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Radius of centripetal force when on an angle.

  • Thread starter Esoremada
  • Start date
  • #1
52
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http://puu.sh/4IjIT.png [Broken]

I don't quite understand why this is correct. My initial try was this... (the only difference is at the end, bolded)

r = 0.154m
period = 0.426s
rev per second = 1/0.426s
w = 2*pi*(1/0.426)

mg = Fn*cosx
Fn = mg/cosx
Fc = Fn*sinx = mg/cosx
= mg*tanx

mg*tanx = mv^2/r
mg*tanx = m(r*sinx)w^2
g*tanx = (r*sinx)*w^2
9.8*tan(x) = 0.154*sinx*[2*pi*(1/0.426)]^2

EDIT: So this way does work, why did the way in the screenshot get the same answer?
EDIT 2: Oh... I just did it wrong and coincidentally got a close enough answer to round to the right answer.
 
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Answers and Replies

  • #2
Doc Al
Mentor
44,892
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So why is the radius of the rotation about the y axis constantly equal to the given radius of the ring, shouldn't the distance from the bead to the centre of rotation depend on the angle of the bead?

If the bead was at a 1 degree angle from the vertical, the circle of motion it makes would be nowhere near 15cm. I don't get why the radius in the equation mv^2/r or mrw^2 wouldn't be r*sinx.
You are correct. The radius of the circular path made by the bead is not the radius of the hoop.
 
  • #3
52
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You are correct. The radius of the circular path made by the bead is not the radius of the hoop.
So why did I get the correct answer by using the full radius of the hoop as the path made by the bead?

When I use my way I get 90 degrees, so it's definitely wrong.

EDIT: Did the math wrong, my way does work.

9.8*tan(x) = 0.154*sinx*[2*pi*(1/0.426)]^2
9.8/cosx = 0.154*[2*pi*(1/0.426)]^2
9.8 = cosx*(0.154*[2*pi*(1/0.426)]^2)
arccos(9.8 / [0.154*[2*pi*(1/0.426)]^2]) = 73
 
Last edited:

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