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Find the position of centre of mass of a uniform disc of radius R from which a hole of radius r is cut out. The centre of the hole is at a distance $\frac{R}{2}$ from the centre of the disc.

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Rr22(R2-r2) towards right of O

Rr22(R2-r2) towards left of O

2Rr2(R2+r2) towards right of O

2Rr2(R2+r2)towards left of O

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detailed solution

Correct option is B

For a disc, mass is proportional to surface area. SoMass of cut portion: m1 = kπr2,Mass of remaining portion: m2 = kπ(R2-r2)Apply m1x1 = m2x2 where x1 = R2, and solve to getx2 = Rr22(R2-r2)

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Find the position of centre of mass of a uniform disc of radius R from which a hole of radius r is cut out. The centre of the hole is at a distance R2 from the centre of the disc.

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Find the position of centre of mass of a uniform disc of radius R from which a hole of radius r is cut out. The centre of the hole is at a distance $\frac{R}{2}$ from the centre of the disc.