## anonymous 4 years ago The rod is massless. At t = 0 rod is pointing up. It is of length R. rod is rotating without friction around the lower end which is hinged to rotation axis. The mass M is attached at the top end of the rod. The mass is pushed at t=0 with initial velocity . g is gravitational acceleration. All these parameters are given and known.

1. anonymous

2. anonymous

WHEN WILL THE MASS REACH THE LOWEST POINT ?

3. anonymous

@mukushla @UnkleRhaukus @experimentex @estudier

4. anonymous

5. anonymous

@sauravshakya

6. anonymous

@Hero @apoorvk @AccessDenied

7. UnkleRhaukus

find the angular velocity

8. anonymous

Well it is not constant and it is not explicit in energy - the linear v is there

9. anonymous

@ganeshie8 @ajprincess

10. anonymous

@ganeshie8 @ajprincess

11. anonymous

@ParthKohli @mathslover

12. anonymous

$0.5mv^2 + mgh = E_0 = constant$

13. anonymous

CAN ANYONE SOLVE IT AROUND HERE ?

14. anonymous

@Omniscience ?

15. anonymous

HINTS ?

16. anonymous

17. anonymous

At least some SIGNIFICAN part of the way forward ?

18. anonymous

19. apoorvk

Hmm I guess you will just have to find the general term for velocity at any instant, and integrate the term in 'dt' - since energy conservation and angular velocity fundae won't work here.

20. apoorvk

let me try this out...

21. anonymous

You ARE 20% forward

22. anonymous

Hi @ajprincess

23. anonymous

help

24. anonymous

Ok , the prize for the sover will be 25 minute-worth field integral from EM that I am ready to do in return !

25. anonymous

OK the beauty of this situation is that it is VERY common and simple setting. Buuut very unusual answer

26. anonymous

27. anonymous

28. anonymous

@demitris @Directrix can you help ?

29. anonymous

Hi @sauravshakya

30. anonymous

@demitris @Directrix can you help ?

31. anonymous

Hi @Mikael ..

32. anonymous

I AM VERY SORRY I STARTED WITHOUT YOU. Complete please

33. anonymous

Wait a minute.... Let me see what I can do.

34. anonymous

@Directrix ?

35. anonymous

36. anonymous

Hello @Callisto - try to help please

37. anonymous

Well imagine my sour expression when a nice timid student asks me this question in the middle of a nice sunny day ....

38. anonymous

Why no rotational energy term?

39. anonymous

Yea , you can use that. Although this is not the only way. Just do it please

40. anonymous

This requires physics too

41. mathslover

why not to use law of conservation of energy?

42. anonymous

Calculus and mechanics are essentially one united universal soul

43. anonymous

Good @mathslover USE it and do the soln. please

44. mathslover

potential energy + kinetic energy = constant. I cn help after 1 n half hours

45. anonymous

Actually I have already pointed that in a hint above. But go forward - when is the bottom reached ?

46. anonymous

@ganeshie8 @UnkleRhaukus help pls

47. anonymous

$\frac{ t ^{2} }{2 } + \frac{ V _{o}}{ R } t =\int\limits_{ }^{ }\int\limits_{ }^{ } \frac {-R }{ g \cos \theta }$ 0..pi

48. anonymous

A) What is the integration variable ? B) Why (suddenly ) double int ???? c) What is the reason you write this so categorically ?

49. anonymous

A) g, of course B) equation of motion C) integral is tough, not bothering

50. anonymous

theta is the angle the rod makes with the vertical 0...pi

51. anonymous

g CANNOT CHANGE - it is CONSTANT !

52. anonymous

it's theta, why would you ask?

53. anonymous

would it be g or R?

54. anonymous

Please Show the FULL argument. Double integral seems, on the face of it, out of place in this problem

55. anonymous

56. anonymous

argument from torque = -mgRcos(theta) I=mR^2 d^2/ dt^2 (theta) =- gRcos(theta)

57. anonymous

$E_0 = 2mgR +0.5 mv^2$

58. anonymous

Giving the solution soon in 3-4 minutes

59. anonymous

E = CONSTANT

60. anonymous

h for you is like the y axis ?

61. anonymous

yes

62. anonymous

OK here goes the solution

I have concluded this problem may be above my pay-grade.

64. anonymous

$t(h) = \int\limits_{h=2R}^{h=0} \frac{dl}{v(h)} dh$

65. anonymous

Now $dl = \frac{dh}{\cos \theta}$

66. anonymous

Forgot to say - thta is the angle with the horizon of the rod.The relation dh to dl is a geometric fact that follows from$h = Sin \theta ====> dh = dl * Cos \theta$

67. anonymous

sorry h = R*Sin theta

68. anonymous

Thus gathering all the above observations until now we obtain the following integral

69. anonymous

$\int\limits_{h=2R}^{h=0} dh \frac{1}{\sqrt{1-\frac{h^2}{R^2}}*\sqrt{\frac{2E_0}{m} -2gh}}$

70. anonymous

Here the 1-st square root in the denominator is simply 1/Cos theta

71. anonymous

The second square root is the velocity as a function of h - of course derived from the law of conservation of energy stated above

72. anonymous

can you please explain how h = rsintheta means dh = dl costheta

73. anonymous

Now by rescaling the distance by R and taking one convenient constant (from the denominator roots) out of the integral we simplify it all to

74. mathslover

I did this

75. anonymous

|dw:1347813111455:dw|

76. mathslover

$\large{t = \frac{-1+\sqrt{1+8\pi R}}{2v_0}}$

77. anonymous

thanks

78. mathslover

did that make sense? @Coolsector and @Mikael

79. anonymous

@mathslover I would be amazed and thankful for COMPLETE explanation why do you claim the first equation - that the angle is a simple quadratic function of time. As you know the angular acceleration IS NOT CONSTANT HERE

80. anonymous

i dont understand why alpha is what you said.. but maybe its just me

81. anonymous

yes .. same problem here

82. anonymous

I think @mathslover is NOT correct in HIS FIRST STATEMENT on that page on which all is based upon

83. anonymous

So we proceed

84. anonymous

there's no g in your soln. @mathslover

85. mathslover

Oh I made it for horizontal plane... vertical plane hmn... well guys wait

86. anonymous

$\int\limits_{\xi_0}^{0} d \xi \frac{1}{\sqrt{(1-\xi^2)(A- \xi)}}$

87. mathslover

Mikael whts ur age? grade? so that I can understand the level of this

88. anonymous

HERE Xi is rescaled variable h, and A is some constant containing fractions of E_0 and g. Outside of the integral is of course a Multiplier (not written) which is physically crucial and mathematically trivial

89. mathslover

I think this is $$\large{\textbf{above}}$$ my level sorry.... I am in 9th grade and did my best.... sorry

90. anonymous

This is the level of a good Physical-Course on analytival mechanics in Cambridge/Berkeley/MIT 1-st or second year. Also Ms. @Algebraic! is probably generally pointed to a more Engineering diection of solution - using torques and angular accelerations. Aand Mr. @radar can I send you a message unrelated to math but related to this site ?

91. mathslover

Sorry.... above my level...

92. anonymous

Ms. I like that snark.

93. anonymous

I disagree with you @mathslover - you saw it is almost school level

94. mathslover

What do you mean mikael? well leave ti whatever do you mean........ continue your work..

95. mathslover

*it

96. anonymous

Well 1) This is all - please help me by some Tables / Wolframalpha to find what type of Integral this belongs to 2) I suspect it is elliptic Integral

97. anonymous

Aand please appreciate the effort by medals - you are welcome. The question itself was supplied by Mr. @henpen here http://openstudy.com/study#/updates/50549e72e4b0a91cdf445d5b

98. anonymous

Here is the answer as supplied by Wolframalph

99. anonymous

Sorry here in complete - as I thought elliptic integral

100. anonymous

Can see the soln @mukushla

101. anonymous

i just came here :)

102. anonymous

@mukushla The original challenge was for 2 sets of people 1) You, @sauravshakya, and other olympics-math types 2) Physicists So try solving without any reading the soln

103. anonymous

So I guesses right - about IMO participation ?

104. anonymous

sorry lost connection Mr.Mikael :) man im a chemical engineer :) what is the math part of question ?

105. anonymous

You need to find how long it takes for a spherical bead of polymer to slied without friction though a perfectly circular half circular tube to the sovent chamber if initially the small bead has v_0 velocity and starts from the top of the tube

106. anonymous

solvent chamber (i meant)

107. anonymous

out of my knowledge :)

108. anonymous

You should use the simple mechanical conservation of energy and definition of velocity. Some geometry too

109. anonymous

Well - see the solution and dissolve your sorrows...

110. anonymous

Actually this question is also out of my reach.

111. anonymous

Is this the question from a high school level?