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18 Dec 2007 10:02:37 IST
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ONLY FOR BRAINY IIT ASPIRANTS
Engineering Entrance
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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
Comments (109)
siddharth saxena
Blazing goIITian
Joined: 2 Mar 2007
Posts: 482
18 Dec 2007 14:45:05 IST
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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
Reply
18 Dec 2007 15:03:16 IST
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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
18 Dec 2007 15:05:02 IST
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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.....
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.....
18 Dec 2007 15:33:52 IST
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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!!!!!!!!!!!!!!!
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!!!!!!!!!!!!!!!
18 Dec 2007 17:46:34 IST
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@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,...........
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,...........
18 Dec 2007 18:20:28 IST
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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
P is density n h is the height of the water column from the surface
hopefully it helps
18 Dec 2007 18:21:55 IST
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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.
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.
18 Dec 2007 19:54:59 IST
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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."
ROOT 2GL." THE BOB DESCRIBES VERTICAL CIRCLE OF RADIUS L IF
"
V
ROOT5GL."
ROOT5GL." THE BOB NEITHER PERFORMS OSCILLATIONS NOR DESCRIBE VERTICAL CIRCLE IF V LIES BETWEEN" ROOT2GL" AND "ROOT5GL"
18 Dec 2007 20:05:07 IST
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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.
ROTATIONAL K.E.
s N .
k N.19 Dec 2007 00:50:33 IST
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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.
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.
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.
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 dtYou 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.
19 Dec 2007 07:04:38 IST
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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
19 Dec 2007 08:59:04 IST
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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..... :)
19 Dec 2007 09:01:23 IST
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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........
19 Dec 2007 09:22:57 IST
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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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