First slide

Methods to locate centre of mass

Question

A square of side a and uniform thickness is divided into four equal parts. If upper right part is removed, then find the coordinates of centre of mass of remaining part.

Moderate

A

$(\frac{5}{12} a, \frac{5}{12} a)$
B

$(\frac{7}{12} a, \frac{7}{12} a)$
C

$(\frac{1}{4} a, \frac{1}{4} a)$
D

$(\frac{1}{3} a, \frac{1}{3} a)$

Solution

Let mass of each part is m. Coordinates of their respective centre of masses are shown in figure.
$x_{cm} = \frac{m \frac{a}{4} + m \frac{3 a}{4} + m \frac{a}{4}}{m + m + m} = \frac{5 a}{12}$
$y_{cm} = \frac{m \frac{a}{4} + m \frac{a}{4} + m \frac{3 a}{4}}{m + m + m} = \frac{5 a}{12}$

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