Question

Four particles of masses 2,2,4,4 kg are arranged at the corners A,B,C,D of a square of side 2m . The perpendicular distance of their center of mass from the side AB

Answer 1

Answer:

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if wrong

Answer 2

The perpendicular distance of their center of mass from the side AB is $\dfrac{4}{3}$.

Explanation:

Given that,

Mass of first particle = 2 kg

Mass of second particle = 2 kg

Mass of third particle = 4 kg

Mass of fourth particle = 4 kg

Side of square = 2 m

We need to calculate the perpendicular distance of their center of mass from the side AB

Using formula of center of mass

$y_{cm}=\dfrac{m_{1}y_{1}+m_{2}y_{2}+m_{3}y_{3}+m_{4}y_{4}}{m_{1}+m_{2}+m_{3}+m_{4}}$

Put the value into the formula

$y_{cm}=\dfrac{2\times0+2\times0+4\times2+4\times2}{2+2+4+4}$

$y_{cm}=\dfrac{16}{12}$

$y_{cm}=\dfrac{4}{3}$

Hence, The perpendicular distance of their center of mass from the side AB is $\dfrac{4}{3}$.

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Topic : center of mass

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