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प्रश्न
Solve the following problem.
A uniform solid sphere of radius R has a hole of radius R/2 drilled inside it. One end of the hole is at the center of the sphere while the other is at the boundary. Locate center of mass of the remaining sphere.
उत्तर
Let the centre of the sphere be origin O. Then, r1 be the position vector of the centre of mass of uniform solid sphere and r2 be the position vector of the centre of mass of the cut-out part of the sphere.
Now, mass of the sphere is given as,
M = `(4/3 pi"R"^3)rho` ....(i)
Hence, the mass of the cut part of the sphere will be,
M' = `[4/3 pi("R"/2)^3]rho` .....(∵ sphere is uniform, ρ = constant)
∴ M' = `(4/3 pi"R"^3)rho xx 1/8`
`= "M"/8` ....[From (i)]
From figure, r1 = 0, r2 = `"R"/2`
∴ Position vector of centre of mass of remaining part,
`"r"_"cm" = ("M" xx "r"_1 - "M'" xx "r"_2)/("M" - "M'")`
`= ("M" xx 0 - ("M"/8) xx "R"/2)/("M" - "M"/8)`
`= ((- "RM")/16 xx 8/"7M")`
`"r"_"cm" = (-"R")/14`
(Negative sign indicates the distance is on left side of the origin.)
Position of centre of mass of remaining sphere is `(-"R")/14`
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