A wire of mass m and length L is bent in the form of a circular ring. The moment of inertia of the ring about its axis is
Let the radius of Ring formed is R. Then
2πR= L ⇒ R = L/2π
Moment of Interia= m (L/2π)²= mL²/ 4π²
Three point masses \(m\), \(2m\) and \(3m\) are located at the vertices of an equilateral triangle of side length \(L\). The moment of inertia of the system about an axis passing through mid-point of the side (connecting \(m\) and \(2m\)) and perpendicular to the plane of the triangle, is
The distance of masses \(m\) and \(2m\) from the midpoint of their side is \(L/2\). The third mass \(3m\) lies at a distance of \(h = \frac{\sqrt{3}}{2}L\) (the height of the triangle). The total moment of inertia is \(I = m\left(\frac{L}{2}\right)^2 + 2m\left(\frac{L}{2}\right)^2 + 3m\left(\frac{\sqrt{3}}{2}L\right)^2 = \frac{mL^2}{4} + \frac{2mL^2}{4} + \frac{9mL^2}{4} = 3mL^2\).