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 \(I = m\left(\frac{L}{2}\right)^2 + 2m\left(\frac{L}{2}\right)^2 + 3m\left(\frac{\sqrt{3}}{2}L\right)^2 = \frac{mL^2}{4} + \frac{2mL^2}{4} + \frac{9mL^2}{4} = 3mL^2\).
The moment of inertia of a uniform circular disc of radius $R$ and mass $M$ about an axis touching the disc at its diameter and normal to the disc is:
(2006, 2005)
Using the parallel axis theorem, $I = I_{cm} + Md^2$. Here $I_{cm} = \frac{1}{2}MR^2$ and $d = R$, so $I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.
A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is $K$. If radius of the ball be $R$, then the fraction of total energy associated with its rotational energy will be:
(2003)
Rotational kinetic energy $E_{rot} = \frac{1}{2}MK^2\omega^2$ and total energy $E = \frac{1}{2}M(K^2+R^2)\omega^2$. The fraction is $E_{rot}/E_{total} = \frac{K^2}{K^2+R^2}$.
Moment of inertia of a uniform circular disc about a diameter is $I$. Its moment of inertia about an axis perpendicular to its plane and passing through a point on its rim will be:
(1990)
Given $I_{\text{diameter}} = \frac{MR^2}{4} = I$, which means $MR^2 = 4I$. Using the parallel axis theorem, the moment of inertia about a perpendicular axis on the rim is $I_{\text{rim}} = \frac{MR^2}{2} + MR^2 = \frac{3}{2} MR^2 = \frac{3}{2} (4I) = 6I$.
The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about an axis passing through its midpoint and perpendicular to its length is $I_{0}$. Its moment of inertia about an axis passing through one of its ends perpendicular to its length is
(2011 Mains)
Using the parallel axis theorem, $I = I_{cm} + Md^{2}$. Here, the center of mass moment of inertia is $I_{cm} = I_{0}$ and the distance to the parallel axis is $d = \frac{L}{2}$. Thus, $I = I_{0} + M(\frac{L}{2})^{2} = I_{0} + \frac{ML^{2}}{4}$.
From a circular disc of radius $R$ and mass $9M$, a small disc of mass $M$ and radius $\frac{R}{3}$ is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its center is:
(2010 Mains)
Initial moment of inertia $I_{original} = \frac{1}{2}(9M)R^{2} = \frac{9}{2}MR^{2}$. Moment of inertia of the removed part is $I_{removed} = \frac{1}{2}M(\frac{R}{3})^{2} = \frac{1}{18}MR^{2}$. The remaining moment of inertia is $I = I_{original} - I_{removed} = \frac{9}{2}MR^{2} - \frac{1}{18}MR^{2} = \frac{40}{9}MR^{2}$.