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Three identical uniform rods each of length 40cm and mass 2kg form an equilateral triangle. Find the moment of inertia of this system about an axis passing through one corner ? Thanks in advance !
let ABC is the triangle formed by these rods then we have to find moment of inertia any about vertex , let vertex be A... then , rod AB & AC will have same moment of inertia about end A , IAC = IAB = mL2/12 + mL2/4 = mL2/3 moment of inertia of BC = mL2/12 + md2 (d is perpendicular distance from axis of rotation) from triangle ABC , d = Lcos30 IBC = mL2/12 + 3mL2/4 = 10mL2/12 net moment of inertia = IAB+IBC+IAC Inet =18mL2/12 = 3mL2 /2 Inet = 0.48kg-m2 approve my ans if u like it
let ABC is the triangle formed by these rods then we have to find moment of inertia any about vertex , let vertex be A...
then , rod AB & AC will have same moment of inertia about end A ,
IAC = IAB = mL2/12 + mL2/4 = mL2/3
moment of inertia of BC = mL2/12 + md2 (d is perpendicular distance from axis of rotation)
from triangle ABC , d = Lcos30
IBC = mL2/12 + 3mL2/4 = 10mL2/12
net moment of inertia = IAB+IBC+IAC
Inet =18mL2/12 = 3mL2 /2
Inet = 0.48kg-m2
approve my ans if u like it
Thanks a Lot
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