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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.

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Other answers:

find the angular velocity
Well it is not constant and it is not explicit in energy - the linear v is there
\[0.5mv^2 + mgh = E_0 = constant\]
Please solve it someone
At least some SIGNIFICAN part of the way forward ?
@radar you can help ?
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.
let me try this out...
You ARE 20% forward
Ok , the prize for the sover will be 25 minute-worth field integral from EM that I am ready to do in return !
OK the beauty of this situation is that it is VERY common and simple setting. Buuut very unusual answer
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@demitris @Directrix can you help ?
@demitris @Directrix can you help ?
Hi @Mikael ..
Wait a minute.... Let me see what I can do.
@asnaseer Please help
Hello @Callisto - try to help please
Well imagine my sour expression when a nice timid student asks me this question in the middle of a nice sunny day ....
Why no rotational energy term?
Yea , you can use that. Although this is not the only way. Just do it please
This requires physics too
why not to use law of conservation of energy?
Calculus and mechanics are essentially one united universal soul
Good @mathslover USE it and do the soln. please
potential energy + kinetic energy = constant. I cn help after 1 n half hours
Actually I have already pointed that in a hint above. But go forward - when is the bottom reached ?
\[\frac{ t ^{2} }{2 } + \frac{ V _{o}}{ R } t =\int\limits_{ }^{ }\int\limits_{ }^{ } \frac {-R }{ g \cos \theta } \] 0..pi
A) What is the integration variable ? B) Why (suddenly ) double int ???? c) What is the reason you write this so categorically ?
A) g, of course B) equation of motion C) integral is tough, not bothering
theta is the angle the rod makes with the vertical 0...pi
it's theta, why would you ask?
would it be g or R?
Please Show the FULL argument. Double integral seems, on the face of it, out of place in this problem
PLEASE SOLVE @mukushla @TuringTest
argument from torque = -mgRcos(theta) I=mR^2 d^2/ dt^2 (theta) =- gRcos(theta)
\[E_0 = 2mgR +0.5 mv^2\]
Giving the solution soon in 3-4 minutes
h for you is like the y axis ?
OK here goes the solution
I have concluded this problem may be above my pay-grade.
\[t(h) = \int\limits_{h=2R}^{h=0} \frac{dl}{v(h)} dh\]
Now \[dl = \frac{dh}{\cos \theta}\]
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\[h = Sin \theta ====> dh = dl * Cos \theta\]
sorry h = R*Sin theta
Thus gathering all the above observations until now we obtain the following integral
\[\int\limits_{h=2R}^{h=0} dh \frac{1}{\sqrt{1-\frac{h^2}{R^2}}*\sqrt{\frac{2E_0}{m} -2gh}}\]
Here the 1-st square root in the denominator is simply 1/Cos theta
The second square root is the velocity as a function of h - of course derived from the law of conservation of energy stated above
can you please explain how h = rsintheta means dh = dl costheta
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
I did this
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\[\large{t = \frac{-1+\sqrt{1+8\pi R}}{2v_0}}\]
did that make sense? @Coolsector and @Mikael
@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
i dont understand why alpha is what you said.. but maybe its just me
yes .. same problem here
I think @mathslover is NOT correct in HIS FIRST STATEMENT on that page on which all is based upon
So we proceed
there's no g in your soln. @mathslover
Oh I made it for horizontal plane... vertical plane hmn... well guys wait
\[\int\limits_{\xi_0}^{0} d \xi \frac{1}{\sqrt{(1-\xi^2)(A- \xi)}}\]
Mikael whts ur age? grade? so that I can understand the level of this
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
I think this is \(\large{\textbf{above}}\) my level sorry.... I am in 9th grade and did my best.... sorry
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 ?
Sorry.... above my level...
Ms. I like that snark.
I disagree with you @mathslover - you saw it is almost school level
What do you mean mikael? well leave ti whatever do you mean........ continue your work..
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
Aand please appreciate the effort by medals - you are welcome. The question itself was supplied by Mr. @henpen here
Here is the answer as supplied by Wolframalph
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Sorry here in complete - as I thought elliptic integral
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Can see the soln @mukushla
i just came here :)
@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
So I guesses right - about IMO participation ?
sorry lost connection Mr.Mikael :) man im a chemical engineer :) what is the math part of question ?
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
solvent chamber (i meant)
out of my knowledge :)
You should use the simple mechanical conservation of energy and definition of velocity. Some geometry too
Well - see the solution and dissolve your sorrows...
Actually this question is also out of my reach.
Is this the question from a high school level?

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