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Plutonium88
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Homework Statement
A ball on top of a circle is acted on by a FG. The ball rolls to a certain point on the circle, and detaches. Ffr Is negligable.
Given info:
Diameter Of the circle, Mass of ball no ffr
x = the distance from the balls posistion to the top of the big circle.
x is what is needed to be found (Teal in diagram)Okay... So i think i might have it... http://s13.postimage.org/fpzmhstnb/New_Bitmap_Image_3.png
So what i did, was assumed was that the distance from the top to the center (1/2D) and the distance from the object to the center (1/2) were the adjacent and opposite sides of an isoceles right angle triangle... ( i don't know if this is okay..)Now I sovled the angle with TanA = 1/2D/1/2D = 1(TanInverse) = 45 degrees
Now with this angle...
I assumed that angle it traveled would be the angle it left at... (45 Degrees) and any corresponding angle woudl be 45 degrees anyway..
So i am left with the forces of FN which = 0 at the point it detaches
and i am left with the Force of gravity...
I have come to conclude that gravity is what allows centripetal accellearation...
So because the angle of FN = 45 degrees, when i make my Frame of reference so that the X direction is parallel to FN, i end up with force of gravity with the angle and 2 components.So..
Fnety = mgsin45
may = mgsin45
ay = gsin45
Fnetx = mgcos45 - FN
Fnetx = mgcos45 - 0
max = mgcos45
ax=gcos45
okay.. so now since i know that gravitys X component is what provides the centripetal force...
R = D/2
Fc = Fnetx
Mv^2/R = mgcos45
V =√[Dgcos45/2]Now.. Comparing total mechanical energy..
ET1 = EG1 = mgD
Et2 = EK2 + EG2 = 1/2mv^2 + mg(d-x)
Et1 = Et2
mgD = 1/2mv^2 + mg(d-x)
gx = 1/2v^2
x = dgcos45/2/2g
x = D (gcos45/2g)
x = D (cos45/2)
x = D([√2]/8)and this is my answer... x = D([√2]/8)
so yea.. I'm hoping this strategy is right.. everything seems to add up mathematically.. I'm still not sure about this angle though.. I'm not sure if that is correct way to solv eit.
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