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wel
3 years ago
14

An object is dropped from rest.

Physics
1 answer:
sesenic [268]3 years ago
3 0

Taha xain malai ..........hhdd

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An object in the shape of a thin ring has radius a and mass M. A uniform sphere with mass m and radius R is placed with its cent
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Answer:

F = GMmx/[√(a² + x²)]³

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Since the ring is symmetrical, the vertical components of this force cancel out leaving the horizontal components to add.

So, the horizontal components add from two symmetrically opposite mass elements dM,

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L² = a² + x² where a = radius of ring and x = distance of axis of ring from sphere.

L = √(a² + x²)

cosФ = x/L

dF' = GmdMcosФ/L²

dF' = GmdMx/L³

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Integrating both sides we have

∫dF' = ∫GmdMx/[√(a² + x²)]³

∫dF' = Gm∫dMx/[√(a² + x²)]³    ∫dM = M

F = GmMx/[√(a² + x²)]³  

F = GMmx/[√(a² + x²)]³

So, the force due to the sphere of mass m is

F = GMmx/[√(a² + x²)]³

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