integration - Moment of inertia of the ring through the diameter - Mathematics Stack Exchange
The moment of inertia of a circular ring with mass M and radius R about an axis passing through its centre and perpendicular to its plane is:A. $\\dfrac{MR^2}{4}$B. $MR^2$C. $\\dfrac{MR^2}{2}$D. $\\dfrac{3}{4}MR^2$
SOLVED:Determine the moment of inertia of the thin ring about the z axis. The ring has a mass m.
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Find the out the moment of inertia of a ring having uniform mass distribution of mass M and radius R about an axis which is tangent ot the ring and a in
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Moment of inertia of a ring of radius R whose mass per unit length varies with parametric angle θ according to the relation λ=λ°cos²θ, about its axis will be
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The moment of inertia of a ring of mass M and radius R about an axis, passing through the center and perpendicular to the plane of the ring is:
Moment of inertia
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Formula: Thin circular ring (moment of inertia)
