The moment of inertia of a circular ring of mass M and diameter r about a tangential axis lying in the plane of the ring is :
Solution
<p>The moment of inertia of a circular ring with mass $ M $ and diameter $ r $ about a tangential axis lying in the plane of the ring is calculated as follows:</p>
<p>Given the diameter $ r $ of the ring, the radius $ R $ is $ \frac{r}{2} $.</p>
<p>The formula for the moment of inertia about a tangential axis in the plane of the ring is:</p>
<p>$ I_{\text{tangential}} = \frac{3}{2} M \left(\frac{R}{2}\right)^2 $</p>
<p>Substitute the value of $ R = \frac{r}{2} $ into the equation:</p>
<p>$ I_{\text{tangential}} = \frac{3}{2} M \left(\frac{r}{2}\right)^2 = \frac{3}{8} Mr^2 $</p>
<p>Therefore, the moment of inertia of the ring about the specified axis is $\frac{3}{8} Mr^2$.</p>
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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