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anonymous
 one year ago
fun ques
anonymous
 one year ago
fun ques

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In a game show, a smooth solid cylinder of mass \[M=50Kg\], radius \[r=77cm\] and density \[\rho=2.683Kgm^{3}\] is placed above the ground at a height \[h=10cm\] A boy is asked to stand on the cylinder and run across the length \[L\] of the cylinder when the game starts When the game starts the cylinder starts moving with a velocity \[v_{o}=\sqrt{2}ms^{1}\] and starts rotating with about it's axis at an angular velocity \[\omega\] to cover a distance of \[d=20\sqrt{2}m\] Given the total energy of the cylinder that is constant throughout as \[E=125J\] Find: 1) \[t\] the time taken by the cylinder to go from starting point to finishing point 2) Length of the cylinder \[L\] 3) angular velocity \[\omega\] 4) the resultant velocity \[v\] with which the boy should move to just reach the end of the cylinder in time \[t\] Use the following value for constants \[g\approx10ms^{2}\] \[\pi\approx\frac{22}{7}\] @Michele_Laino @IrishBoy123 @ganeshie8 @Empty @Robert136 @iambatman

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is something I made

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1you invented the question?!?!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1cool!! i certainly will do when i get a chance a drawing would be good, just a sketch of course!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441545598411:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.11) t = 20s 2) L = 10m 3) i'm getting an unbelievable answer for \(\omega\), 2.61 r/s but that assumes you include the potential energy of the cylinder using h = 0.1m in mgh......which is an interesting idea 4) 0.5m/s

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmm your 1st and 2nd part are correct, but I think 3rd and 4th are not correct

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1well for 3 \[E = \frac{1}{2} I \omega^2 + \frac{1}{2} M v^2 + Mgh\] \[ = \frac{1}{2} (\frac{1}{2}MR^2) \omega^2 + \frac{1}{2} M v^2 + Mgh\] \[\frac{4E}{M} = R^2 \omega^2+2 v^2 +2gh\] \[\omega =\sqrt{\frac{\frac{4E}{M}  2 v^2 2gh}{R} }\] http://www.wolframalpha.com/input/?i=%5Csqrt+%28++++%28+%284*125%29%2F50++2*2++2*9.8*%280.1%29+%29++%2F+%280.77%5E2%29++++++++++%29

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1for 4) i gave velocity relative to the cylinder would need to add in the cylinder's own motion to give velocity relative to ground

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{4E}{M}=r^2\omega^2+2v_{o}^2+4gh\] not 2gh :P

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1lol head in hands blessed algebra http://www.wolframalpha.com/input/?i=%5Csqrt+%28++++%28+%284*125%29%2F50++2*2++4*9.8*%280.1%29+%29++%2F+%280.77%5E2%29++++++++++%29 better?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Absolutely sir, but I've asked to use 10m/s^2 for g in the question itself so it can cancel out with the height for easier calculation, your result is more accurate however

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If you look at the energy distribution It's like \[E_{rotational}=25J\] \[E_{kinetic}=50J\] \[E_{potential}=50J\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1aha! i never do calculations until the last minute, i stuff them in a calculator and let it build up so i miss the niceties 4) is that 1.5m/s, boy and cylinder combined?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yep!!! you've done it complete

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1that's a great question to test someone'e ability to think because you cannot put it in a box some of the information looks redundant, it also looks as if some info is missing, until you wade through it it also makes me think about potential energy. the centre of mass of the cylinder is not 0.1m above the ground and i bet a lot of people would put PE in as m x g x (0.1+0.77), which is wrong good stuff

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0making an angle of 135 with the tangential velocitydw:1441548497286:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks, I thought about it, took me some time (I didn't notice how much time went by). I wanted to make a question that used different concepts combined

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1exactly my point! you mixed it up and it is quite disorientating. most times, you know what a question is looking for pretty early.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It took me most of the time in selecting values for quantities such that calculation would be easier while at the same time trying not to make the quantities look absurdly large or small

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@IrishBoy123 Interesting, I never thought about that, I think it is more accurate to say the centre of mass is located at the centre of the axis passing through the centre which is of course at a height of (0.1+0.77) that's a significant amount of error we are ignoring, but then again in questions like these we consider the most favourable conditions, point like objects and so on and one would intuitively think of the height as 0.1m !!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1indeed potential energy is mostly a relative term
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