anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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anonymous
  • anonymous
WHEN WILL THE MASS REACH THE LOWEST POINT ?
anonymous
  • anonymous
@mukushla @UnkleRhaukus @experimentex @estudier

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anonymous
  • anonymous
@ParthKohli @radar @RaphaelFilgueiras
anonymous
  • anonymous
@sauravshakya
anonymous
  • anonymous
@Hero @apoorvk @AccessDenied
UnkleRhaukus
  • UnkleRhaukus
find the angular velocity
anonymous
  • anonymous
Well it is not constant and it is not explicit in energy - the linear v is there
anonymous
  • anonymous
@ganeshie8 @ajprincess
anonymous
  • anonymous
@ganeshie8 @ajprincess
anonymous
  • anonymous
@ParthKohli @mathslover
anonymous
  • anonymous
\[0.5mv^2 + mgh = E_0 = constant\]
anonymous
  • anonymous
CAN ANYONE SOLVE IT AROUND HERE ?
anonymous
  • anonymous
@Omniscience ?
anonymous
  • anonymous
HINTS ?
anonymous
  • anonymous
Please solve it someone
anonymous
  • anonymous
At least some SIGNIFICAN part of the way forward ?
anonymous
  • anonymous
@radar you can help ?
apoorvk
  • 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.
apoorvk
  • apoorvk
let me try this out...
anonymous
  • anonymous
You ARE 20% forward
anonymous
  • anonymous
Hi @ajprincess
anonymous
  • anonymous
help
anonymous
  • anonymous
Ok , the prize for the sover will be 25 minute-worth field integral from EM that I am ready to do in return !
anonymous
  • anonymous
OK the beauty of this situation is that it is VERY common and simple setting. Buuut very unusual answer
anonymous
  • anonymous
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anonymous
  • anonymous
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anonymous
  • anonymous
@demitris @Directrix can you help ?
anonymous
  • anonymous
Hi @sauravshakya
anonymous
  • anonymous
@demitris @Directrix can you help ?
anonymous
  • anonymous
Hi @Mikael ..
anonymous
  • anonymous
I AM VERY SORRY I STARTED WITHOUT YOU. Complete please
anonymous
  • anonymous
Wait a minute.... Let me see what I can do.
anonymous
  • anonymous
@Directrix ?
anonymous
  • anonymous
@asnaseer Please help
anonymous
  • anonymous
Hello @Callisto - try to help please
anonymous
  • anonymous
Well imagine my sour expression when a nice timid student asks me this question in the middle of a nice sunny day ....
anonymous
  • anonymous
Why no rotational energy term?
anonymous
  • anonymous
Yea , you can use that. Although this is not the only way. Just do it please
anonymous
  • anonymous
This requires physics too
mathslover
  • mathslover
why not to use law of conservation of energy?
anonymous
  • anonymous
Calculus and mechanics are essentially one united universal soul
anonymous
  • anonymous
Good @mathslover USE it and do the soln. please
mathslover
  • mathslover
potential energy + kinetic energy = constant. I cn help after 1 n half hours
anonymous
  • anonymous
Actually I have already pointed that in a hint above. But go forward - when is the bottom reached ?
anonymous
  • anonymous
@ganeshie8 @UnkleRhaukus help pls
anonymous
  • anonymous
\[\frac{ t ^{2} }{2 } + \frac{ V _{o}}{ R } t =\int\limits_{ }^{ }\int\limits_{ }^{ } \frac {-R }{ g \cos \theta } \] 0..pi
anonymous
  • anonymous
A) What is the integration variable ? B) Why (suddenly ) double int ???? c) What is the reason you write this so categorically ?
anonymous
  • anonymous
A) g, of course B) equation of motion C) integral is tough, not bothering
anonymous
  • anonymous
theta is the angle the rod makes with the vertical 0...pi
anonymous
  • anonymous
g CANNOT CHANGE - it is CONSTANT !
anonymous
  • anonymous
it's theta, why would you ask?
anonymous
  • anonymous
would it be g or R?
anonymous
  • anonymous
Please Show the FULL argument. Double integral seems, on the face of it, out of place in this problem
anonymous
  • anonymous
PLEASE SOLVE @mukushla @TuringTest
anonymous
  • anonymous
argument from torque = -mgRcos(theta) I=mR^2 d^2/ dt^2 (theta) =- gRcos(theta)
anonymous
  • anonymous
\[E_0 = 2mgR +0.5 mv^2\]
anonymous
  • anonymous
Giving the solution soon in 3-4 minutes
anonymous
  • anonymous
E = CONSTANT
anonymous
  • anonymous
h for you is like the y axis ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
OK here goes the solution
radar
  • radar
I have concluded this problem may be above my pay-grade.
anonymous
  • anonymous
\[t(h) = \int\limits_{h=2R}^{h=0} \frac{dl}{v(h)} dh\]
anonymous
  • anonymous
Now \[dl = \frac{dh}{\cos \theta}\]
anonymous
  • 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\[h = Sin \theta ====> dh = dl * Cos \theta\]
anonymous
  • anonymous
sorry h = R*Sin theta
anonymous
  • anonymous
Thus gathering all the above observations until now we obtain the following integral
anonymous
  • anonymous
\[\int\limits_{h=2R}^{h=0} dh \frac{1}{\sqrt{1-\frac{h^2}{R^2}}*\sqrt{\frac{2E_0}{m} -2gh}}\]
anonymous
  • anonymous
Here the 1-st square root in the denominator is simply 1/Cos theta
anonymous
  • 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
anonymous
  • anonymous
can you please explain how h = rsintheta means dh = dl costheta
anonymous
  • 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
mathslover
  • mathslover
I did this
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anonymous
  • anonymous
|dw:1347813111455:dw|
mathslover
  • mathslover
\[\large{t = \frac{-1+\sqrt{1+8\pi R}}{2v_0}}\]
anonymous
  • anonymous
thanks
mathslover
  • mathslover
did that make sense? @Coolsector and @Mikael
anonymous
  • 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
anonymous
  • anonymous
i dont understand why alpha is what you said.. but maybe its just me
anonymous
  • anonymous
yes .. same problem here
anonymous
  • anonymous
I think @mathslover is NOT correct in HIS FIRST STATEMENT on that page on which all is based upon
anonymous
  • anonymous
So we proceed
anonymous
  • anonymous
there's no g in your soln. @mathslover
mathslover
  • mathslover
Oh I made it for horizontal plane... vertical plane hmn... well guys wait
anonymous
  • anonymous
\[\int\limits_{\xi_0}^{0} d \xi \frac{1}{\sqrt{(1-\xi^2)(A- \xi)}}\]
mathslover
  • mathslover
Mikael whts ur age? grade? so that I can understand the level of this
anonymous
  • 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
mathslover
  • mathslover
I think this is \(\large{\textbf{above}}\) my level sorry.... I am in 9th grade and did my best.... sorry
anonymous
  • 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 ?
mathslover
  • mathslover
Sorry.... above my level...
anonymous
  • anonymous
Ms. I like that snark.
anonymous
  • anonymous
I disagree with you @mathslover - you saw it is almost school level
mathslover
  • mathslover
What do you mean mikael? well leave ti whatever do you mean........ continue your work..
mathslover
  • mathslover
*it
anonymous
  • 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
anonymous
  • 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
anonymous
  • anonymous
Here is the answer as supplied by Wolframalph
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anonymous
  • anonymous
Sorry here in complete - as I thought elliptic integral
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anonymous
  • anonymous
Can see the soln @mukushla
anonymous
  • anonymous
i just came here :)
anonymous
  • 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
anonymous
  • anonymous
So I guesses right - about IMO participation ?
anonymous
  • anonymous
sorry lost connection Mr.Mikael :) man im a chemical engineer :) what is the math part of question ?
anonymous
  • 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
anonymous
  • anonymous
solvent chamber (i meant)
anonymous
  • anonymous
out of my knowledge :)
anonymous
  • anonymous
You should use the simple mechanical conservation of energy and definition of velocity. Some geometry too
anonymous
  • anonymous
Well - see the solution and dissolve your sorrows...
anonymous
  • anonymous
Actually this question is also out of my reach.
anonymous
  • anonymous
Is this the question from a high school level?

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