TS EAMCET 14-Sep-2020 Shift 2 Solved Paper

(*) Given, mass of solid cylinder, M=10kg
Radius, R=40cm=0.4m
L=9R=9×0.4=36m
Moment of inertia of the solid cylinder about the line passing through its centre and perpendicular to its axis is given as ICOM‌‌=M[‌

L2

12

+‌

R2

4

]=10[‌

(36)2

12

+‌

(0.4)2

4

]
‌‌=10[1.08+0.04]=10×112
⇒ ICOM=11.2kg−m2
According to parallel axis theorem, moment of inertia of cylinder about aline passing through its edge and perpendicular to its axis is given as
I′‌‌=ICOM+M(‌

L

2

)2=11.2+10(‌

36

2

)2
‌‌=11.2+10(324)=11.2+324=436kg−m2