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 $K = \frac{3}{4}$.
Three objects, A: (a solid sphere), B: (a thin circular disk) and C: (a circular ring), each have the same mass M and radius R. They all spin with the same angular speed $\omega$ about their own symmetry axes. The amounts of work (W) required to bring them to rest, would satisfy the relation
(2018)
Work required equals rotational kinetic energy $\frac{1}{2}I\omega^2$. Comparing moments of inertia, $I_C = MR^2 > I_B = 0.5MR^2 > I_A = 0.4MR^2$, leading to $W_C > W_B > W_A$.
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 $I = \mu \ell^2 = \frac{m_1 m_2}{m_1 + m_2}\ell^2$.
A solid sphere of mass m and radius R is rotating about its diameter. A solid cylinder of the same mass and same radius is also rotating about its geometrical axis with an angular speed twice that of the sphere. The ratio of their kinetic energies of rotation ($E_{sphere} / E_{cylinder}$) will be:
(2016 – II)
Kinetic energy $E = \frac{1}{2}I\omega^2$. For sphere, $E_1 = \frac{1}{2}(\frac{2}{5}mR^2)\omega^2 = \frac{1}{5}mR^2\omega^2$. For cylinder, $E_2 = \frac{1}{2}(\frac{1}{2}mR^2)(2\omega)^2 = mR^2\omega^2$. Ratio is $\frac{1/5}{1} = 1:5$.
A thin rod of length $L$ and mass $M$ is bent at its midpoint into two halves so that the angle between them is $90^\circ$. The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is: (2008)
Treat the bent rod as two rods of length $L/2$ and mass $M/2$ connected at the origin. Using the moment of inertia formula for a rod about its end $I = \frac{m l^2}{3}$, total $I = 2 \times \frac{(M/2)(L/2)^2}{3} = \frac{ML^2}{12}$.
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 $I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.
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 $E_{rot}/E_{total} = \frac{K^2}{K^2+R^2}$.