## ParthKohli one year ago Physics (continued)

1. Empty

nice art m8

2. ParthKohli

|dw:1443169743112:dw|

3. ganeshie8

in 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

4. ParthKohli

$g(d) = g\left(1 - \frac{d}{R}\right)$If $$d = R$$, $$g=0$$ so $$F=0$$.

5. ParthKohli

Firstly, 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.

6. ganeshie8

why is that so ?

7. ganeshie8

"string becomes slack" is that same as "velocity of bob =0" ?

8. ParthKohli

Initial 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.

9. ganeshie8

Ohk, it is in circular motion till the tension becoems 0, afterwards the string is not pulling the bob

10. ParthKohli

If 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 pendulum-kind 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 first-quadrant of the circle.

11. ParthKohli

Yes. 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.

12. ganeshie8

Interesting, 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..

13. ParthKohli

14. ParthKohli

15. ganeshie8

Awesome! so the string wont become slack at all if the bob is released from a height below the peg

16. ParthKohli

ya

17. ganeshie8

Since 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 : $mgL = mgL' + \frac{1}{2}m{v_{L'}}^2\tag{1}$

18. ganeshie8

two unknowns in that equation, $$L'$$ and $$v_{L'}$$ so need to cookup another equation hmm..

19. ParthKohli

yeah, one with the forces

20. ganeshie8

I'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..

21. ParthKohli

|dw:1443173019236:dw|

22. ganeshie8

Oh we can consider that bit anchored to peg as another circular motion is it |dw:1443173202385:dw|

23. ParthKohli

Let's draw the free-body diagram of the bob in the frame of the bob itself.|dw:1443173207901:dw|

24. ParthKohli

Since we're in the bob's frame, it's not accelerating towards or away from the string at this moment, so $$mg \sin \theta = 2mv^2/L$$

25. DanJS

is 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

26. ParthKohli

no, we can't use conservation of momentum here.

27. DanJS

ok

28. ParthKohli

The height at this moment can be expressed as $$L' = L + L\sin\theta$$

29. ParthKohli

so now we have two equations that we can solve! yay!

30. ParthKohli

$2mgL = mgL(1 + \sin \theta) + \frac{mv^2}{2}$$mg\sin \theta = \frac{2mv^2}{L}$

31. ganeshie8

I may be wrong, but don't we also have another new variable, $$\theta$$ ?

32. ParthKohli

Two variables: v and theta.

33. ganeshie8

nvm, i see..

34. ParthKohli

Using 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.

35. ganeshie8
36. ParthKohli

ohh!!

37. ParthKohli

height is $$L/2 ( 1 + \sin \theta)$$

38. ganeshie8

$$v= 2\sqrt{L}$$ plugging this in first equation gives $$L' = \dfrac{4L}{5}$$

39. ParthKohli

heyyy the first equation is $$2mgL = mgL/2 (1 + \sin \theta) +1/2 mv^2$$

40. ganeshie8

Nice 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.

41. ganeshie8

Oh was there some mistake ?

42. ParthKohli

yes, I forgot that the radius was L/2 in the second part of the motion so the height is $$L/2 + L/2 \sin \theta$$

43. ParthKohli

I have a huge collection of mechanics problems so if circular motion bores you

44. ganeshie8

after fixing : $$v = \sqrt{\dfrac{10L}{3}}\implies L' = \dfrac{5L}{6}$$

45. ParthKohli

That is correct!`

46. ParthKohli

Ganeshie, have you ever studied gaseous state/thermodynamics/calorimetry and the like?

47. DanJS

the initial energy is mgL right, sorry

48. DanJS

$2mgL = mgL/2 (1 + \sin \theta) +1/2 mv^2$ what is that 2

49. ParthKohli

yes 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...

50. ParthKohli

whenever 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 :|

51. DanJS

how would you do that without using conservation of energy, with angular velocities? will that double as soon as you shorten the radius by 2?

52. ParthKohli

yes it does... you can try that too

53. ganeshie8

@ParthKohli physics and me don't get along that well, Never had to study thermodynamics before... I'm just a dumb engineer..

54. ParthKohli

wait so you didn't study that in high school?

55. ganeshie8

It was more than 10 years ago, can't say I remember much...

56. ParthKohli

CBSE? Or state board?

57. ganeshie8

AP state board, it is tougher than CBSE maybe, lets work a question together..

58. DanJS

So 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?

59. DanJS

I took thermo about 8 yrs ago,

60. DanJS

i can still picture those bell curve looking state diagrams in my mind

61. Jhannybean

thermo was really difficult for me :\

62. Jhannybean

I need to review that stuff all over again sometime haha

63. DanJS

lots of tables, change of state of something

64. DanJS

internal energies and entropy at a state, blah blah, The 3 space PVT survace is nice to remember, the other curves are slices of it

65. DanJS

I would go through that again sometime if you want, so we both are to ask questions

66. DanJS

i still have all my practice and hw probs and stuff

67. ganeshie8

I 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..

68. ParthKohli

Yes. Mechanics is wayyy better.

69. ganeshie8

classical 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

70. DanJS

hah right, but with feynman i can follow along great, then think what the heck did i learn again a few days later..

71. DanJS

You 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

72. DanJS

given state 1 of a system at these conditions, and state 2 at some other conditions, find some value that is missing

73. DanJS

incompressible fluids, or, ideal gasses

74. DanJS

Energy transfers through heat and work and mass

75. DanJS

usually ignore the mechanical energy in those systems, mostly internal energy for closed systems