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ParthKohli
 one year ago
Physics (continued)
ParthKohli
 one year ago
Physics (continued)

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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1dw:1443169743112:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2in that image, let \(d = L/2\) and \(\theta = 90^{\circ}\) now if I release it that way, find the maximum height reached by the pendulum bob before the string becomes slack

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1\[g(d) = g\left(1  \frac{d}{R}\right)\]If \(d = R\), \(g=0\) so \(F=0\).

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Firstly, you can see that if the initial height of the pendulum is less than that of the peg, then it'd reach the same height.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2"string becomes slack" is that same as "velocity of bob =0" ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Initial energy is \(mgh_1\) Final energy is \(mgh_2\) \(h_1 = h_2\) The reason why this doesn't apply when the bob is at a higher height than that of the peg is that the string becomes slack before reaching its initial height. Slack meaning that \(T = 0\). When the string becomes slack, then the path carried out by the bob is a projectile as you would imagine.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Ohk, it is in circular motion till the tension becoems 0, afterwards the string is not pulling the bob

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1If you want to delve deeper into the analysis of vertical circular motion, here it is: The minimum velocity at the bottom to complete the whole circle is \(\sqrt{5gR}\). The maximum velocity at the bottom to do a pendulumkind motion like on a clock is \(\sqrt{2gR}\) (use conservation of energy for this too) But if the velocity is between those two, then the string would become slack at one point in the firstquadrant of the circle.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Yes. And the maximum height reached by the bob is not when the string becomes slack. If the velocity points upwards at that moment, then it'd go further up in parabolic motion.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Interesting, I think I have understood the question and why the heights must be same when the bob is released below that peg. Let me try working what happens when the bob is released from 90 degrees..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Awesome! so the string wont become slack at all if the bob is released from a height below the peg

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Since the forces involved here are conservative, I think conservation of energy can be applied : PE + KE = constant Let us define \(0\) at the bottom most of pendulum. Then total mechanical energy at \(\theta=90\) is \(mgL +0\) Total mechanical energy at the position where the string becomes slack is \(mgL' + \frac{1}{2}m{v_{L'}}^2\) These muse be equal : \[mgL = mgL' + \frac{1}{2}m{v_{L'}}^2\tag{1}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2two unknowns in that equation, \(L'\) and \(v_{L'}\) so need to cookup another equation hmm..

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1yeah, one with the forces

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2I'm not good with these tensions and stuff, so im not so sure what happens when the string loses tension and how to take advantage of that..

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1dw:1443173019236:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Oh we can consider that bit anchored to peg as another circular motion is it dw:1443173202385:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Let's draw the freebody diagram of the bob in the frame of the bob itself.dw:1443173207901:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Since we're in the bob's frame, it's not accelerating towards or away from the string at this moment, so \(mg \sin \theta = 2mv^2/L\)

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1is this like tha tprob where a bullet strikes a block on a string, and the block moves a certain angle to a height depending on the mass and velocity of the bullet

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1no, we can't use conservation of momentum here.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1The height at this moment can be expressed as \(L' = L + L\sin\theta \)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1so now we have two equations that we can solve! yay!

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1\[2mgL = mgL(1 + \sin \theta) + \frac{mv^2}{2}\]\[mg\sin \theta = \frac{2mv^2}{L}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2I may be wrong, but don't we also have another new variable, \(\theta\) ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Two variables: v and theta.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Using that, we can find that the height at which the string becomes slack. To find the maximum height in the path, we know that the angle is \(90  \theta \) to the horizontal.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1height is \(L/2 ( 1 + \sin \theta)\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\(v= 2\sqrt{L}\) plugging this in first equation gives \(L' = \dfrac{4L}{5}\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1heyyy the first equation is \(2mgL = mgL/2 (1 + \sin \theta) +1/2 mv^2 \)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Nice problem, I think the key thing was observing that the "centripetal acceleration" and the component of "acceleration due to gravity" along string are equal just at the instant of string losing tension.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Oh was there some mistake ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1yes, I forgot that the radius was L/2 in the second part of the motion so the height is \(L/2 + L/2 \sin \theta\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1I have a huge collection of mechanics problems so if circular motion bores you

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2after fixing : \(v = \sqrt{\dfrac{10L}{3}}\implies L' = \dfrac{5L}{6}\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Ganeshie, have you ever studied gaseous state/thermodynamics/calorimetry and the like?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1the initial energy is mgL right, sorry

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1\[2mgL = mgL/2 (1 + \sin \theta) +1/2 mv^2\] what is that 2

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1yes true the last time I did this question I just multiplied the equation by \(2\) so it became \(2mgL = mgL(1 + \sin \theta ) + mv^2\) which is why I also made the mistake earlier...

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1whenever you do a question the second time you don't treat it the same way you do when you do it the first time ... your memory sorta reminds you of the solution and then you do wrong things :

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1how would you do that without using conservation of energy, with angular velocities? will that double as soon as you shorten the radius by 2?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1yes it does... you can try that too

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2@ParthKohli physics and me don't get along that well, Never had to study thermodynamics before... I'm just a dumb engineer..

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1wait so you didn't study that in high school?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2It was more than 10 years ago, can't say I remember much...

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1CBSE? Or state board?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2AP state board, it is tougher than CBSE maybe, lets work a question together..

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1So th ething will have an angular accel up to some tangential velocity at the bottom, then double that velocity in the smaller circle, and with that will reach a certain height until the pull of the rope, or the radial accel is zero?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1I took thermo about 8 yrs ago,

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1i can still picture those bell curve looking state diagrams in my mind

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thermo was really difficult for me :\

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I need to review that stuff all over again sometime haha

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1lots of tables, change of state of something

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1internal energies and entropy at a state, blah blah, The 3 space PVT survace is nice to remember, the other curves are slices of it

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1I would go through that again sometime if you want, so we both are to ask questions

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1i still have all my practice and hw probs and stuff

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2I doubt if I ever get hands on thermodynamics... I seriously wish sometimes that it is possible to learn without putting any effort by directly downloading to brain, like in matrix..

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Yes. Mechanics is wayyy better.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2classical mechanics is really enjoyable.. with professors like walter lewin / leonord suskind, they almost spoonfeed you.. there is absolutely no chance of finding it tough by following their lectures

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1hah right, but with feynman i can follow along great, then think what the heck did i learn again a few days later..

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1You already know half of it, the first law is a more general energy conservation equation that includes more things that is like 1/3 the class

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1given state 1 of a system at these conditions, and state 2 at some other conditions, find some value that is missing

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1incompressible fluids, or, ideal gasses

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1Energy transfers through heat and work and mass

DanJS
 one year ago
Best ResponseYou've already chosen the best response.1usually ignore the mechanical energy in those systems, mostly internal energy for closed systems
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