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Which one of the options in the pic is correct and why are the others incorrect?

Which one of the options in the pic is correct and why are the others incorrect?

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Grade:11

19 Answers

Neeti
571 Points
8 years ago
I’ve answered in the previous post of the same question. Check it out :)
Raisinten
12 Points
8 years ago
It would be better if you explain why one is correct and the others incorrect, please.
Raisinten
12 Points
8 years ago
It`s not the same question as @ is not max. A
Neeti
571 Points
8 years ago
Absolutely, see the thing is your diagram is incorrect that is why all of your equations are wrong.
 
See, mv2/ l is the centripetal force which is present along the radial direction along the center of rotation. Now since net force is centripetal (towards O which is the point of attachment of the thread) The difference bwteen the two forces must also be towards the point O .
Since mv2/l is positive and directed towards the center, the difference should also be such that it’s direction is towards O , meaning it should be T – mgcos theta = mv2/l 
The mistake which you’ve made is, you’ve got the direction for mv2/l wrong, it isn’t directed outwards but it is directed inwards towards O. 
 
And even if it’s not max at A. The basic concept is the same for every point in circular motion, whether it is maximum or not. See, if you’ve solved questions related to pulley, when a body of mass m which is hanging by a thread, is undergoing accelaration UPWARDS, the equation is, T – mg = ma  and not mg  – T , because the NET FORCE is directed upwards so the equation should be Upward force minus downward force = net force which is upwards. 
Had the net accelaration been downwards, the equation would have been downward force minus upward force = net force which is acting downwards.
 
In circular motion., the circular path followed by the body is due to a net force which acts CENTRIPETALLY, (directed towards center or inwards) that is why the equation is inward force minus ouward force = net force which is inwards . In this case the inward force is Tension, the outward force is component of mg in radial direction and net force is mv2/l acting INWARDS.
 
Did you get it? 
Raisinten
12 Points
8 years ago
Y
Raisinten
12 Points
8 years ago
I got it. But if I change all the signs of centrifugal forces which one’s correct and why is it correct and others wrong?
Raisinten
12 Points
8 years ago
Anyway, don't we consider centrifugal forces in place of centripetal force while doing numericals?
Neeti
571 Points
8 years ago
No, see centrifugal and centripetal are equal in magnitude but they’re opposite in direction. And depends on the topic under consideration. For rotational motion, the rotation occurs due to a force which pulls the body innwards.that is a centripetal force. so in rotation we consider centripetal force. When considering a body moving on an accelarating frame, we consider a centrifugal force which is pseudo force. so do you get the difference? depends on what you are considering and the force under consideration changes. :) 
anything else?
Raisinten
12 Points
8 years ago
 
I got it. But if I change all the signs of centrifugal forces which one’s correct and why is it correct and others wrong?
Neeti
571 Points
8 years ago
Well if you do that, the first equation will be correct with a little modification. since the force is inwards, for the first equation when you change the sign of mv2/l it becomes mv2/l = mgcos theta – T but that’s wrong because net force is inwards so it should be inward force minus outward force so T – mgcos theta in place of mgcos theta minus T .
And the other equations are wrong because the equation formed by you should be along the radius since that’s the force you are considering.
In the second equation, first of all you’ve ignored the mg component secondly it’s not along the radius so you cannot simply equate them. 
Third equation, same reason, the equation being considered is for the centripetal force which is radial so you can’t take components and equate them. For forces to simply be equal, the body has to be in equilibrium.
 
 
Raisinten
12 Points
8 years ago
Why do we take the radial component always?
Neeti
571 Points
8 years ago
Well because that’s the direction of the force causing the circular motion, it’s not like we assume that the force is in radial direction. The force actually in reality acts in the radial direction hence we take the radial component. I’m sorry i was a little busy so it took me some time to get back :) 
 
So what else? any other doubt? is something still unclear?
Raisinten
12 Points
8 years ago
But we are taking the components in other directions
 
Neeti
571 Points
8 years ago
Yes but you can’t equate forces in random directions. 
like, see we know that the forces in that particular direction can be equated, in any other direction, the same forces may not act, like see, let me give you another example. the second equation in your doubt, it is incorrect because mv2/l has a wrong direction, but also, even if the direction was correct, 
the equation would still be wrong because you are unable to include mg in the equation because mg has no component in the horizontal direction but then you can’t simple ignore mg because it does play a part in the circular motion.
Now are you getting it? you won’t get all the forces right in every direction. But it has specifically been mentioned in the radial direction because that’s why the circular motion is happening.
 
In the question regardin the bob and pendulum, at point A when the thread had not been cut, At that point you can use this concept because it is mentioned in the question that at point A the bob is in EQUILIBRIUM so you can equation forces in EVERY DIRECTION because the body is in equilibrium. 
You can only do it (equate in any random direction) when the body is in equilibrium, not otherwise. 
 
In your doubt in this question, the body is not in equilibrium so you cannot assume the sum of different forces to be equal to one another.
 
 
Raisinten
12 Points
8 years ago
Is this kind of concept applied only for circular motions, if not, then what shall I consider in other cases?
Neeti
571 Points
8 years ago
Well it is allplied whenever different forces are acting on a body in different directions. Remember that you can equate them only when the body is in equilibrium, irrespective of the type of motion. 
But yeah, any other case where more than one force acts on a body and in different directions with different angles between them, then you must apply this concept. :)
Raisinten
12 Points
8 years ago
Thanks a tonne!!! You helped me a lot. But I have another confusion. Check it out at http://www.askiitians.com/forums/Mechanics/using-2-light-cords-a-bob-is-held-in-equilibrium-a_129777.htm
Neeti
571 Points
8 years ago
umm i have no idea about that. See it literaly took me 30mins to solve the question with the steps. But that’s the only approach i could come up with. I don’t know how that would give 5:7
You said the answer was (cos \Theta)2 that’s still understandable and can be derived from my equations ( i just don’t know how :P ) but i can’t think how 5:7 would be possible. 
Wait for the faculty member to reply and clear your doubt, meanwhile i’ll try to come up with an explanation and a different approach.
 
And you are most welcome. You can get your doubts cleared here i’ll try to solve as many as possible :) 
vedant patil
40 Points
8 years ago
all are wrong

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