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Mikael
 3 years ago
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.
Mikael
 3 years ago
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.

This Question is Closed

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3WHEN WILL THE MASS REACH THE LOWEST POINT ?

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3@mukushla @UnkleRhaukus @experimentex @estudier

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3@ParthKohli @radar @RaphaelFilgueiras

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3@Hero @apoorvk @AccessDenied

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0find the angular velocity

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

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

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3CAN ANYONE SOLVE IT AROUND HERE ?

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3At least some SIGNIFICAN part of the way forward ?

apoorvk
 3 years ago
Best ResponseYou've already chosen the best response.0Hmm 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.

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

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

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3@demitris @Directrix can you help ?

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3@demitris @Directrix can you help ?

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3I AM VERY SORRY I STARTED WITHOUT YOU. Complete please

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.1Wait a minute.... Let me see what I can do.

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3Hello @Callisto  try to help please

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

Algebraic!
 3 years ago
Best ResponseYou've already chosen the best response.1Why no rotational energy term?

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

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.1This requires physics too

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.1why not to use law of conservation of energy?

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

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3Good @mathslover USE it and do the soln. please

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.1potential energy + kinetic energy = constant. I cn help after 1 n half hours

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

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3@ganeshie8 @UnkleRhaukus help pls

Algebraic!
 3 years ago
Best 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

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

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

Algebraic!
 3 years ago
Best ResponseYou've already chosen the best response.1theta is the angle the rod makes with the vertical 0...pi

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3g CANNOT CHANGE  it is CONSTANT !

Algebraic!
 3 years ago
Best ResponseYou've already chosen the best response.1it's theta, why would you ask?

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

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3PLEASE SOLVE @mukushla @TuringTest

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

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3\[E_0 = 2mgR +0.5 mv^2\]

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3Giving the solution soon in 34 minutes

Coolsector
 3 years ago
Best ResponseYou've already chosen the best response.0h for you is like the y axis ?

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3OK here goes the solution

radar
 3 years ago
Best ResponseYou've already chosen the best response.0I have concluded this problem may be above my paygrade.

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

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

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3Forgot 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
 3 years ago
Best ResponseYou've already chosen the best response.3Thus gathering all the above observations until now we obtain the following integral

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

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

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

Coolsector
 3 years ago
Best ResponseYou've already chosen the best response.0can you please explain how h = rsintheta means dh = dl costheta

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3Now 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
 3 years ago
Best ResponseYou've already chosen the best response.1\[\large{t = \frac{1+\sqrt{1+8\pi R}}{2v_0}}\]

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.1did that make sense? @Coolsector and @Mikael

Mikael
 3 years ago
Best 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

Coolsector
 3 years ago
Best ResponseYou've already chosen the best response.0i dont understand why alpha is what you said.. but maybe its just me

Coolsector
 3 years ago
Best ResponseYou've already chosen the best response.0yes .. same problem here

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

Algebraic!
 3 years ago
Best ResponseYou've already chosen the best response.1there's no g in your soln. @mathslover

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.1Oh I made it for horizontal plane... vertical plane hmn... well guys wait

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

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.1Mikael whts ur age? grade? so that I can understand the level of this

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3HERE 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
 3 years ago
Best ResponseYou've already chosen the best response.1I think this is \(\large{\textbf{above}}\) my level sorry.... I am in 9th grade and did my best.... sorry

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3This 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 ?

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.1Sorry.... above my level...

Algebraic!
 3 years ago
Best ResponseYou've already chosen the best response.1Ms. I like that snark.

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

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.1What do you mean mikael? well leave ti whatever do you mean........ continue your work..

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3Well 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
 3 years ago
Best ResponseYou've already chosen the best response.3Aand please appreciate the effort by medals  you are welcome. The question itself was supplied by Mr. @henpen here http://openstudy.com/study#/updates/50549e72e4b0a91cdf445d5b

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3Here is the answer as supplied by Wolframalph

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3Sorry here in complete  as I thought elliptic integral

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3Can see the soln @mukushla

Mikael
 3 years ago
Best 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

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3So I guesses right  about IMO participation ?

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

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3You 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
 3 years ago
Best ResponseYou've already chosen the best response.3solvent chamber (i meant)

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.0out of my knowledge :)

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

Mikael
 3 years ago
Best ResponseYou've already chosen the best response.3Well  see the solution and dissolve your sorrows...

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.1Actually this question is also out of my reach.

sauravshakya
 3 years ago
Best ResponseYou've already chosen the best response.1Is this the question from a high school level?
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