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## ParthKohli one year ago this is incomplete argh

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1. ParthKohli

|dw:1443611451350:dw|

2. ParthKohli

|dw:1443611647543:dw|

3. ParthKohli

Find the minimum value of $$v$$ given all that.

4. ParthKohli

In the vertical direction,$\frac{a}{\sqrt{2}} + b\frac{\sqrt{3}}{2} = 5g$

5. ParthKohli

In the horizontal direction,$\frac{a}{\sqrt{2}} + \frac{b}2 = \frac{5v^2}{1.6}$

6. ParthKohli

$v = \sqrt{0.32 \left(\frac{a}{\sqrt 2} + \frac{b}{2}\right)}$ http://www.wolframalpha.com/input/?i=minimise+sqrt%280.32*%28a%2F2+%2B+b%2Fsqrt%282%29%29%29+constrained+to+sqrt%283%29%2F2*a+%2B+b%2Fsqrt%282%29+%3D+50

7. anonymous

Can someone please help me??????

8. ganeshie8
9. anonymous

seriously people?!?!?!

10. ganeshie8

we want to find the minimum value of $$v$$, not the exact value of $$v$$ so the problem is fine i think

11. ParthKohli

yes, and it's turning out to be zero.

12. ganeshie8

the greater the linear velocity, the greater will be the tension in wires in which wire do you think the tension will be more ?

13. IrishBoy123

the minimum value will occur when the tension in the lower string/rod = 0

14. ganeshie8

for a given linear velocity, which wire experiences more tension, AC or BC ?

15. ParthKohli

@IrishBoy123 why?!

16. IrishBoy123
17. ParthKohli

ooo, I forgot to add the fact that a>0, b>0

18. ParthKohli
19. IrishBoy123

if you start it spinning from zero, all the tension will be in the longer string so in the plot follow that red circle out to where it meets the blue circle. that is the cross over

20. mathmate

@IrishBoy123 Agree! By intuition, I believe that any one time, only one wire will be in tension. There is (perhaps) exactly one value of v that will give T1>0 and T2>0 on both wires because of the physical arrangement. (assuming the wires don't take compression).

21. IrishBoy123

yeah, @mathmate if these are rods ....

22. ParthKohli

Yes, the minimum occurs when the tension in the lower string is zero.

23. ParthKohli

In other words, you mean that if we want to decrease velocity, we increase the radius.

24. ParthKohli

Thanks guys.

25. IrishBoy123

this is what i meant to share in that link...when i was talking about red and blue circles |dw:1443613483068:dw|

26. ParthKohli

wait, I'm confused again.

27. ParthKohli

what are those red and blue circles?

28. IrishBoy123

|dw:1443613751302:dw|

29. IrishBoy123

circles show the radii of the taut strings

30. ParthKohli

oh.

31. IrishBoy123

assuming these are strings, if you started at v = 0 and slowly increased the rotat speed, it would be supported by the red string [the higher one] until you reach crossover

32. IrishBoy123

so analyse it as a 1 string prpblem where you know that the one string has to reach an angle of 30 degrees $$T = \dfrac{mv^2}{r \sin 30} = \dfrac{mg}{ \cos 30}$$

33. ParthKohli

this is great! thanks!

34. mathmate

@irishboy Exactly. I have done just that, and there is no point where both wires are taut. Analyzing separately, the long wire will attain 30 degrees at about 3 m/s. The short wire will attain 45 degrees at almost 4 m/s. So there is no point (velocity-wise) where both wires will be taut.

35. IrishBoy123

cool!

36. anonymous

Can someone please help me??

37. anonymous

Help Me ...Easy Medal ^^^

38. anonymous

Determine algebraically whether the function is even, odd, or neither even nor odd. f(x) = -3x4 - 2x - 5 Neither Even Odd

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