , its = (2/3) gsin@

solid sphere , its = (5/7) gsin@

spherical shell , its (3.5) gsin@

3rd time taken to reach at bottom....

for

circular ring or cyl shell , its { (4s) / gsin@ }1/2

circular disc or solid cyl , its { (3s) / gsin@ }1/2

solid sphere , its { (14s) / 5gsin@ }1/2

spherical shell , its  { (10s) / 3gsin@ }1/2

hope this helps.....

Forum Expert
Joined: 11 Jun 2007
Posts: 1048
18 Dec 2007 15:33:52 IST
5 people liked this

Here are some more important things to "Remember" :

1)When a body is projected vertically upwards with Velocity
having the magnitude equal to the magnitude of Acceleration
due to Gravity then time taken to reach max Height is 1sec
and max height reached is g/2.

2)The Ratio of Distances travelled in 1st,2nd,3rd second
of freely falling body is 1 : 3 : 5.

3)The ratio of maxim. height reached by different bodies
projected upwards with Velocities U1,U2,U3 is equal to
U12 : U22 : U32 .

4)For a body projected upwards,the distance covered by
the body in the last second of its Upward journey remains
4.9m irrespective of Velocity of Projection.

Hope you find it useful.
Cheers!!!!!!!!!!!!!!!

Blazing goIITian

Joined: 12 Jun 2007
Posts: 1078
18 Dec 2007 17:11:43 IST
3 people liked this

Tension in the string

Guess whats the tension in the string????

Blazing goIITian

Joined: 12 Jun 2007
Posts: 1078
18 Dec 2007 17:12:26 IST
0 people liked this

Ans: 49 N

Blazing goIITian

Joined: 7 Mar 2007
Posts: 784
18 Dec 2007 17:46:34 IST
2 people liked this

@siddarth
Due to global warming, ice at the pole, will melt into water
and will flow into sea . thus mass from the axis of rotation will get away from it,therefore moment of inertia sud increase
I=MR2
BECAUSE MASS IS CONSTANT BUT R WILL INCREASE???

and hance the day and night sud be longer????

PLS crct if i am wrong,...........

Blazing goIITian

Joined: 1 Apr 2007
Posts: 1084
18 Dec 2007 17:50:09 IST
3 people liked this

no friction acts on a ring rolling on a rough surface....

work done by friction in rolling of any object is zero...

Scorching goIITian

Joined: 21 Oct 2007
Posts: 271
18 Dec 2007 18:20:28 IST
4 people liked this

The common error i found is when water or any fluid is filled in a container of irregular shape then the pressure actin at the bottom only depends on the height of the fluid from the surface tat is Pgh
P is density n h is the height of the water column from the surface

hopefully it helps

Blazing goIITian

Joined: 22 Apr 2007
Posts: 2537
18 Dec 2007 18:21:55 IST
3 people liked this

We resolve a force depending on the convenience of the problem.

In case of a body sliding down an inclined plane, we usually resolve mg into mg sin   along the inline and mg cos   perpendicular to incline. we do not resolve normal reaction.
ex. A body slides down an incline at an angle 30 degree with acceleration g /4 . Find the kinetic friction coefficient. ( HCV I , pg. 92 )

In case of circular motion, say banking of track, we keep mg as it is. Here we resolve normal reaction N into  N cos   in the vertical dirn and  N sin in the horizontal direction.
ex. A circular track of radius 600 m is to be designed for cars at an average speed of 180 km / hr. What should be the angle of banking of the track? ( HCV I , pg 107 )

We can do also reverse, but then the maths is complicated.

Cool goIITian

Joined: 4 Jun 2007
Posts: 72
18 Dec 2007 19:19:37 IST
2 people liked this

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

Scorching goIITian

Joined: 10 Dec 2007
Posts: 214
18 Dec 2007 19:54:59 IST
2 people liked this

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
"
VROOT5GL."
THE BOB NEITHER PERFORMS OSCILLATIONS NOR DESCRIBE VERTICAL CIRCLE IF  V LIES BETWEEN" ROOT2GL" AND "ROOT5GL"

Scorching goIITian

Joined: 10 Dec 2007
Posts: 214
18 Dec 2007 20:05:07 IST
4 people liked this

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.

Scorching goIITian

Joined: 14 Dec 2007
Posts: 266
18 Dec 2007 21:52:20 IST
3 people liked this

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

Scorching goIITian

Joined: 14 Dec 2007
Posts: 266
19 Dec 2007 00:50:33 IST
6 people liked this

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 + at2
b)  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.

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.

Hot goIITian

Joined: 27 Jun 2007
Posts: 178
19 Dec 2007 06:53:53 IST
0 people liked this

com on more
more contributions awaited
good job done by everyone

cheers

Hot goIITian

Joined: 27 Jun 2007
Posts: 178
19 Dec 2007 07:04:38 IST
3 people liked this

HERE IS ONE FROM ME ALSO

DURING CIRCULAR MOTION
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

Blazing goIITian

Joined: 10 Sep 2007
Posts: 1319
19 Dec 2007 08:59:04 IST
5 people liked this

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..... :)

Blazing goIITian

Joined: 10 Jan 2007
Posts: 391
19 Dec 2007 09:01:23 IST
2 people liked this

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........

Blazing goIITian

Joined: 10 Sep 2007
Posts: 1319
19 Dec 2007 09:22:57 IST
5 people liked this

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

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