The way I understand it, you're swinging this thing as a pendulum, so you have to take into account the velocity and radius of the line.
This is the equation I came up with:
Angle (with respect to the vertical) = cos^-1{(m((v^2)/r) + g) / 44.5}
Where v is the velocity of the ball, g is acceleration due to gravity, r is the radius.
Cheers!
2007-03-12 15:37:38
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answer #1
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answered by pedros2008 3
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Forget it, I have it now. You don't need the speed, and I am gonna show it to you.
Draw the line in a horizontal position, and then draw the line making and angle with the vertical, that is the angle you have been asked.
Then, you will find that the altitude from where the ball was released, is : H
H = Lcos(angle)
Now, when the line breaks, is because of the resultant force in the line, but this force, to find it, we need to considere the centripetal force and the weight of the ball, but be careful.
Draw the weight of the ball at the second point I mentioned, then decomposed the weight, you will find a radial direction for it : mgcos(angle)
angle = angle made by the line and the vertical
The centripetal force : m*v^2 / R
but in this case : R = L >>> lenght of the line
then : the maximum tension the line can support is 44.5 N, so when the reaches that force, it will break :
T - mgcos(angle) = m*v^2 / L
T = 44.5, >>> On the second point, when the line makes the angle with the vertical.
44.5 - mgcos(angle) = m*v^2 / L
Now, you have the altitude where the ball was released :
H = Lcos(angle)
Using conservation of energy :
m*g*Lcos(angle) = m*v^2 / 2
v^2 = 2*g*Lcos(angle)
using this equation and the other equation we found :
44.5 - mgcos(angle) = m*v^2 / L
44.5 - mgcos(angle) = m*2*g*L*cos(angle) / L
44.5 - mgcos(angle) = 2mgcos(angle)
m = 20
g = 9.8
44.5 = 3*20*9.8*cos(angle)
cos(angle) = 0.08
angle = 85º
That's it, hope that might help you
2007-03-12 15:43:03
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answer #2
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answered by anakin_louix 6
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