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