Moment of Inertia of Compound Bodies and Theorem of Moment of Inertia Questions in English

(B) The moment of inertia of a ring about its diameter is $I_{diam} = \frac{1}{2} MR^2$,where $R$ is the radius.
Given the diameter is $r$,the radius $R = \frac{r}{2}$.
Substituting $R$ in the formula,$I_{diam} = \frac{1}{2} M(\frac{r}{2})^2 = \frac{1}{8} Mr^2$.
According to the parallel axis theorem,$I_{tangent} = I_{CM} + Md^2$,where $I_{CM} = I_{diam} = \frac{1}{8} Mr^2$ and $d = R = \frac{r}{2}$.
$I_{tangent} = \frac{1}{8} Mr^2 + M(\frac{r}{2})^2 = \frac{1}{8} Mr^2 + \frac{1}{4} Mr^2 = \frac{3}{8} Mr^2$.

(A) The moment of inertia of the complete disc of mass $M$ and radius $R$ about an axis passing through its centre $O$ and perpendicular to its plane is $I_1 = \frac{1}{2} MR^2$.
The mass of the removed disc of radius $r = R/3$ is $m = \frac{M}{\pi R^2} \times \pi (R/3)^2 = \frac{M}{9}$.
The moment of inertia of the removed disc about the same axis passing through $O$ is calculated using the parallel axis theorem: $I_2 = I_{cm} + md^2$,where $I_{cm} = \frac{1}{2} m r^2$ and $d = R - r = R - R/3 = 2R/3$.
$I_2 = \frac{1}{2} \left(\frac{M}{9}\right) \left(\frac{R}{3}\right)^2 + \left(\frac{M}{9}\right) \left(\frac{2R}{3}\right)^2 = \frac{MR^2}{162} + \frac{4MR^2}{81} = \frac{MR^2 + 8MR^2}{162} = \frac{9MR^2}{162} = \frac{MR^2}{18}$.
The moment of inertia of the remaining part is $I = I_1 - I_2 = \frac{1}{2} MR^2 - \frac{1}{18} MR^2 = \frac{9-1}{18} MR^2 = \frac{8}{18} MR^2 = \frac{4}{9} MR^2$.
Comparing this with $\frac{4}{x} MR^2$,we get $x = 9$.

(C) Let the density of the sphere be $\rho$. The mass of the larger sphere of radius $2R$ is $M = \rho \cdot \frac{4}{3} \pi (2R)^3 = 8 \rho \cdot \frac{4}{3} \pi R^3$. Let $m_0 = \rho \cdot \frac{4}{3} \pi R^3$ be the mass of the smaller sphere. Then $M = 8m_0$,so $m_0 = M/8$.
The moment of inertia of the larger sphere about the $Y$-axis (passing through its center) is $I_1 = \frac{2}{5} M (2R)^2 = \frac{8}{5} MR^2$.
The smaller sphere has radius $R$ and its center is at $x = R$. Its moment of inertia about the $Y$-axis is calculated using the parallel axis theorem: $I_2 = I_{cm} + m_0 d^2 = \frac{2}{5} m_0 R^2 + m_0 R^2 = \frac{7}{5} m_0 R^2$.
Substituting $m_0 = M/8$,we get $I_2 = \frac{7}{5} (M/8) R^2 = \frac{7}{40} MR^2$.
The moment of inertia of the remaining part is $I_{rem} = I_1 - I_2 = \frac{8}{5} MR^2 - \frac{7}{40} MR^2 = \frac{64 - 7}{40} MR^2 = \frac{57}{40} MR^2$.
The ratio of the moment of inertia of the smaller sphere to that of the rest part is $\frac{I_2}{I_{rem}} = \frac{7/40 MR^2}{57/40 MR^2} = \frac{7}{57}$.

(C) The moment of inertia of a solid sphere about its diameter is given by $I = \frac{2}{5} mR^2$,which implies $mR^2 = \frac{5}{2} I$.
For the system,the total moment of inertia about axis $AB$ is $I_{AB} = I_1 + I_2$.
For the first sphere,the axis $AB$ passes through its center,so its moment of inertia is $I_1 = I$.
For the second sphere,the axis $AB$ is tangent to it. Using the parallel axis theorem,$I_2 = I_{com} + mR^2 = I + mR^2$.
Substituting $mR^2 = \frac{5}{2} I$,we get $I_2 = I + \frac{5}{2} I = \frac{7}{2} I$.
Therefore,the total moment of inertia is $I_{AB} = I + \frac{7}{2} I = \frac{9}{2} I$.

(A) The mass per unit area of the disc is $\sigma = \frac{9M}{\pi R^2}$.
The mass of the removed small disc of radius $r = \frac{R}{3}$ is $m = \sigma \cdot \pi r^2 = \frac{9M}{\pi R^2} \cdot \pi \left(\frac{R}{3}\right)^2 = M$.
The moment of inertia $(I)$ of the complete disc about the axis passing through $O$ is $I_{total} = \frac{1}{2}(9M)R^2 = \frac{9}{2}MR^2$.
The moment of inertia of the removed disc about its own center is $I_{cm} = \frac{1}{2}m r^2 = \frac{1}{2}M(\frac{R}{3})^2 = \frac{1}{18}MR^2$.
Using the parallel axis theorem,the moment of inertia of the removed disc about the axis passing through $O$ is $I_{removed} = I_{cm} + md^2$,where $d$ is the distance between the centers,$d = R - \frac{R}{3} = \frac{2R}{3}$.
$I_{removed} = \frac{1}{18}MR^2 + M(\frac{2R}{3})^2 = \frac{1}{18}MR^2 + \frac{4}{9}MR^2 = \frac{1+8}{18}MR^2 = \frac{9}{18}MR^2 = \frac{1}{2}MR^2$.
The moment of inertia of the remaining disc is $I_{remaining} = I_{total} - I_{removed} = \frac{9}{2}MR^2 - \frac{1}{2}MR^2 = 4MR^2$.

(B) The moment of inertia $(I)$ of a thin circular disc about a tangent in the plane of the disc is given by the parallel axis theorem: $I = I_{cm} + MR^2 = \frac{1}{4}MR^2 + MR^2 = \frac{5}{4}MR^2$.
From this,we get $MR^2 = \frac{4}{5}I$.
The moment of inertia of the disc about an axis perpendicular to the plane and passing through its centre is $I_z = \frac{1}{2}MR^2$.
Substituting the value of $MR^2$,we get $I_z = \frac{1}{2} \times (\frac{4}{5}I) = \frac{2}{5}I$.

(B) Let the two spheres be $S_1$ and $S_2$. The axis of rotation passes through the centre of $S_1$ and is perpendicular to the rod.
For sphere $S_1$,the moment of inertia about its own centre is $I_1 = \frac{2}{5} MR^2$.
For sphere $S_2$,the distance of its centre from the axis of rotation is $d = R + 4R + R = 6R$ is incorrect based on the diagram. The rod length is $4R$ between the surfaces,or the distance between centres is $4R$. Given the diagram shows the rod length $4R$ between the spheres,the distance between the centres is $d = R + 4R + R = 6R$. However,if the rod length $4R$ is the distance between centres,then $d = 4R$.
Assuming the distance between centres is $4R$:
For $S_1$,$I_1 = \frac{2}{5} MR^2$.
For $S_2$,using the parallel axis theorem,$I_2 = I_{cm} + Md^2 = \frac{2}{5} MR^2 + M(4R)^2 = \frac{2}{5} MR^2 + 16MR^2 = \frac{82}{5} MR^2$.
Total moment of inertia $I = I_1 + I_2 = \frac{2}{5} MR^2 + \frac{82}{5} MR^2 = \frac{84}{5} MR^2$.

(C) The moment of inertia of a solid sphere of mass $m$ and radius $r$ about its diameter is $I_{cm} = \frac{2}{5} mr^2$.
For sphere $1$,the axis $YY^1$ passes through its center (diameter). Thus,$I_1 = \frac{2}{5} mr^2$.
For spheres $2$ and $3$,the axis $YY^1$ is tangent to them. The distance of the center of these spheres from the axis $YY^1$ is $r$. Using the parallel axis theorem,$I = I_{cm} + md^2$,where $d = r$.
So,$I_2 = I_3 = \frac{2}{5} mr^2 + mr^2 = \frac{7}{5} mr^2$.
The total moment of inertia of the system is $I_{total} = I_1 + I_2 + I_3 = \frac{2}{5} mr^2 + \frac{7}{5} mr^2 + \frac{7}{5} mr^2 = \frac{16}{5} mr^2$.

(C) The moment of inertia of a solid sphere about its diametric axis is given by $I = \frac{2}{5} mR^2$.
According to the parallel axis theorem,the moment of inertia about a tangent is $I_{tangent} = I_{cm} + mR^2$.
Substituting the value of $I_{cm} = I = \frac{2}{5} mR^2$,we get:
$I_{tangent} = \frac{2}{5} mR^2 + mR^2 = \frac{7}{5} mR^2$.
Since $I = \frac{2}{5} mR^2$,we can write $mR^2 = \frac{5}{2} I$.
Substituting this into the expression for $I_{tangent}$:
$I_{tangent} = \frac{7}{5} \times (\frac{5}{2} I) = \frac{7}{2} I = 3.5 I$.

(D) The moment of inertia of a thin uniform rod about an axis passing through its center of mass and perpendicular to its length is $I_{cm} = \frac{ML^2}{12}$.
According to the parallel axis theorem,$I = I_{cm} + Md^2$,where $d$ is the distance between the center of mass and the given axis.
The center of mass of the rod is at a distance $L/2$ from one end.
The given axis is at a distance $L/4$ from the same end.
Therefore,the distance $d$ between the center of mass and the axis is $d = |L/2 - L/4| = L/4$.
Substituting these values into the parallel axis theorem:
$I = \frac{ML^2}{12} + M(L/4)^2$
$I = \frac{ML^2}{12} + \frac{ML^2}{16}$
Taking the least common multiple of $12$ and $16$,which is $48$:
$I = \frac{4ML^2 + 3ML^2}{48} = \frac{7ML^2}{48}$.

(C) For a disc of mass $M$ and radius $R$,the moment of inertia about its diameter $(I_D)$ is given by:
$I_D = \frac{MR^2}{4}$
For a ring of mass $M$ and radius $R$,the moment of inertia about a tangent in its plane $(I_T)$ is calculated using the parallel axis theorem:
$I_T = I_{CM} + MR^2$
Since the moment of inertia of a ring about its diameter is $I_{CM} = \frac{MR^2}{2}$,we have:
$I_T = \frac{MR^2}{2} + MR^2 = \frac{3}{2} MR^2$
The ratio of the moment of inertia of the disc about its diameter to that of the ring about a tangent in its plane is:
$\frac{I_D}{I_T} = \frac{\frac{MR^2}{4}}{\frac{3}{2} MR^2} = \frac{1}{4} \times \frac{2}{3} = \frac{1}{6}$
Thus,the ratio is $1: 6$.

(D) Let $I$ be the moment of inertia of the square plate about an axis passing through the centre '$O$' and perpendicular to the plane of the plate.
According to the perpendicular axis theorem,for any two mutually perpendicular axes in the plane of the plate intersecting at the centre,the sum of the moments of inertia about these axes equals the moment of inertia about the axis perpendicular to the plane.
For axes $1$ and $2$ (diagonals),$I = I_1 + I_2$.
For axes $3$ and $4$ (mid-lines),$I = I_3 + I_4$.
Since the plate is a square,by symmetry,$I_1 = I_2$ and $I_3 = I_4$.
Thus,$I = 2I_1$ and $I = 2I_3$,which implies $I_1 = I_3$.
Therefore,the moment of inertia about the perpendicular axis is $I = I_1 + I_3$ (or $I_2 + I_4$,or $I_1 + I_4$,etc.).
Comparing with the given options,the correct expression is $I_1 + I_3$.

(B) According to the parallel axis theorem,the moment of inertia $I$ about any axis is given by $I = I_{CM} + Md^2$,where $I_{CM}$ is the moment of inertia about the center of mass,$M$ is the mass of the disc,and $d$ is the perpendicular distance of the axis from the center of mass.
Since $I_{CM}$ and $M$ are constant for the disc,the moment of inertia $I$ is directly proportional to the square of the distance $d$ from the center of mass $(A)$.
Comparing the distances of points $A, B, C,$ and $D$ from the center of mass $A$:
- For point $A$,$d = 0$.
- For point $B$,$d = R$ (radius of the disc).
- For point $C$,$d < R$.
- For point $D$,$d < R$.
Since point $B$ is at the maximum distance $(d = R)$ from the center of mass,the moment of inertia is maximum about the axis passing through point $B$.

(A) The moment of inertia of a disc about an axis passing through its centre of mass and perpendicular to its plane is $I = \frac{MR^2}{2}$.
Using the parallel axis theorem,$I_{axis} = I_{cm} + Mh^2$.
For an axis tangential to the rim,the distance from the centre is $h = R$. Therefore,the moment of inertia $I_1$ is:
$I_1 = \frac{MR^2}{2} + MR^2 = \frac{3}{2} MR^2$.
For an axis passing through a point midway between the centre and the rim,the distance from the centre is $h = R/2$. Therefore,the moment of inertia $I_2$ is:
$I_2 = \frac{MR^2}{2} + M(R/2)^2 = \frac{MR^2}{2} + \frac{MR^2}{4} = \frac{3}{4} MR^2$.
The ratio of the two moments of inertia is:
$\frac{I_1}{I_2} = \frac{\frac{3}{2} MR^2}{\frac{3}{4} MR^2} = \frac{3}{2} \times \frac{4}{3} = \frac{2}{1}$.
Thus,the ratio is $2:1$.

(D) To find the moment of inertia of the square frame about an axis passing through its center $O$ and perpendicular to its plane,we use the parallel axis theorem for each rod.
For a single rod of mass $M$ and length $L$,the moment of inertia about an axis passing through its center of mass and perpendicular to its length is $I_{cm} = \frac{ML^2}{12}$.
The distance from the center of the square to the center of each rod is $d = \frac{L}{2}$.
Using the parallel axis theorem,the moment of inertia of one rod about the axis passing through the center of the square is $I_{rod} = I_{cm} + Md^2 = \frac{ML^2}{12} + M(\frac{L}{2})^2 = \frac{ML^2}{12} + \frac{ML^2}{4} = \frac{ML^2 + 3ML^2}{12} = \frac{4ML^2}{12} = \frac{ML^2}{3}$.
Since there are four such rods,the total moment of inertia of the system is $I_{total} = 4 \times I_{rod} = 4 \times \frac{ML^2}{3} = \frac{4ML^2}{3}$.

(B) The moment of inertia of a square lamina of mass $M$ and side $b$ about an axis perpendicular to its plane and passing through its centre is given by $I_{\text{lamina}} = \frac{M(b^2 + b^2)}{12} = \frac{Mb^2}{6}$.
The moment of inertia of a disc of mass $M$ and radius $R$ about an axis perpendicular to its plane and passing through its centre is given by $I_{\text{disc}} = \frac{MR^2}{2}$.
According to the problem,the moments of inertia are equal:
$\frac{Mb^2}{6} = \frac{MR^2}{2}$.
Canceling $M$ from both sides,we get:
$\frac{b^2}{6} = \frac{R^2}{2}$.
Rearranging the terms to find the ratio $\frac{b^2}{R^2}$:
$\frac{b^2}{R^2} = \frac{6}{2} = 3$.
Taking the square root of both sides:
$\frac{b}{R} = \sqrt{3} = \frac{\sqrt{3}}{1}$.

(D) The moment of inertia $(I)$ of a single circular disc about an axis passing through its center and perpendicular to its plane is $I_{cm} = \frac{MR^2}{2}$.
For the system of seven discs,the total moment of inertia about the axis passing through the center of the central disc (point $O$) and normal to the plane is the sum of the moment of inertia of the central disc and the six surrounding discs.
$1$. For the central disc: The axis passes through its center,so its moment of inertia is $I_1 = \frac{MR^2}{2}$.
$2$. For the six surrounding discs: The distance between the center of the central disc and the center of any surrounding disc is $d = 2R$. Using the parallel axis theorem,the moment of inertia of each surrounding disc about the central axis is $I_2 = I_{cm} + Md^2 = \frac{MR^2}{2} + M(2R)^2 = \frac{MR^2}{2} + 4MR^2 = \frac{9MR^2}{2}$.
Since there are six such surrounding discs,their total moment of inertia is $6 \times I_2 = 6 \times \frac{9MR^2}{2} = 27MR^2$.
Total moment of inertia $I = I_1 + 6I_2 = \frac{MR^2}{2} + 27MR^2 = \frac{MR^2 + 54MR^2}{2} = \frac{55MR^2}{2}$.

(B) Let $I_0$ be the moment of inertia about the axis passing through the center $O$ and perpendicular to the plane,given by $I_0 = \frac{Ma^2}{6}$.
The distance $h$ from the center $O$ to any corner $A$ is half the length of the diagonal of the square.
The diagonal of the square is $\sqrt{a^2 + a^2} = \sqrt{2}a$.
Therefore,the distance $h = \frac{\sqrt{2}a}{2} = \frac{a}{\sqrt{2}}$.
Using the parallel axis theorem,the moment of inertia $I_A$ about an axis passing through the corner $A$ and parallel to the axis through the center is:
$I_A = I_0 + Mh^2$
$I_A = \frac{Ma^2}{6} + M\left(\frac{a}{\sqrt{2}}\right)^2$
$I_A = \frac{Ma^2}{6} + \frac{Ma^2}{2}$
$I_A = \frac{Ma^2 + 3Ma^2}{6} = \frac{4Ma^2}{6} = \frac{2}{3}Ma^2$.

(D) Let the axis pass through the centre of sphere $1$. The distance between the centres of the two spheres is $2R$.
Using the parallel axis theorem for sphere $1$,the moment of inertia about its own centre is $I_{c1} = \frac{2}{5} M (\frac{R}{2})^2 = \frac{1}{10} MR^2$.
For sphere $2$,the axis is at a distance $d = 2R$ from its centre. Using the parallel axis theorem,$I_2 = I_{c2} + Md^2 = \frac{2}{5} M (\frac{R}{2})^2 + M(2R)^2 = \frac{1}{10} MR^2 + 4MR^2 = \frac{41}{10} MR^2$.
The rod is massless,so its moment of inertia is $0$.
The total moment of inertia $I = I_{c1} + I_2 = \frac{1}{10} MR^2 + \frac{41}{10} MR^2 = \frac{42}{10} MR^2 = \frac{21}{5} MR^2$.

(A) The moment of inertia of the remaining part of the disc $(I_r)$ is given by the principle of superposition:
$I_r = I_{\text{disc}} - I_{\text{hole}}$
where $I_{\text{disc}}$ is the moment of inertia of the original disc about the central axis,and $I_{\text{hole}}$ is the moment of inertia of the removed circular part about the same axis.
$1$. Moment of inertia of the original disc:
$I_{\text{disc}} = \frac{1}{2} MR^2$
$2$. Properties of the removed hole:
Radius of the hole $r = \frac{R}{2}$.
Since the surface mass density $\sigma = \frac{M}{\pi R^2}$ is uniform,the mass of the hole $M_h$ is:
$M_h = \sigma \cdot \pi r^2 = \left(\frac{M}{\pi R^2}\right) \cdot \pi \left(\frac{R}{2}\right)^2 = \frac{M}{4}$.
$3$. Moment of inertia of the hole about the central axis of the original disc:
Using the parallel axis theorem,$I_{\text{hole}} = I_{\text{cm}} + M_h d^2$,where $I_{\text{cm}} = \frac{1}{2} M_h r^2$ and $d = \frac{R}{2}$ is the distance between the center of the hole and the center of the original disc.
$I_{\text{hole}} = \frac{1}{2} \left(\frac{M}{4}\right) \left(\frac{R}{2}\right)^2 + \left(\frac{M}{4}\right) \left(\frac{R}{2}\right)^2$
$I_{\text{hole}} = \frac{MR^2}{32} + \frac{MR^2}{16} = \frac{MR^2 + 2MR^2}{32} = \frac{3MR^2}{32}$.
$4$. Moment of inertia of the remaining part:
$I_r = \frac{1}{2} MR^2 - \frac{3MR^2}{32} = \frac{16MR^2 - 3MR^2}{32} = \frac{13MR^2}{32}$.

(D) Using the parallel axis theorem,$I_2 = I_{CM} + Mh^2$.
For a circular disc,the moment of inertia about an axis passing through its centre and perpendicular to its plane is $I_1 = I_{CM} = \frac{1}{2}MR^2$.
The distance between the parallel axes is $h = \frac{2R}{3}$.
Substituting these values into the parallel axis theorem:
$I_2 = \frac{1}{2}MR^2 + M\left(\frac{2R}{3}\right)^2$
$I_2 = \frac{1}{2}MR^2 + M\left(\frac{4R^2}{9}\right)$
$I_2 = MR^2 \left(\frac{1}{2} + \frac{4}{9}\right) = MR^2 \left(\frac{9 + 8}{18}\right) = \frac{17}{18}MR^2$.
Now,find the ratio $\frac{I_2}{I_1}$:
$\frac{I_2}{I_1} = \frac{\frac{17}{18}MR^2}{\frac{1}{2}MR^2} = \frac{17}{18} \times 2 = \frac{17}{9}$.
Comparing this with the given ratio $\frac{x}{9}$,we get $x = 17$.

(A) The total mass of the wire is $M = \lambda L$. Since the wire is bent into a circular coil of radius $R$,the circumference is $2 \pi R = L$,so $R = \frac{L}{2 \pi}$.
The moment of inertia of a circular ring about its diameter is $I_{diam} = \frac{1}{2} M R^2$.
Using the parallel axis theorem,the moment of inertia about a tangential axis in the plane of the coil is $I = I_{diam} + M R^2$.
$I = \frac{1}{2} M R^2 + M R^2 = \frac{3}{2} M R^2$.
Substituting $M = \lambda L$ and $R = \frac{L}{2 \pi}$:
$I = \frac{3}{2} (\lambda L) \left( \frac{L}{2 \pi} \right)^2 = \frac{3}{2} \lambda L \left( \frac{L^2}{4 \pi^2} \right) = \frac{3 \lambda L^3}{8 \pi^2}$.

(B) The moment of inertia of a single solid sphere about its own central axis is $I_{cm} = \frac{2}{5} MR^2$. Using the parallel axis theorem,$I = I_{cm} + Md^2$,where $d$ is the distance from the axis of rotation.
For the axis passing through the center of sphere $A$:
$I_A = I_A' + I_B' + I_C' + I_D'$
$I_A = \frac{2}{5} MR^2 + (\frac{2}{5} MR^2 + M(2R)^2) + (\frac{2}{5} MR^2 + M(4R)^2) + (\frac{2}{5} MR^2 + M(6R)^2)$
$I_A = \frac{8}{5} MR^2 + M(4R^2 + 16R^2 + 36R^2) = 1.6 MR^2 + 56 MR^2 = 57.6 MR^2$.
For the axis passing through the center of sphere $B$:
$I_B = I_A' + I_B' + I_C' + I_D'$
$I_B = (\frac{2}{5} MR^2 + M(2R)^2) + \frac{2}{5} MR^2 + (\frac{2}{5} MR^2 + M(2R)^2) + (\frac{2}{5} MR^2 + M(4R)^2)$
$I_B = \frac{8}{5} MR^2 + M(4R^2 + 4R^2 + 16R^2) = 1.6 MR^2 + 24 MR^2 = 25.6 MR^2$.
The difference is $|I_A - I_B| = 57.6 MR^2 - 25.6 MR^2 = 32 MR^2$.

(A) The moment of inertia of a ring about its central axis perpendicular to its plane is $I_{cm} = MR^2 = 4 \,kg \,m^2$.
By the perpendicular axis theorem, the moment of inertia about a diameter in its plane is $I_d = \frac{1}{2} MR^2 = \frac{1}{2} \times 4 = 2 \,kg \,m^2$.
Using the parallel axis theorem, the moment of inertia about a tangent in the plane is $I_t = I_d + MR^2$.
Substituting the values: $I_t = 2 \,kg \,m^2 + 4 \,kg \,m^2 = 6 \,kg \,m^2$.

(D) The moment of inertia of a solid sphere about an axis passing through its centre of mass is $I_{cm} = \frac{2}{5} MR^2$.
According to the parallel axis theorem,the moment of inertia $I$ about an axis at a distance $d$ from the centre is given by $I = I_{cm} + Md^2$.
Here,$d = \frac{R}{2}$.
Substituting the values,we get $I = \frac{2}{5} MR^2 + M\left(\frac{R}{2}\right)^2$.
$I = \frac{2}{5} MR^2 + M\left(\frac{R^2}{4}\right)$.
$I = \left(\frac{2}{5} + \frac{1}{4}\right) MR^2$.
$I = \left(\frac{8 + 5}{20}\right) MR^2 = \frac{13}{20} MR^2$.

(D) The moment of inertia of a disc of mass $m$ and radius $r$ about an axis passing through its center and perpendicular to its plane is $I_{cm} = \frac{1}{2} m r^2$.
For disc $A$,the axis passes through its center,so its moment of inertia is $I_A = \frac{1}{2} m r^2$.
For disc $B$,the axis is at a distance $d = 2r$ from its center. Using the parallel axis theorem,$I_B = I_{cm} + m d^2 = \frac{1}{2} m r^2 + m(2r)^2 = \frac{1}{2} m r^2 + 4 m r^2 = \frac{9}{2} m r^2$.
The total moment of inertia of the system is $I_{total} = I_A + I_B = \frac{1}{2} m r^2 + \frac{9}{2} m r^2 = \frac{10}{2} m r^2 = 5 m r^2$.

(D) Concept: Application of the parallel axis theorem.
The moment of inertia of a solid sphere about an axis passing through its centre is:
$I_{\text{cm}} = \frac{2}{5} MR^2$
According to the parallel axis theorem,the moment of inertia $I$ about an axis parallel to the one passing through the centre of mass at a distance $d$ is given by:
$I = I_{\text{cm}} + Md^2$
Here,the distance $d = \frac{R}{2}$.
Substituting the values:
$I = \frac{2}{5} MR^2 + M\left(\frac{R}{2}\right)^2$
$I = \frac{2}{5} MR^2 + \frac{1}{4} MR^2$
$I = \left(\frac{8 + 5}{20}\right) MR^2$
$I = \frac{13}{20} MR^2$

(A) The moment of inertia of a disc about an axis passing through its center of mass and perpendicular to its plane is given by $I_{cm} = \frac{1}{2} MR^2$.
Given $M = 1 \,kg$ and $R = 2 \,m$,$I_{cm} = \frac{1}{2} \times 1 \times (2)^2 = 2 \,kg \cdot m^2$.
According to the parallel axis theorem,the moment of inertia about an axis parallel to the $XY$ axis and passing through the edge of the disc is $I' = I_{cm} + Md^2$,where $d = R$ is the distance between the two axes.
Substituting the values,$I' = 2 + (1 \times 2^2) = 2 + 4 = 6 \,kg \cdot m^2$.

(B) The axis $YY'$ passes through the center of the top ring and is a diameter of that ring. The moment of inertia of the top ring about its diameter is $I_1 = \frac{1}{2} MR^2$.
For the two lower rings,the axis $YY'$ is a tangent to each ring in its own plane. According to the parallel axis theorem,the moment of inertia of a ring about a tangent in its plane is $I_{tangent} = I_{cm} + MR^2 = \frac{1}{2} MR^2 + MR^2 = \frac{3}{2} MR^2$.
Since there are two such lower rings,their combined moment of inertia is $I_2 + I_3 = \frac{3}{2} MR^2 + \frac{3}{2} MR^2 = 3 MR^2$.
The total moment of inertia of the system about the axis $YY'$ is $I = I_1 + I_2 + I_3 = \frac{1}{2} MR^2 + \frac{3}{2} MR^2 + \frac{3}{2} MR^2 = \frac{7}{2} MR^2$.

(A) The moment of inertia of the upper sphere about the axis $YY'$ passing through its center is $I_1 = \frac{2}{5} MR^2$.
For each of the two lower spheres,the axis $YY'$ is at a distance $R$ from their centers. Using the parallel axis theorem,the moment of inertia for each lower sphere is $I_2 = I_{cm} + MR^2 = \frac{2}{5} MR^2 + MR^2 = \frac{7}{5} MR^2$.
Therefore,the total moment of inertia of the system is $I = I_1 + 2 \times I_2 = \frac{2}{5} MR^2 + 2 \times \frac{7}{5} MR^2 = \frac{2}{5} MR^2 + \frac{14}{5} MR^2 = \frac{16}{5} MR^2$.

(B) The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about an axis passing through its center of mass and perpendicular to its length is $I_{cm} = ML^2/12$.
Using the parallel axis theorem,$I = I_{cm} + Md^2$,where $d$ is the distance between the center of mass and the new axis.
The center of mass is at $L/2$ from one end. The new axis is at $L/4$ from the same end.
Therefore,the distance $d = |L/2 - L/4| = L/4$.
Substituting these values into the parallel axis theorem:
$I = ML^2/12 + M(L/4)^2$
$I = ML^2/12 + ML^2/16$
Taking the least common multiple of $12$ and $16$,which is $48$:
$I = (4ML^2 + 3ML^2) / 48 = 7ML^2/48$.

(A) The mass of the original disc is $M_{total} = 9M$ and its radius is $R$. The moment of inertia of the complete disc about an axis passing through its centre and perpendicular to its plane is $I_{1} = \frac{1}{2} M_{total} R^{2} = \frac{1}{2} (9M) R^{2} = \frac{9 MR^{2}}{2}$.
Since the mass is proportional to the area,the mass of the removed disc $m$ is given by $m = M_{total} \times \frac{\pi (R/3)^{2}}{\pi R^{2}} = 9M \times \frac{1}{9} = M$.
The moment of inertia of the removed disc about the same axis is calculated using the parallel axis theorem: $I^{\prime} = I_{cm} + m d^{2}$,where $I_{cm} = \frac{1}{2} m (R/3)^{2}$ and $d = \frac{2R}{3}$.
$I^{\prime} = \frac{1}{2} M (R/3)^{2} + M (2R/3)^{2} = \frac{MR^{2}}{18} + \frac{4MR^{2}}{9} = \frac{MR^{2} + 8MR^{2}}{18} = \frac{9MR^{2}}{18} = \frac{MR^{2}}{2}$.
The moment of inertia of the remaining disc is $I_{remaining} = I_{1} - I^{\prime} = \frac{9 MR^{2}}{2} - \frac{MR^{2}}{2} = \frac{8 MR^{2}}{2} = 4 MR^{2}$.

(C) To find the moment of inertia about an axis passing through a point at a distance $\frac{L}{3}$ from one end,we use the parallel axis theorem: $I = I_{cm} + Mh^2$.
Here,$I_{cm}$ is the moment of inertia about the center of mass,which is $\frac{ML^2}{12}$.
The distance $h$ between the center of mass (at $\frac{L}{2}$ from the end) and the given axis (at $\frac{L}{3}$ from the end) is $h = |\frac{L}{2} - \frac{L}{3}| = \frac{L}{6}$.
Substituting these values into the theorem:
$I = \frac{ML^2}{12} + M(\frac{L}{6})^2$
$I = \frac{ML^2}{12} + \frac{ML^2}{36}$
$I = \frac{3ML^2 + ML^2}{36} = \frac{4ML^2}{36} = \frac{ML^2}{9}$.

(D) Given,the moment of inertia about the center $O$ is $I_{o} = \frac{Ma^{2}}{6}$.
To find the moment of inertia about an axis passing through a corner (say $A$),we use the parallel axis theorem: $I_{A} = I_{o} + Mh^{2}$,where $h$ is the distance between the center and the corner.
The diagonal of the square is $d = \sqrt{a^{2} + a^{2}} = a\sqrt{2}$.
The distance $h$ from the center to the corner is half the diagonal: $h = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.
Substituting the values into the theorem:
$I_{A} = \frac{Ma^{2}}{6} + M\left(\frac{a}{\sqrt{2}}\right)^{2}$
$I_{A} = \frac{Ma^{2}}{6} + \frac{Ma^{2}}{2}$
$I_{A} = \frac{Ma^{2} + 3Ma^{2}}{6} = \frac{4Ma^{2}}{6} = \frac{2Ma^{2}}{3}$.

(A) The total mass of the wire is $m = Q \cdot L$.
Since the wire is bent into a circular coil of radius $R$,the circumference is $2 \pi R = L$,which gives $R = L / (2 \pi)$.
The moment of inertia of a circular ring about its diameter is $I_{diam} = (1/2) m R^2$.
Using the parallel axis theorem,the moment of inertia about the tangent axis $XX'$ is $I_{XX'} = I_{cm} + m R^2$,where $I_{cm}$ is the moment of inertia about the diameter passing through the center.
$I_{XX'} = (1/2) m R^2 + m R^2 = (3/2) m R^2$.
Substituting $m = Q L$ and $R = L / (2 \pi)$ into the expression:
$I_{XX'} = (3/2) \cdot (Q L) \cdot (L / (2 \pi))^2$
$I_{XX'} = (3/2) \cdot Q L \cdot (L^2 / (4 \pi^2))$
$I_{XX'} = 3 Q L^3 / (8 \pi^2)$.

(B) The moment of inertia of a thin uniform rod of length $L$ and mass $M$ about an axis passing through its center of mass $(CM)$ and perpendicular to the rod is $I_{CM} = \frac{M L^2}{12}$.
The distance of the given axis from the center of mass is $x = \frac{L}{2} - \frac{L}{3} = \frac{L}{6}$.
Using the parallel axis theorem,$I = I_{CM} + M x^2$.
Substituting the values,we get $I = \frac{M L^2}{12} + M \left( \frac{L}{6} \right)^2$.
$I = \frac{M L^2}{12} + \frac{M L^2}{36}$.
Taking the least common multiple,$I = \frac{3 M L^2 + M L^2}{36} = \frac{4 M L^2}{36} = \frac{M L^2}{9}$.

(A) The moment of inertia of a disc of mass $M$ and radius $R$ about its diameter is given by $I = \frac{1}{4} M R^2$.
From this,we get $M R^2 = 4 I$.
To find the moment of inertia about an axis perpendicular to the plane and passing through the rim,we use the parallel axis theorem.
First,the moment of inertia about an axis perpendicular to the plane and passing through the center (center of mass) is $I_{cm} = \frac{1}{2} M R^2$.
By the parallel axis theorem,$I_{rim} = I_{cm} + M R^2$.
Substituting $I_{cm} = \frac{1}{2} M R^2$,we get $I_{rim} = \frac{1}{2} M R^2 + M R^2 = \frac{3}{2} M R^2$.
Substituting $M R^2 = 4 I$ into the expression,we get $I_{rim} = \frac{3}{2} (4 I) = 6 I$.

(A) Let $\sigma$ be the surface mass density of the disc. The mass of the original disc of radius $R$ is $M_1 = \sigma \pi R^2$. The mass of the removed portion of radius $r$ is $M_2 = \sigma \pi r^2$. The mass of the annular disc is $M = M_1 - M_2 = \sigma \pi (R^2 - r^2)$,so $\sigma = \frac{M}{\pi(R^2 - r^2)}$.
The moment of inertia of the original disc is $I_1 = \frac{1}{2} M_1 R^2 = \frac{1}{2} (\sigma \pi R^2) R^2 = \frac{1}{2} \sigma \pi R^4$.
The moment of inertia of the removed portion is $I_2 = \frac{1}{2} M_2 r^2 = \frac{1}{2} (\sigma \pi r^2) r^2 = \frac{1}{2} \sigma \pi r^4$.
The moment of inertia of the annular disc is $I = I_1 - I_2 = \frac{1}{2} \sigma \pi (R^4 - r^4)$.
Substituting $\sigma = \frac{M}{\pi(R^2 - r^2)}$,we get $I = \frac{1}{2} \left( \frac{M}{\pi(R^2 - r^2)} \right) \pi (R^4 - r^4) = \frac{1}{2} M \frac{(R^2 - r^2)(R^2 + r^2)}{(R^2 - r^2)} = \frac{1}{2} M(R^2 + r^2)$.

(D) To find the moment of inertia about an axis passing through a point midway between the center and the end,we use the parallel axis theorem: $I = I_{CM} + M d^2$.
Here,$I_{CM} = \frac{M L^2}{12}$ is the moment of inertia about the center of mass.
The distance $d$ between the center of mass and the new axis is $d = \frac{L}{4}$.
Substituting these values into the theorem:
$I = \frac{M L^2}{12} + M \left( \frac{L}{4} \right)^2$
$I = \frac{M L^2}{12} + \frac{M L^2}{16}$
Taking the least common multiple $(LCM)$ of $12$ and $16$,which is $48$:
$I = \frac{4 M L^2 + 3 M L^2}{48} = \frac{7 M L^2}{48}$.

(C) The moment of inertia of a uniform circular disc about an axis passing through its center of mass and perpendicular to its plane is given by $I_{CM} = \frac{1}{2} M R^{2}$.
According to the parallel axis theorem,the moment of inertia $I$ about an axis parallel to the central axis at a distance $d = R$ is given by $I = I_{CM} + M d^{2}$.
Substituting $d = R$ into the formula,we get:
$I = \frac{1}{2} M R^{2} + M R^{2} = \frac{3}{2} M R^{2}$.

(A) The system consists of one central disc and six outer discs. We need to find the moment of inertia of the six outer discs about an axis passing through the centre of the central disc and perpendicular to the plane.
For each outer disc,the distance of its centre from the central axis is $d = 2R$.
The moment of inertia of a single disc about its own central axis (normal to the plane) is $I_{cm} = \frac{1}{2} m R^2$.
Using the parallel axis theorem,the moment of inertia of one outer disc about the central axis is:
$I_{outer} = I_{cm} + m d^2 = \frac{1}{2} m R^2 + m(2R)^2 = \frac{1}{2} m R^2 + 4 m R^2 = \frac{9}{2} m R^2$.
Since there are six such outer discs,the total moment of inertia is:
$I_{total} = 6 \times I_{outer} = 6 \times \frac{9}{2} m R^2 = 27 m R^2$.

(B) According to the perpendicular axis theorem, for a planar body (lamina), the moment of inertia about an axis perpendicular to the plane $(I_Z)$ is equal to the sum of the moments of inertia about two mutually perpendicular axes ($I_X$ and $I_Y$) lying in the plane of the body and intersecting at the same point.
Given:
$I_X = 20 \text{ kg m}^2$
$I_Y = 25 \text{ kg m}^2$
Using the theorem: $I_Z = I_X + I_Y$
$I_Z = 20 \text{ kg m}^2 + 25 \text{ kg m}^2 = 45 \text{ kg m}^2$
Therefore, the moment of inertia about the axis perpendicular to the plane is $45 \text{ kg m}^2$.

(D) According to the parallel axis theorem,the moment of inertia $I$ about an axis parallel to the axis passing through the center of mass is given by:
$I = I_{cm} + Md^{2}$
Here,$I_{cm} = 2 \,kg \,m^{2}$ is the moment of inertia about the axis passing through the center of mass.
$M = 1 \,kg$ is the mass of the disc.
$d = R = 2 \,m$ is the distance between the two parallel axes (which is equal to the radius of the disc).
Substituting these values into the formula:
$I = 2 + (1)(2)^{2}$
$I = 2 + (1)(4)$
$I = 2 + 4 = 6 \,kg \,m^{2}$
Therefore,the moment of inertia about the axis passing through the edge of the disc is $6 \,kg \,m^{2}$.

(D) The moment of inertia of the system is the sum of the moments of inertia of the two discs about the given axis.
Let $I_B$ be the moment of inertia of disc $B$ about an axis perpendicular to its plane and passing through its centre. $I_B = \frac{1}{2} Mr^2$.
Let $I_A$ be the moment of inertia of disc $A$ about the same axis. Since the axis is parallel to the axis passing through the centre of disc $A$ at a distance $d = 2r$,we use the parallel axis theorem: $I_A = I_{cm} + Md^2$.
Here,$I_{cm} = \frac{1}{2} Mr^2$ and $d = 2r$.
So,$I_A = \frac{1}{2} Mr^2 + M(2r)^2 = \frac{1}{2} Mr^2 + 4Mr^2 = 4.5 Mr^2$.
The total moment of inertia $I = I_B + I_A = 0.5 Mr^2 + 4.5 Mr^2 = 5 Mr^2$.

(C) Let $M$ be the mass,$R$ be the radius,and $L$ be the length of the uniform solid cylinder.
The moment of inertia of the cylinder about its central axis is $I_1 = \frac{1}{2}MR^2$.
The moment of inertia of the cylinder about an axis passing through its center and perpendicular to its length is $I_2 = \frac{MR^2}{4} + \frac{ML^2}{12}$.
According to the problem,$I_1 = \frac{1}{n} I_2$,which implies $n I_1 = I_2$.
Substituting the expressions: $n \left( \frac{1}{2} MR^2 \right) = \frac{MR^2}{4} + \frac{ML^2}{12}$.
Dividing by $M$: $\frac{nR^2}{2} = \frac{R^2}{4} + \frac{L^2}{12}$.
Multiply by $12$: $6nR^2 = 3R^2 + L^2$.
Rearranging for $L^2$: $L^2 = 6nR^2 - 3R^2 = 3R^2(2n - 1)$.
Therefore,$\frac{L^2}{R^2} = 3(2n - 1)$,which gives $\frac{L}{R} = \sqrt{3(2n - 1)}$.

(A) Let $M$ be the mass and $R$ be the radius of the thin circular ring.
The moment of inertia of the ring about an axis passing through its center and perpendicular to its plane is $I_{cm} = MR^2$.
According to the parallel axis theorem,the moment of inertia about an axis passing through its edge and perpendicular to its plane is $I = I_{cm} + MR^2 = MR^2 + MR^2 = 2MR^2$.
Given that this value is $I$,we have $2MR^2 = I$,which implies $MR^2 = \frac{I}{2}$.
The moment of inertia of the ring about its diameter is $I_{diameter} = \frac{1}{2}MR^2$.
Substituting the value of $MR^2$,we get $I_{diameter} = \frac{1}{2} \times \frac{I}{2} = \frac{I}{4}$.

(C) The moment of inertia of a solid sphere about an axis passing through its center is $I_{cm, sphere} = \frac{2}{5} MR^2$. Using the parallel axis theorem,the moment of inertia about the point of contact is $I_1 = I_{cm, sphere} + MR^2 = \frac{2}{5} MR^2 + MR^2 = \frac{7}{5} MR^2$.
The moment of inertia of a disc about an axis passing through its center and perpendicular to its plane is $I_{cm, disc} = \frac{1}{2} MR^2$. Using the parallel axis theorem,the moment of inertia about the point of contact is $I_2 = I_{cm, disc} + MR^2 = \frac{1}{2} MR^2 + MR^2 = \frac{3}{2} MR^2$.
The total moment of inertia of the system about the axis passing through the point of contact is $I = I_1 + I_2 = \left(\frac{7}{5} + \frac{3}{2}\right) MR^2 = \left(\frac{14 + 15}{10}\right) MR^2 = \frac{29 MR^2}{10}$.