## Saturday, March 26, 2011

### Fox 336

Let's mess around little bit of what we got, with a little twist of Physics. Don't blame us college kids! It is the truth that's calling. What an overwhelming call that is!

Do we need to know the object's mass? Probably not.

http://www.8foxes.com/

1. if v=0 M reaches the first point with y=1
but ymax=pi/2
high difference (pi/2-1)
we know mgh=1/2mv²
v=sqr(g(pi-2))

2. With this initial velocity value, when object reaches maximum height, velocity comes to 0. Since rope tension disappears, object falls before getting to A.

Actually, rope tension disappears before object gets to maximum, but it does not get to A either. Solution must be greater than sqrt(g(pi-2)).

bleaug

3. I concur with bleaug. There should be angular speed on the object (or tension on the rope) when it reaches to the maximum height. Otherwise the object will fall short of point A.

- Newton (just kidding :)

4. v>sqrt(g(pi-2))
if v=sqrt(g(pi-2))+e
the velocity at the maximum is e
the motion is localy circular with radius pi/2
the tension of the rope is 2e²/pi

5. I assume you mean that e is such that centrifugal force at maximum equals gravity (i.e. rope tension=0). Hence, object reaches A because rope tension is always >= 0 for every angle. In that case e=sqrt(g*pi/2).

This is an upper bound for initial velocity. Optimum is somewhere between sqrt(g*(pi-2)) and sqrt(g*(pi-2))+sqrt(g*pi/2). With g=9.81, that is between 3.346 and 7.272.

Optimum is somewhere in the middle... Around 4.81 actually.

bleaug

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