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Mikael
Group Title
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.
 one year ago
 one year ago
Mikael Group Title
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.
 one year ago
 one year ago

This Question is Closed

Mikael Group TitleBest ResponseYou've already chosen the best response.3
WHEN WILL THE MASS REACH THE LOWEST POINT ?
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
@mukushla @UnkleRhaukus @experimentex @estudier
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
@ParthKohli @radar @RaphaelFilgueiras
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
@sauravshakya
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
@Hero @apoorvk @AccessDenied
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
find the angular velocity
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
Well it is not constant and it is not explicit in energy  the linear v is there
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
@ganeshie8 @ajprincess
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
@ganeshie8 @ajprincess
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
@ParthKohli @mathslover
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
\[0.5mv^2 + mgh = E_0 = constant\]
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
CAN ANYONE SOLVE IT AROUND HERE ?
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
@Omniscience ?
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
Please solve it someone
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
At least some SIGNIFICAN part of the way forward ?
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
@radar you can help ?
 one year ago

apoorvk Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

apoorvk Group TitleBest ResponseYou've already chosen the best response.0
let me try this out...
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
You ARE 20% forward
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
Hi @ajprincess
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
Ok , the prize for the sover will be 25 minuteworth field integral from EM that I am ready to do in return !
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
OK the beauty of this situation is that it is VERY common and simple setting. Buuut very unusual answer
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
@demitris @Directrix can you help ?
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
Hi @sauravshakya
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
@demitris @Directrix can you help ?
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
Hi @Mikael ..
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
I AM VERY SORRY I STARTED WITHOUT YOU. Complete please
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
Wait a minute.... Let me see what I can do.
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
@Directrix ?
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
@asnaseer Please help
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
Hello @Callisto  try to help please
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
Well imagine my sour expression when a nice timid student asks me this question in the middle of a nice sunny day ....
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
Why no rotational energy term?
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
Yea , you can use that. Although this is not the only way. Just do it please
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
This requires physics too
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.1
why not to use law of conservation of energy?
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
Calculus and mechanics are essentially one united universal soul
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
Good @mathslover USE it and do the soln. please
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.1
potential energy + kinetic energy = constant. I cn help after 1 n half hours
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
Actually I have already pointed that in a hint above. But go forward  when is the bottom reached ?
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
@ganeshie8 @UnkleRhaukus help pls
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{ t ^{2} }{2 } + \frac{ V _{o}}{ R } t =\int\limits_{ }^{ }\int\limits_{ }^{ } \frac {R }{ g \cos \theta } \] 0..pi
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
A) What is the integration variable ? B) Why (suddenly ) double int ???? c) What is the reason you write this so categorically ?
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
A) g, of course B) equation of motion C) integral is tough, not bothering
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
theta is the angle the rod makes with the vertical 0...pi
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
g CANNOT CHANGE  it is CONSTANT !
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
it's theta, why would you ask?
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
would it be g or R?
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
Please Show the FULL argument. Double integral seems, on the face of it, out of place in this problem
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
PLEASE SOLVE @mukushla @TuringTest
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
argument from torque = mgRcos(theta) I=mR^2 d^2/ dt^2 (theta) = gRcos(theta)
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
\[E_0 = 2mgR +0.5 mv^2\]
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
Giving the solution soon in 34 minutes
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
E = CONSTANT
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
h for you is like the y axis ?
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
OK here goes the solution
 one year ago

radar Group TitleBest ResponseYou've already chosen the best response.0
I have concluded this problem may be above my paygrade.
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
\[t(h) = \int\limits_{h=2R}^{h=0} \frac{dl}{v(h)} dh\]
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
Now \[dl = \frac{dh}{\cos \theta}\]
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
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\[h = Sin \theta ====> dh = dl * Cos \theta\]
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
sorry h = R*Sin theta
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
Thus gathering all the above observations until now we obtain the following integral
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
\[\int\limits_{h=2R}^{h=0} dh \frac{1}{\sqrt{1\frac{h^2}{R^2}}*\sqrt{\frac{2E_0}{m} 2gh}}\]
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
Here the 1st square root in the denominator is simply 1/Cos theta
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
The second square root is the velocity as a function of h  of course derived from the law of conservation of energy stated above
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
can you please explain how h = rsintheta means dh = dl costheta
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
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
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.1
I did this
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
dw:1347813111455:dw
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.1
\[\large{t = \frac{1+\sqrt{1+8\pi R}}{2v_0}}\]
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.1
did that make sense? @Coolsector and @Mikael
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
@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
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
i dont understand why alpha is what you said.. but maybe its just me
 one year ago

Coolsector Group TitleBest ResponseYou've already chosen the best response.0
yes .. same problem here
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
I think @mathslover is NOT correct in HIS FIRST STATEMENT on that page on which all is based upon
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
So we proceed
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
there's no g in your soln. @mathslover
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.1
Oh I made it for horizontal plane... vertical plane hmn... well guys wait
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
\[\int\limits_{\xi_0}^{0} d \xi \frac{1}{\sqrt{(1\xi^2)(A \xi)}}\]
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.1
Mikael whts ur age? grade? so that I can understand the level of this
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
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
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.1
I think this is \(\large{\textbf{above}}\) my level sorry.... I am in 9th grade and did my best.... sorry
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
This is the level of a good PhysicalCourse on analytival mechanics in Cambridge/Berkeley/MIT 1st 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 ?
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.1
Sorry.... above my level...
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
Ms. I like that snark.
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
I disagree with you @mathslover  you saw it is almost school level
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.1
What do you mean mikael? well leave ti whatever do you mean........ continue your work..
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
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
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
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
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
Here is the answer as supplied by Wolframalph
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
Sorry here in complete  as I thought elliptic integral
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
Can see the soln @mukushla
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
i just came here :)
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
@mukushla The original challenge was for 2 sets of people 1) You, @sauravshakya, and other olympicsmath types 2) Physicists So try solving without any reading the soln
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
So I guesses right  about IMO participation ?
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
sorry lost connection Mr.Mikael :) man im a chemical engineer :) what is the math part of question ?
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
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
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
solvent chamber (i meant)
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
out of my knowledge :)
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
You should use the simple mechanical conservation of energy and definition of velocity. Some geometry too
 one year ago

Mikael Group TitleBest ResponseYou've already chosen the best response.3
Well  see the solution and dissolve your sorrows...
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
Actually this question is also out of my reach.
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
Is this the question from a high school level?
 one year ago
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