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        Ques no.14.Plzz ans urgentIMAGE IS ATTACHEDIt is an example with solution but I can't understand the solution.plzz explain
one year ago

## Answers : (1)

Kapil Khare
80 Points
							Let us assume that the disc lies in x-y plane                      Ix + Iy = IzWe 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$\pi$r*dr$\implies$         Iz = ∫2$\pi$r*dr*r2]On integrating above function from r=R1 to r=R2, we will get                 Iz = M(R12 + R22)/2Ix = Iy because the distribution of mass is same for all the axis passing through centre and lies in the plane of paperSo,           2Ix =  Iz = M(R12 + R22)/2$\implies$          Ix = M(R12 + R22)/4This 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]

one year ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions