Three balls of masses 1, 2 and 3 kg respectively are arranged at the corners of an equilateral triangle of side 1m. What will be the M. I. Of the system about an axis through the centroid and perpendicular to the plane of triangle?

1 Answer
P dilip_k
Jan 5, 2018

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Height of the equilateral triangle having each side 1 m is
#h=sqrt3/2xx1^2=sqrt3/2#m
So its circumradius will be #r=sqrt3/2xx2/3=1/sqrt3#m. Since in case of equilateral triangle heights or medians are divided in the ratio of #2:1# at the center of gravity (G).
Hence the M. I. Of the system about an axis through the centroid (G) and perpendicular to the plane of triangle will be given by

#I=(m_1+m_2+m_3)r^2=(1+2+3)xx(1/sqrt3)^2=2kgm^2#

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