anonymous
  • anonymous
fun ques
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
In 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
  • anonymous
This is something I made
IrishBoy123
  • IrishBoy123
you invented the question?!?!

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anonymous
  • anonymous
yep
anonymous
  • anonymous
Give it a try
IrishBoy123
  • IrishBoy123
cool!! i certainly will do when i get a chance a drawing would be good, just a sketch of course!!
anonymous
  • anonymous
|dw:1441545598411:dw|
IrishBoy123
  • IrishBoy123
1) 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
  • anonymous
Hmm your 1st and 2nd part are correct, but I think 3rd and 4th are not correct
IrishBoy123
  • IrishBoy123
well 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
  • IrishBoy123
for 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
  • anonymous
\[\frac{4E}{M}=r^2\omega^2+2v_{o}^2+4gh\] not 2gh :P
IrishBoy123
  • IrishBoy123
lol 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
  • anonymous
Absolutely 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
  • anonymous
If you look at the energy distribution It's like \[E_{rotational}=25J\] \[E_{kinetic}=50J\] \[E_{potential}=50J\]
IrishBoy123
  • IrishBoy123
aha! 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
  • anonymous
yep!!! you've done it complete
IrishBoy123
  • IrishBoy123
that'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
  • anonymous
making an angle of 135 with the tangential velocity|dw:1441548497286:dw|
IrishBoy123
  • IrishBoy123
nice!
anonymous
  • anonymous
Thanks, 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
  • IrishBoy123
exactly my point! you mixed it up and it is quite disorientating. most times, you know what a question is looking for pretty early.
anonymous
  • anonymous
It 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
  • anonymous
@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
  • IrishBoy123
indeed potential energy is mostly a relative term

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