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18 Dec 2007 10:02:37 IST
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ONLY FOR BRAINY IIT ASPIRANTS
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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)
20 Dec 2007 22:07:08 IST
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discussing a little theory
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
2glSo we see that if velocity at lowest position should be equal to 2gl such that the string becomes slack at the horizontal diameter...
2gl such that the string becomes slack at the horizontal diameter...![[Thumb - 1.JPG]](http://www.goiit.com/upload/2007/9/7/77881f980bcb6008134c65cce3ab74fa_19520.JPG_thumb)
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 .
5gl .So we conclude that....
if velocity is less than 2gl, then string becomes slack below the horizontal diameter.
2gl, then string becomes slack below the horizontal diameter.if velocity = 2gl, then string becomes slack at horizontal diameter..
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....
2gl but less than 5gl, then string becomes slack in the region between the horizontal diameter and the topmost position....
23 Dec 2007 12:24:58 IST
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well this one from HCV(actually last few problems of HCV use this concept),
suppose a sphere rolling towards right with angular velocity w in clockwise direction,
when the sphere collides with the wall, then after collision, the linear velocity of sphere is leftwards (that is , linear velocity chages its direction) but the angular velocity w is still in clockwise direction
..
suppose a sphere rolling towards right with angular velocity w in clockwise direction,
when the sphere collides with the wall, then after collision, the linear velocity of sphere is leftwards (that is , linear velocity chages its direction) but the angular velocity w is still in clockwise direction
..
23 Dec 2007 18:15:56 IST
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this is for all those who admire rotational dynamics...
here it is ....
1) we can use instantaneous point of rest for writing the torque equation only when the body is symmetrical and it should be on ground....for fast results
it can be used for non symmetrical bodies also i e when the ipr (instantaneous point of rest) is not in the same vertical line of the centre of mass...
but be careful of writing the torque of pseudo force...
for those who do not know this method earlier...we can use ipr for writing the torque equation because the pseudo force's torque will lie in the same vertical line so it wouldnt produce any torque as such ....
2) during a rod and a mass collision...most favourite questions of iit
conserve angular momentum about the point of collision...
the most amazing part of this solution is that whatever the collision might be ...elastic inelastic or semi....where it is quite complicated in getting the relation betn v and omega....it comes directly....using this and the only thing that will vary is the distance of point of collision..the same relation comes if the distance of point of collision is same and different ....collisions are taking place
3) and choose always a point which is not accelrated ( so as to avoid further errors...like writing pseudo force's torque ) so that atleast we get...relieved of unwanted ...forces ...that more and more equations ...to solve..
4) beware guys ....torque = i alpha can be written only using two points ..one is cm...and the other where it is hinged....
i gave the explanation why we use ipr in some cases above...
hope this helps guys...
go on iit is awating ....!!!
here it is ....
1) we can use instantaneous point of rest for writing the torque equation only when the body is symmetrical and it should be on ground....for fast results
it can be used for non symmetrical bodies also i e when the ipr (instantaneous point of rest) is not in the same vertical line of the centre of mass...
but be careful of writing the torque of pseudo force...
for those who do not know this method earlier...we can use ipr for writing the torque equation because the pseudo force's torque will lie in the same vertical line so it wouldnt produce any torque as such ....
2) during a rod and a mass collision...most favourite questions of iit
conserve angular momentum about the point of collision...
the most amazing part of this solution is that whatever the collision might be ...elastic inelastic or semi....where it is quite complicated in getting the relation betn v and omega....it comes directly....using this and the only thing that will vary is the distance of point of collision..the same relation comes if the distance of point of collision is same and different ....collisions are taking place
3) and choose always a point which is not accelrated ( so as to avoid further errors...like writing pseudo force's torque ) so that atleast we get...relieved of unwanted ...forces ...that more and more equations ...to solve..
4) beware guys ....torque = i alpha can be written only using two points ..one is cm...and the other where it is hinged....
i gave the explanation why we use ipr in some cases above...
hope this helps guys...
go on iit is awating ....!!!
23 Dec 2007 21:16:37 IST
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This is a usual mistake. I've seen many people ( including some on this site commit this mistake. I hope this will make it clear)
As known by all, friction acts to prevent relative motion between two bodies.
This doesn't imply that if there's no motion, there is no friction.
Imagine this, when you push a block on a rough surface, there is no motion initially, but still friction acts
Some people feel that since there's no motion (relative) that implies there is no friction.
Then,
Why do we say that friction acts?[ Even though there is no relative motion]
It is because there 's a tendency of relative motion. Had there been no friction, the body would have moved.
I guess it is clearer now. At least, I have succeeded in bringing the mistake under light.
Nudge me more explanation if reqd.
PS : rate me if I'm correct
As known by all, friction acts to prevent relative motion between two bodies.
This doesn't imply that if there's no motion, there is no friction.
Imagine this, when you push a block on a rough surface, there is no motion initially, but still friction acts
Some people feel that since there's no motion (relative) that implies there is no friction.
Then,
Why do we say that friction acts?[ Even though there is no relative motion]
It is because there 's a tendency of relative motion. Had there been no friction, the body would have moved.
I guess it is clearer now. At least, I have succeeded in bringing the mistake under light.
Nudge me more explanation if reqd.
PS : rate me if I'm correct
28 Dec 2007 00:03:14 IST
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in pure rolling direction of friction is difficult to find out.
for eg:
A body is rolling with linear velocity v and rotating with angular velocity
case1
if we take friction is acting front.
then it will decrease and increase v
and it will slip
case2
if it acts backwards
then it will increase and decrease v.
and again it will slip
i hope this was useful
for eg:
A body is rolling with linear velocity v and rotating with angular velocity
case1
if we take friction is acting front.
then it will decrease
and increase vand it will slip
case2
if it acts backwards
then it will increase
and decrease v.and again it will slip
i hope this was useful
28 Dec 2007 11:50:37 IST
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For questions concerning friction,
whenever it becomes difficult to find the direction of friction, what do you do.
Skip the question. Well, why should you skip a question if that is the only problem you have. I hope the following will help you all.
To find the direction of friction, imagine the case when no friction was acting and then analyze the situation. For example, if a block slips over a rough surface, then ( as you already know) friction acts backwards. But lets try this method.
Imagine there was no friction. Then in which direction is the relative motion. With respect to the ground it is in the forward direction ( hope you know that much).
Now, remember, FRICTION ACTS TO RESIST RELATIVE MOTION. This is in itself the basic and the best formula to get the direction of friction.
Now apply it on the given case. Hence the friction acts backwards on the block.
I know that this was a simple example. Well, you have a book with you ( H.C. Verma ) that can give you various questions to practice this method. So I guess, I need not give any more examples.
Did you find this useful????
If the answer is yes, I'll be waiting for my reward (rating, for those who don't get my point)
whenever it becomes difficult to find the direction of friction, what do you do.
Skip the question. Well, why should you skip a question if that is the only problem you have. I hope the following will help you all.
To find the direction of friction, imagine the case when no friction was acting and then analyze the situation. For example, if a block slips over a rough surface, then ( as you already know) friction acts backwards. But lets try this method.
Imagine there was no friction. Then in which direction is the relative motion. With respect to the ground it is in the forward direction ( hope you know that much).
Now, remember, FRICTION ACTS TO RESIST RELATIVE MOTION. This is in itself the basic and the best formula to get the direction of friction.
Now apply it on the given case. Hence the friction acts backwards on the block.
I know that this was a simple example. Well, you have a book with you ( H.C. Verma ) that can give you various questions to practice this method. So I guess, I need not give any more examples.
Did you find this useful????
If the answer is yes, I'll be waiting for my reward (rating, for those who don't get my point)
30 Dec 2007 17:10:48 IST
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Mechanics rocks
Important in Laws of motion::::::
1>> It is not neccessary that friction always opposes relative motion of bodies...
2>>The original Newtons law states that change in momentum is proportional to the force and not ma
3>>ma is not a force,,,its change in momentum????
4>>The centrifugal force is a pseudo force ....
5>>Pseudo force is just something which we apply to make a non-inertial frame inertial.
6>>Centripetal force is not a new force ...
7>>Centripetal force is due to already existing forces(component of them)...
8>>The point of application of static frictional force always remains at rest relative to both the bodies..
Hope it is useful...
30 Dec 2007 21:11:11 IST
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2 people liked this
I would like to add a comment on message posted by
nishantsingh89
As he said, when a ball collides with a wall, the linear velocity changes in the opposite direction, while the angular velocity remains in the same direction.
This happens as the normal acts through the centre (of mass). Hence no torque is produced about the com.
Hope u find it useful. Please Rate
nishantsingh89
As he said, when a ball collides with a wall, the linear velocity changes in the opposite direction, while the angular velocity remains in the same direction.
This happens as the normal acts through the centre (of mass). Hence no torque is produced about the com.
Hope u find it useful. Please Rate
31 Dec 2007 01:39:59 IST
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2 people liked this
while solving few problems in work,power and energy,it is generally considered that work done by internal forces is always zero.
But that is not the right concept.In some problems, we have to take the work done by internal forces into consideration.Hence,always proceed by havinf a F.B.D in mind else not in paper and then use newton's 2nd law F(net)=Ma
But that is not the right concept.In some problems, we have to take the work done by internal forces into consideration.Hence,always proceed by havinf a F.B.D in mind else not in paper and then use newton's 2nd law F(net)=Ma
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in +icap direction.
what all forces are acting on 'm' whan it is viewed in ground frame as well as car frame
the body is pushed back due to motion of car.
Forces are:-
1) mg ( -j^)
2)reaction(N) by floor (+j^)
and N = mg
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
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.
figure
the ball is pushed back due to motion of car.
Forces are:-
1) mg ( -j^)
2)tension component Tsin
Now the tension component Tcos
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
3)pseudo force = ma(-i^)
Since the ball is at rest as seen from the car,
so Tcos
In both cases , Tsin
But from ground .... its a single force which actually is Tsin
and from car.... they are two different forces in opposite direction making the body stay at rest.