## anonymous one year ago fun ques

1. 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

2. anonymous

3. IrishBoy123

you invented the question?!?!

4. anonymous

yep

5. anonymous

Give it a try

6. IrishBoy123

cool!! i certainly will do when i get a chance a drawing would be good, just a sketch of course!!

7. anonymous

|dw:1441545598411:dw|

8. 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

9. anonymous

Hmm your 1st and 2nd part are correct, but I think 3rd and 4th are not correct

10. 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

11. 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

12. anonymous

$\frac{4E}{M}=r^2\omega^2+2v_{o}^2+4gh$ not 2gh :P

13. 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?

14. 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

15. anonymous

If you look at the energy distribution It's like $E_{rotational}=25J$ $E_{kinetic}=50J$ $E_{potential}=50J$

16. 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?

17. anonymous

yep!!! you've done it complete

18. 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

19. anonymous

making an angle of 135 with the tangential velocity|dw:1441548497286:dw|

20. IrishBoy123

nice!

21. 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

22. 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.

23. 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

24. 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 !!

25. IrishBoy123

indeed potential energy is mostly a relative term