The ratio of the radius of gyration of a thin uniform disc about an axis passing through it centre and normal to its plane to the radius of gyration of the disc about its diameter is:
(2022)
Radius of gyration about center normal axis is $k_1 = R/\sqrt{2}$ and about diameter is $k_2 = R/2$. The ratio $k_1/k_2$ simplifies to $\sqrt{2}:1$.
From a circular ring of mass ‘M’ and radius ‘R’ an arc corresponding to a $90^\circ$ sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is ‘K’ times $MR^2$. Then the value of ‘K’ is:
(2021)
Since mass is uniformly distributed, removing a $90^\circ$ sector removes $1/4$ of the mass, leaving $3/4$ of the mass at distance $R$. Thus $I = \frac{3}{4}MR^2$, making $K = \frac{3}{4}$.
A light rod of length $\ell$ has two masses $m_1$ and $m_2$ attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is:
(2016 – II)
The moment of inertia about the center of mass uses reduced mass $\mu = \frac{m_1 m_2}{m_1 + m_2}$, resulting in $I = \mu \ell^2 = \frac{m_1 m_2}{m_1 + m_2}\ell^2$.
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 $I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.
The ratio of the radii of gyration of a circular disc about a tangential axis in the plane of the disc and of a circular ring of the same radius about a tangential axis in the plane of the ring is:
(2004)
For disc about in-plane tangent, $I_d = \frac{5}{4}MR^2 \implies k_d = \frac{\sqrt{5}}{2}R$. For ring about in-plane tangent, $I_r = \frac{3}{2}MR^2 \implies k_r = \sqrt{\frac{3}{2}}R$. Ratio is $\sqrt{5}:\sqrt{6}$.
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 $E_{rot}/E_{total} = \frac{K^2}{K^2+R^2}$.
A circular disc is to be made by using iron and aluminium so that it acquired maximum moment of inertia about geometrical axis. It is possible with:
(2002)
Moment of inertia is $I = \int r^2 dm$. To maximize $I$, denser material (iron) should be placed at the outer periphery and lighter material (aluminium) at the interior.
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 fly wheel rotating about fixed axis has a kinetic energy of $360 \text{ joule}$ when its angular speed is $30 \text{ rad/sec}$. The moment of inertia of the wheel about the axis of rotation is:
(1990)
The formula for rotational kinetic energy is $K = \frac{1}{2} I \omega^2$. Substituting the given values, $360 = \frac{1}{2} I (30)^2 = 450 I$. Solving for $I$ gives $I = \frac{360}{450} = 0.8 \text{ kgm}^2$.
The moment of the force, $\vec{F} = 4\hat{i} + 5\hat{j} – 6\hat{k}$ at $(2, 0, -3)$, about the point $(2, -2, -2)$ is given by
(2018)
The relative position vector is $\vec{r} = (2-2)\hat{i} + (0 - (-2))\hat{j} + (-3 - (-2))\hat{k} = 2\hat{j} - \hat{k}$. Torque is $\vec{\tau} = \vec{r} \times \vec{F} = (2\hat{j} - \hat{k}) \times (4\hat{i} + 5\hat{j} - 6\hat{k}) = -7\hat{i} - 4\hat{j} - 8\hat{k}$.