another m3 question please :D II Watch

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RObTRIP
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#1
Report Thread starter 14 years ago
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hey another m3 question i haven't been able to solve, its one of those nasty horizontal circles in bowls! thanks for your time!

A particle P, of mass M, moves on the smooth inner surface of a fixed hollow spherical bowl, centre O, and inner radius r, describing a horiontal circle at a constant speed. the centre C of this circle is at a depth of 1r/3 vertically below I. determine:
a) the magnitude of the force exerted by the surface of the sphere on P
b)the speed of P

thanks a lot!!!
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SsEe
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Report 14 years ago
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(Original post by RObTRIP)
hey another m3 question i haven't been able to solve, its one of those nasty horizontal circles in bowls! thanks for your time!

A particle P, of mass M, moves on the smooth inner surface of a fixed hollow spherical bowl, centre O, and inner radius r, describing a horiontal circle at a constant speed. the centre C of this circle is at a depth of 1r/3 vertically below O. determine:
a) the magnitude of the force exerted by the surface of the sphere on P
b)the speed of P

thanks a lot!!!


a) Call the reaction R. Mg = R sin(t) where t is the angle between the radius of the bowl and the radius of the circle of motion. From simple trig sin(t) = 1/3. Hence R = 3Mg

b) radius of circle of motion = [√(8)/3]r
The force causing the particle to accelerate in a circular path is Rcos(t) = Mg√(8) = 3Mv²/r√(8)
v = √(8gr/3)
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RObTRIP
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Report Thread starter 14 years ago
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ooh we like it, we sure do !!!
thanks a lot its a greeaaat help

didnt know it could be so simple lol :rolleyes:
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RObTRIP
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Report Thread starter 14 years ago
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btw i think you got your simple trig wrong lol
if the adjacent is r and the opposite is 1r/3 then you dont get sin(t)=1/3, you got the angles mixed up i get R=root10.mg

please feel free to embarass me if i'm wrong
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