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Ques no.14.
Plzz ans urgent
IMAGE IS ATTACHED
It is an example with solution but I can't understand the solution.plzz explain
4 months ago

Answers : (1)

Kapil Khare
79 Points
							
Let us assume that the disc lies in x-y plane
                      I+ I= Iz
We will calculate I about COM and perpendicular to plane of disc, i.e., Iz
\sigma(Mass/Area) = M/[\pi(R22 _ R12)]
So            dm = \sigma * 2\pir*dr
\implies         Iz = ∫2\pir*dr*r2]
On integrating above function from r=R1 to r=R2, we will get
                 Iz = M(R12 + R22)/2
Ix = Iy because the distribution of mass is same for all the axis passing through centre and lies in the plane of paper
So,           2Ix =  Iz = M(R12 + R22)/2
\implies          Ix = M(R12 + R22)/4
This is the moment of inertia about axis passing through COM and in the plane of paper
 
a)  For axis tangent to inner circle = [M(R12 + R22)/4] + [MR12]
​b)  For axis tangent to outer circle = [M(R12 + R22)/4] + [MR22]
4 months ago
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