Question

A square plate of side 20 cm has uniform thickness and density. A circular part of diameter 8 cm is cut out symmetrically as shown in figure. The position of centre of mass of the remaining portion is :​

Answer 1

Answer:

area of square=20×20

=400

area of circle=πrSquare

=22/7×8×8

=201.14

Answer 2

The position of centre of mass is on the left side of O .

Step-by-step explanation:

let the total mass be M

mass of the circular part be m

diameter of circle = 8 cm

radius = 4 cm

let Xcm be the center of the mass

then,

$X_c_m = \frac{m(0)+M(6)}{M+m}$

$X_c_m = \frac{6m}{M+m}$

taking m as negative mass

$X_c_m = \frac{-6m}{M+(-m)}$

$M = \frac{\sigma (-20)^2}{\sigma (\pi (4)^2))}$

$X_c_m = \frac{-6 (\sigma\times \pi (16) }{\sigma (400-16\pi)}$

$X_c_m = \frac{-96 \pi}{400-16\pi}$

on solving we get

$X_c_m = -0.736 cm$

here ,

negative sign indicates that center of mass is on the left side of O .

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