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WHEN WILL THE MASS REACH THE LOWEST POINT ?

find the angular velocity

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

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

CAN ANYONE SOLVE IT AROUND HERE ?

HINTS ?

Please solve it someone

At least some SIGNIFICAN part of the way forward ?

let me try this out...

You ARE 20% forward

help

@demitris @Directrix can you help ?

@demitris @Directrix can you help ?

I AM VERY SORRY I STARTED WITHOUT YOU.
Complete please

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

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 ?

@ganeshie8 @UnkleRhaukus help pls

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

g CANNOT CHANGE - it is CONSTANT !

it's theta, why would you ask?

would it be g or R?

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

E = CONSTANT

h for you is like the y axis ?

yes

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

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

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

I did this

|dw:1347813111455:dw|

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

thanks

did that make sense?
@Coolsector and @Mikael

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

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..

*it

Here is the answer as supplied by Wolframalph

Sorry here in complete - as I thought elliptic integral

i just came here :)

So I guesses right - about IMO participation ?

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

solvent chamber (i meant)

out of my knowledge :)

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?