AP EAPCET 23-Aug-2021 Shift 1 Paper

Given, mass of rods, m=1‌kg
Length of rod, l=2m
Let moment of inertia of rod about centre of rod, ICM=

ml2

12

D,E and F are mid points of side AC,AB and BC,O is centroid.
∴tan‌30°=

OE

EB

⇒

1

√3

=OE‌

1

2

⇒OE=

1

2√3

=OD=OF=d
By using parallel axis theorem,I=ICM+md2
and for three rods,
I=3[

ml2

12

+m(

l

2√3

)2] =3[

ml2

12

+

ml2

12

] =

ml2

2

=

1×22

2

=2‌kg‌m2