# 3particals of masses 100gmeach kept at corners of equilatral triangle of side 2m the M.I.of systemabout atrancevers axis passing through it's centroid is....? A.0.4kg-m2 B.0.6 kg-m2 C.0.8kg-m2 D.1.2kg-m2

509 Points
13 years ago

distance of any corner from center is d =a/sqrt3...            (a is side of triangle)

moment of inertia of system is I=(I1+I2+I3)

mass and distance from axis of rotation is same for all hree particles  so I1=I2=I3

total moment of inertia    I = 3I1 = md2 =3(0.1. a2 /3)=0.4kg-m2

hence option A is correct

Devasish Bindani
45 Points
13 years ago

all the masses are kept at a distance of 2/31/2 m from the axis at the centroid.

there fore m.i. of each mass at the axis willl be same i.e.=0.4/3kgm2 on adding all the three moments of inertia it =0.4kgm2

thus according to me the answer is <a> if its incorrect please send me the correct explanation if possible

Priya Gorakshnath Anandkar
18 Points
13 years ago

plz solve my problem &give me list of reagents in org chem with there spatial feachers.  !!!!!!!!!!!!!!!!!!!

plz help me!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!