Centre of mass(com)

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Question

The disc of mass M and radius R with uniform surface mass density $σ$ as shown in the figure. The centre of mass of quarter disc (the shaded area) is at the position $\frac{x}{3} \frac{R}{π}, \frac{x}{3} \frac{R}{π}$ where x is _____, (Round off to the Nearest integer)
[a is the shaded area as shown in the figure]

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Solution

The distance of centre of mass of certain symmetrical shape of disc from centre is
$d = \frac{4 R \sin (θ / 2)}{3 θ} F o r q u a r t e r, p u t θ = \frac{π}{2} ∴ d = \frac{8 R}{3 \sqrt{2} π}$

$∴ x (c o o r d i a n t e) = d \cos 45 = \frac{8 R}{3 \sqrt{2} π} \times \frac{1}{\sqrt{2}} = \frac{4 R}{3 π}$
$∴ y (c o o r d i a n t e) = d \sin 45 = \frac{8 R}{3 \sqrt{2} π} \times \frac{1}{\sqrt{2}} = \frac{4 R}{3 π}$

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Centre of mass(com)

Remember concepts with our Masterclasses.

The disc of mass M and radius R with uniform surface mass density σ as shown in the figure. The centre of mass of quarter disc (the shaded area) is at the position x3Rπ,x3Rπ where x is _____, (Round off to the Nearest integer)[a is the shaded area as shown in the figure]

The distance of centre of mass of certain symmetrical shape of disc from centre is d=4Rsinθ/23θ For quarter, put θ=π2 ∴d=8R32π ∴xcoordiante=dcos45=8R32π×12=4R3π ∴ycoordiante=dsin45=8R32π×12=4R3π

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