The ratio of the radii of gyration of a circular disc to that of a circular ring, each of same mass and radius, around their respective axes is:
(2008)
Radius of gyration $k = \sqrt{I/M}$. For a disc, $I = \frac{1}{2}MR^2 \implies k_d = R/\sqrt{2}$. For a ring, $I = MR^2 \implies k_r = R$. Ratio $k_d/k_r = 1/\sqrt{2}$.
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 $I_{\text{rim}} = \frac{MR^2}{2} + MR^2 = \frac{3}{2} MR^2 = \frac{3}{2} (4I) = 6I$.
A solid cylinder of mass $2 \text{ kg}$ and radius $4 \text{ cm}$ is rotating about its axis at the rate of $3 \text{ rpm}$. The torque required to stop after $2\pi$ revolutions is
(2019)
Here $I = \frac{1}{2}MR^2 = 1.6 \times 10^{-3} \text{ kg m}^2$, $\omega_0 = 3 \times \frac{2\pi}{60} = \frac{\pi}{10} \text{ rad/s}$, and $\theta = 4\pi^2 \text{ rad}$. Using $\omega^2 = \omega_0^2 + 2\alpha\theta$, $\alpha = -\frac{1}{800} \text{ rad/s}^2$. The required torque magnitude is $\tau = I\alpha = 2 \times 10^{-6} \text{ N m}$.
A rope is wound around a hollow cylinder of mass $3 \text{ kg}$ and radius $40 \text{ cm}$. What is the angular acceleration of the cylinder if the rope is pulled with a force of $30 \text{ N}$?
(2017-Delhi)
For a hollow cylinder, the moment of inertia is $I = MR^2$. The torque provided by the rope is $\tau = F \times R = I\alpha$. Substituting $I$, we get $F \times R = MR^2 \alpha$, which gives $\alpha = \frac{F}{MR} = \frac{30}{3 \times 0.4} = 25 \text{ rad/s}^2$.
Two rotating bodies $A$ and $B$ of masses $m$ and $2m$ with moments of inertia $I_A$ and $I_B$ ($I_B > I_A$) have equal kinetic energy of rotation. If $L_A$ and $L_B$ be their angular momenta respectively, then:
(2016-II)
Rotational kinetic energy is related to angular momentum by the formula $K = \frac{L^2}{2I}$, which yields $L = \sqrt{2KI}$. Since $K$ is the same for both bodies and it is given that $I_B > I_A$, it directly follows that $L_B > L_A$.
A force $\vec{F} = \alpha\hat{i} + 3\hat{j} + 9\hat{k}$ is acting at a point $\vec{r} = 2\hat{i} – 6\hat{j} – 12\hat{k}$. The value of $\alpha$ for which angular momentum about origin is conserved is: (2015 Re)
For angular momentum to be conserved, torque $\vec{\tau} = \vec{r} \times \vec{F}$ must be zero, meaning $\vec{r}$ and $\vec{F}$ are collinear. Taking the ratio of their components: $\frac{2}{\alpha} = \frac{-6}{3} = \frac{-12}{9}$, which simplifies to $\frac{2}{\alpha} = -2$, giving $\alpha = -1$.
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 $I = I_{original} - I_{removed} = \frac{9}{2}MR^{2} - \frac{1}{18}MR^{2} = \frac{40}{9}MR^{2}$.