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.

- anonymous

- katieb

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- anonymous

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- anonymous

WHEN WILL THE MASS REACH THE LOWEST POINT ?

- anonymous

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## More answers

- anonymous

- anonymous

- anonymous

- UnkleRhaukus

find the angular velocity

- anonymous

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

- anonymous

- anonymous

- anonymous

- anonymous

\[0.5mv^2 + mgh = E_0 = constant\]

- anonymous

CAN ANYONE SOLVE IT AROUND HERE ?

- anonymous

- anonymous

HINTS ?

- anonymous

Please solve it someone

- anonymous

At least some SIGNIFICAN part of the way forward ?

- anonymous

@radar you can help ?

- 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

let me try this out...

- anonymous

You ARE 20% forward

- anonymous

Hi @ajprincess

- anonymous

help

- 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

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

- anonymous

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- anonymous

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- anonymous

@demitris @Directrix can you help ?

- anonymous

- anonymous

@demitris @Directrix can you help ?

- anonymous

Hi @Mikael ..

- anonymous

I AM VERY SORRY I STARTED WITHOUT YOU.
Complete please

- anonymous

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

- anonymous

- anonymous

@asnaseer Please help

- anonymous

Hello @Callisto - try to help please

- anonymous

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

- anonymous

Why no rotational energy term?

- anonymous

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

- anonymous

This requires physics too

- mathslover

why not to use law of conservation of energy?

- anonymous

Calculus and mechanics are essentially one united universal soul

- anonymous

Good @mathslover USE it and do the soln. please

- mathslover

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

- anonymous

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

- anonymous

@ganeshie8 @UnkleRhaukus help pls

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

g CANNOT CHANGE - it is CONSTANT !

- anonymous

it's theta, why would you ask?

- anonymous

would it be g or R?

- anonymous

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

- anonymous

PLEASE SOLVE @mukushla @TuringTest

- anonymous

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

- anonymous

\[E_0 = 2mgR +0.5 mv^2\]

- anonymous

Giving the solution soon in 3-4 minutes

- anonymous

E = CONSTANT

- anonymous

h for you is like the y axis ?

- anonymous

yes

- anonymous

OK here goes the solution

- radar

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

- anonymous

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

- anonymous

Now \[dl = \frac{dh}{\cos \theta}\]

- 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

sorry h = R*Sin theta

- anonymous

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

- anonymous

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

- anonymous

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

- 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

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

- 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

I did this

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- anonymous

|dw:1347813111455:dw|

- mathslover

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

- anonymous

thanks

- mathslover

did that make sense?
@Coolsector and @Mikael

- 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

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

- anonymous

yes .. same problem here

- anonymous

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

- anonymous

So we proceed

- anonymous

there's no g in your soln. @mathslover

- mathslover

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

- anonymous

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

- mathslover

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

- 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

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

- 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

Sorry.... above my level...

- anonymous

Ms. I like that snark.

- anonymous

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

- mathslover

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

- mathslover

*it

- 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

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

Here is the answer as supplied by Wolframalph

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- anonymous

Sorry here in complete - as I thought elliptic integral

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- anonymous

Can see the soln @mukushla

- anonymous

i just came here :)

- 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

So I guesses right - about IMO participation ?

- anonymous

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

- 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

solvent chamber (i meant)

- anonymous

out of my knowledge :)

- anonymous

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

- anonymous

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

- anonymous

Actually this question is also out of my reach.

- anonymous

Is this the question from a high school level?

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