Moment of Inertia and Radius of gyration Questions in English

(A) The moment of inertia of a circular wire (ring) about an axis passing through its center and perpendicular to its plane is $I_z = MR^2$.
By the perpendicular axis theorem,$I_z = I_x + I_y$,where $I_x$ and $I_y$ are the moments of inertia about the diameters.
Since the ring is symmetric,$I_x = I_y = I_{diameter}$.
Therefore,$I_z = 2I_{diameter}$.
Substituting the value of $I_z$,we get $MR^2 = 2I_{diameter}$.
Thus,the moment of inertia about the diameter is $I_{diameter} = \frac{1}{2}MR^2$.

(B) The moment of inertia of a single rod of mass $m$ and length $\ell$ about an axis passing through its center and perpendicular to its length is $I = \frac{m\ell^2}{12}$.
Since the two rods are joined at their centers and are perpendicular to each other,the total moment of inertia about the axis passing through the common center and perpendicular to the plane is the sum of the moments of inertia of both rods.
$I_{total} = I_1 + I_2 = \frac{m\ell^2}{12} + \frac{m\ell^2}{12}$.
$I_{total} = \frac{2m\ell^2}{12} = \frac{m\ell^2}{6}$.

(C) The moment of inertia of a uniform ring of mass $M$ and radius $r$ about an axis passing through its center and perpendicular to its plane is given by $I = Mr^2$.
Assuming the rings are made of the same material and have the same thickness,the mass $M$ is proportional to the circumference,so $M \propto R$.
Let $M_1$ be the mass of the first ring with radius $R_1 = R$,and $M_2$ be the mass of the second ring with radius $R_2 = nR$.
Then $M_1 = kR$ and $M_2 = k(nR) = nkR$,where $k$ is a constant.
The ratio of the moments of inertia is given by:
$\frac{I_1}{I_2} = \frac{M_1 R_1^2}{M_2 R_2^2} = \frac{(kR)(R^2)}{(nkR)(nR)^2} = \frac{kR^3}{n^3 k R^3} = \frac{1}{n^3}$.
Given that $\frac{I_1}{I_2} = \frac{1}{8}$,we have $\frac{1}{n^3} = \frac{1}{8}$.
Therefore,$n^3 = 8$,which implies $n = 2$.

(A) The moment of inertia of a complete uniform circular disc of mass $M_{total}$ and radius $R$ about an axis passing through its center and perpendicular to its plane is $I = \frac{1}{2} M_{total} R^2$.
Since the disc is uniform,the mass is proportional to the area. $A$ quarter of the disc has an area equal to one-fourth of the total area.
Therefore,the mass of the full disc would be $M_{total} = 4M$.
The moment of inertia of the full disc about the given axis is $I_{full} = \frac{1}{2} (4M) R^2 = 2 M R^2$.
Since the moment of inertia is an additive property for parts of a rigid body,the moment of inertia of the quarter disc $(I_{quarter})$ is one-fourth of the moment of inertia of the full disc about the same axis.
$I_{quarter} = \frac{1}{4} I_{full} = \frac{1}{4} (2 M R^2) = \frac{1}{2} M R^2$.

(B) The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about an axis passing through one of its ends and perpendicular to its length is $I = \frac{ML^2}{3}$.
For the given system:
$1$. The rod along the $x$-axis lies in the $xz$-plane. Its moment of inertia about the $z$-axis is $I_x = \frac{ML^2}{3}$.
$2$. The rod along the $y$-axis lies in the $yz$-plane. Its moment of inertia about the $z$-axis is $I_y = \frac{ML^2}{3}$.
$3$. The rod along the $z$-axis lies along the axis of rotation itself. Since the rod is thin,the distance of every mass element from the $z$-axis is zero. Therefore,its moment of inertia about the $z$-axis is $I_z = 0$.
The total moment of inertia of the system about the $z$-axis is $I_{total} = I_x + I_y + I_z = \frac{ML^2}{3} + \frac{ML^2}{3} + 0 = \frac{2ML^2}{3}$.

(B) The moment of inertia $(I)$ of a solid cylinder about its geometric axis (longitudinal axis) is given by the formula:
$I = \frac{1}{2} M R^2$
Given:
Mass $(M)$ = $20 \ kg$
Radius $(R)$ = $0.2 \ m$
Length $(L)$ = $1 \ m$ (Note: The moment of inertia about the geometric axis is independent of the length).
Substituting the values:
$I = \frac{1}{2} \times 20 \times (0.2)^2$
$I = 10 \times 0.04$
$I = 0.4 \ kg \cdot m^2$

(B) The rod is bent at the center,so each half has mass $m = M/2$ and length $l = L/2$.
The axis of rotation passes through the point of bending $O$ and is perpendicular to the plane of the bent rod.
For a rod of mass $m$ and length $l$ rotating about an axis passing through one of its ends,the moment of inertia is $I = \frac{1}{3} ml^2$.
Here,we have two such rods,each of mass $M/2$ and length $L/2$,rotating about the common end $O$.
The total moment of inertia $I_{total}$ is the sum of the moments of inertia of the two halves:
$I_{total} = I_1 + I_2 = \frac{1}{3} (M/2) (L/2)^2 + \frac{1}{3} (M/2) (L/2)^2$
$I_{total} = 2 \times \left[ \frac{1}{3} \cdot \frac{M}{2} \cdot \frac{L^2}{4} \right]$
$I_{total} = 2 \times \left[ \frac{ML^2}{24} \right] = \frac{ML^2}{12}$.

(A) The moment of inertia of a uniform circular disc of mass $M_{total}$ and radius $R$ about an axis passing through its center and perpendicular to its plane is $I = \frac{1}{2} M_{total} R^2$.
For a uniform disc,the mass is proportional to the area. $A$ quarter disc has an area equal to $1/4$ of the full disc. If the mass of the quarter disc is $M$,then the mass of the full disc would be $4M$.
The moment of inertia of the quarter disc about the same axis is the sum of the moments of inertia of its constituent particles. Since the axis of rotation passes through the center of the original disc,the distance of every mass element $dm$ of the quarter disc from the axis is the same as it would be in the full disc.
Thus,the moment of inertia $I$ of the quarter disc is given by $I = \int r^2 dm$. Since the quarter disc is a part of the full disc,its moment of inertia is simply $1/4$ of the moment of inertia of the full disc of mass $4M$.
$I = \frac{1}{4} \times (\frac{1}{2} \times (4M) \times R^2) = \frac{1}{2} M R^2$.

(B) The moment of inertia of the disc about an axis passing through its center and perpendicular to its plane is $I_{disc} = \frac{1}{2}MR^2$.
Since the four point masses $m$ are placed on the circumference,each is at a distance $R$ from the axis of rotation.
The moment of inertia of each point mass is $I_{point} = mR^2$.
Since there are four such point masses,their total moment of inertia is $I_{total\,points} = 4 \times mR^2 = 4mR^2$.
The total moment of inertia of the system is the sum of the moment of inertia of the disc and the four point masses:
$I = I_{disc} + I_{total\,points} = \frac{1}{2}MR^2 + 4mR^2$.

(A) The moment of inertia $I$ of a system of point masses about an axis is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the $i$-th mass from the axis.
For the $x$-axis,the perpendicular distance of a point $(x, y, z)$ is $r = \sqrt{y^2 + z^2}$.
$1$. For mass $m_1 = 1 \text{ kg}$ at $(0,0,0)$: $r_1 = \sqrt{0^2 + 0^2} = 0$. So,$I_1 = 1(0)^2 = 0$.
$2$. For mass $m_2 = 2 \text{ kg}$ at $(2,0,0)$: $r_2 = \sqrt{0^2 + 0^2} = 0$. So,$I_2 = 2(0)^2 = 0$.
$3$. For mass $m_3 = 3 \text{ kg}$ at $(0,3,0)$: $r_3 = \sqrt{3^2 + 0^2} = 3$. So,$I_3 = 3(3)^2 = 27$.
$4$. For mass $m_4 = 4 \text{ kg}$ at $(-2,-2,0)$: $r_4 = \sqrt{(-2)^2 + 0^2} = 2$. So,$I_4 = 4(2)^2 = 16$.
The total moment of inertia $I = I_1 + I_2 + I_3 + I_4 = 0 + 0 + 27 + 16 = 43 \text{ kg m}^2$.

(B) The rod is bent at its midpoint $A$ into two segments,each of length $L/2$ and mass $M/2$.
Each segment acts as a rod of length $l = L/2$ and mass $m = M/2$ rotating about one of its ends.
The moment of inertia of a rod of mass $m$ and length $l$ about an axis passing through one end and perpendicular to its length is given by $I = \frac{ml^2}{3}$.
Here,for each segment,$m = M/2$ and $l = L/2$.
So,the moment of inertia for one segment is $I_1 = \frac{(M/2)(L/2)^2}{3} = \frac{(M/2)(L^2/4)}{3} = \frac{ML^2}{24}$.
Since there are two such segments,the total moment of inertia about the axis passing through $A$ is $I_A = I_1 + I_1 = 2 \times \frac{ML^2}{24} = \frac{ML^2}{12}$.

(B) The moment of inertia of the system about axis $AX$ is the sum of the moments of inertia of individual particles about the same axis.
$I_{AX} = I_A + I_B + I_C$
Particle $A$ is on the axis,so its distance $r_A = 0$,hence $I_A = m(0)^2 = 0$.
Particle $B$ is at a distance $\ell$ from the axis,so $I_B = m\ell^2$.
Particle $C$ is at a perpendicular distance of $\ell/2$ from the axis $AX$,so $I_C = m(\ell/2)^2 = m\ell^2/4$.
Therefore,$I_{AX} = 0 + m\ell^2 + m\ell^2/4 = \frac{5}{4}m\ell^2$.

(A) The system consists of three rods $A$,$B$,and $C$. Rod $B$ acts as the axis of rotation.
$1$. For rod $B$: Since the axis of rotation passes through its longitudinal center,the moment of inertia of rod $B$ about this axis is $I_B = 0$ (assuming the rod is a thin line).
$2$. For rods $A$ and $C$: The axis of rotation passes through the center of rod $B$. Since rods $A$ and $C$ are perpendicular to rod $B$ and attached at its ends,the axis of rotation passes through the center of rods $A$ and $C$ perpendicular to their length.
The moment of inertia of a rod of mass $M$ and length $L$ about an axis passing through its center and perpendicular to its length is $I = \frac{M L^2}{12}$.
Since there are two such rods ($A$ and $C$),the total moment of inertia of the system is:
$I_{total} = I_A + I_B + I_C$
$I_{total} = \frac{M L^2}{12} + 0 + \frac{M L^2}{12}$
$I_{total} = \frac{2 M L^2}{12} = \frac{M L^2}{6}$

(D) Let the mass of each sphere be $m = 5 \ kg$ and the length of the rod be $L = 1 \ m$.
Case $1$: Axis passing through $A$ and perpendicular to the rod.
The moment of inertia $I_1$ is given by:
$I_1 = m_A \cdot (0)^2 + m_B \cdot (L)^2 = 0 + m \cdot (1)^2 = m$
Case $2$: Axis passing through the center of the rod and perpendicular to it.
The moment of inertia $I_2$ is given by:
$I_2 = m_A \cdot (L/2)^2 + m_B \cdot (L/2)^2 = m \cdot (0.5)^2 + m \cdot (0.5)^2 = 0.25m + 0.25m = 0.5m = m/2$
Ratio of moments of inertia:
$I_1 / I_2 = m / (m/2) = 2 / 1$
Thus,the ratio is $2 : 1$.

(A) For a complete circular disc of mass $M'$ and radius $r$,the moment of inertia about an axis passing through its center and perpendicular to its plane is $I = \frac{1}{2}M'r^2$.
Since the semi-circular disc is half of the complete disc,its mass is $M = \frac{M'}{2}$,which implies $M' = 2M$.
Substituting $M' = 2M$ into the formula for the moment of inertia:
$I = \frac{1}{2}(2M)r^2 = Mr^2$.
Therefore,the moment of inertia of the semi-circular disc is $Mr^2$.

(B) The moment of inertia of a ring about its central axis is given by $I = MR^2$.
Given the ratio of masses $M_1 : M_2 = 1 : 2$ and the ratio of diameters $D_1 : D_2 = 2 : 1$.
Since the ratio of radii $R_1 : R_2$ is the same as the ratio of diameters,we have $R_1 : R_2 = 2 : 1$.
The ratio of moments of inertia is $\frac{I_1}{I_2} = \frac{M_1 R_1^2}{M_2 R_2^2} = \left( \frac{M_1}{M_2} \right) \left( \frac{R_1}{R_2} \right)^2$.
Substituting the given values: $\frac{I_1}{I_2} = \left( \frac{1}{2} \right) \times (2)^2 = \frac{1}{2} \times 4 = 2$.
Thus,the ratio is $2 : 1$.

(B) The moment of inertia of a cylinder of mass $M$,radius $R$,and length $L$ about an axis passing through its center of mass and perpendicular to its length is given by the formula:
$I = M \left( \frac{L^2}{12} + \frac{R^2}{4} \right)$
Given that the diameter is $D$,the radius $R = \frac{D}{2}$.
Substituting $R$ into the formula:
$I = M \left[ \frac{L^2}{12} + \frac{(D/2)^2}{4} \right]$
$I = M \left[ \frac{L^2}{12} + \frac{D^2/4}{4} \right]$
$I = M \left[ \frac{L^2}{12} + \frac{D^2}{16} \right]$

(C) The moment of inertia $I$ of a ring about its central axis is given by $I = MR^2$,where $M$ is the mass and $R$ is the radius.
Given the ratio of moments of inertia $\frac{I_1}{I_2} = \frac{4}{1}$ and the ratio of diameters $\frac{D_1}{D_2} = \frac{4}{1}$.
Since the ratio of diameters is equal to the ratio of radii,$\frac{R_1}{R_2} = \frac{4}{1}$.
Using the formula $\frac{I_1}{I_2} = \frac{M_1 R_1^2}{M_2 R_2^2}$,we can write $\frac{M_1}{M_2} = \frac{I_1}{I_2} \times \left( \frac{R_2}{R_1} \right)^2$.
Substituting the given values: $\frac{M_1}{M_2} = \frac{4}{1} \times \left( \frac{1}{4} \right)^2 = 4 \times \frac{1}{16} = \frac{1}{4}$.

(B) The moment of inertia $I$ of a system of particles about an axis is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the $i$-th particle from the axis of rotation.
For the $Y$-axis,the perpendicular distance of a point $(x, y)$ is $|x|$.
The coordinates of the four masses are $(a, 0)$,$(0, a)$,$(-a, 0)$,and $(0, 2a)$.
The perpendicular distances from the $Y$-axis are $x_1 = a$,$x_2 = 0$,$x_3 = -a$,and $x_4 = 0$.
Thus,$I_y = m(a)^2 + m(0)^2 + m(-a)^2 + m(0)^2$.
$I_y = ma^2 + 0 + ma^2 + 0 = 2ma^2$.

(C) Let the equilateral triangle have vertices at $A, B,$ and $C$. Let the axis of rotation pass through the side $AB$.
The moment of inertia $I$ of a system of particles is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the $i$-th mass from the axis.
$1$. The two masses at vertices $A$ and $B$ lie on the axis of rotation,so their perpendicular distances are $r_1 = 0$ and $r_2 = 0$.
$2$. The third mass at vertex $C$ is at a perpendicular distance $h$ from the side $AB$. In an equilateral triangle of side $\ell$,the height $h$ is given by $h = \ell \sin(60^\circ) = \frac{\sqrt{3}}{2}\ell$.
$3$. The moment of inertia of the system is $I = m(0)^2 + m(0)^2 + m(h)^2 = m\left(\frac{\sqrt{3}}{2}\ell\right)^2 = m\left(\frac{3}{4}\ell^2\right) = \frac{3}{4}m\ell^2$.

(C) The moment of inertia of a solid sphere about its diameter is given by $I_A = \frac{2}{5} MR^2$.
The moment of inertia of a hollow sphere about its diameter is given by $I_B = \frac{2}{3} MR^2$.
Since the mass $M$ and radius $R$ are the same for both spheres,we compare the coefficients:
$\frac{I_A}{I_B} = \frac{\frac{2}{5} MR^2}{\frac{2}{3} MR^2} = \frac{3}{5} = 0.6$.
Since $0.6 < 1$,it follows that $I_A < I_B$.

(A) The distance of each particle of mass $m$ from the center of the square is $r = \frac{\text{diagonal}}{2} = \frac{\ell\sqrt{2}}{2} = \frac{\ell}{\sqrt{2}}$.
The moment of inertia $I$ of the system about an axis passing through the center and perpendicular to the plane is given by $I = \sum mr^2 = 4 \times m \times \left(\frac{\ell}{\sqrt{2}}\right)^2$.
$I = 4 \times m \times \frac{\ell^2}{2} = 2m\ell^2$.
The radius of gyration $K$ is defined by $I = MK^2$, where $M$ is the total mass of the system $(M = 4m)$.
$4mK^2 = 2m\ell^2$.
$K^2 = \frac{2m\ell^2}{4m} = \frac{\ell^2}{2}$.
$K = \frac{\ell}{\sqrt{2}}$.

(B) The length of the wire is $\ell = \pi r$,which implies $r = \frac{\ell}{\pi}$.
The moment of inertia of a semicircular wire about the axis passing through its diameter $(XX')$ is given by $I = Mr^2$.
Substituting the value of $r$ into the formula:
$I_{XX'} = M \left( \frac{\ell}{\pi} \right)^2 = \frac{M\ell^2}{\pi^2}$.
Wait,let's re-evaluate the moment of inertia of a semicircular wire about its diameter. For a thin ring of mass $M$ and radius $r$,the moment of inertia about a diameter is $\frac{1}{2}Mr^2$. Since a semicircle is half of a ring,the moment of inertia of the semicircular wire about the diameter $XX'$ is also $\frac{1}{2}Mr^2$.
Therefore,$I_{XX'} = \frac{1}{2} M \left( \frac{\ell}{\pi} \right)^2 = \frac{M\ell^2}{2\pi^2}$.

(A) The moment of inertia $I$ of a system of particles about an axis is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the $i$-th particle from the axis.
Here,the axis passes through the $100 \ cm$ mark.
The distance of the $1 \ kg$ mass (at $20 \ cm$) from the $100 \ cm$ mark is $r_1 = |100 \ cm - 20 \ cm| = 80 \ cm = 0.8 \ m$.
The distance of the $4 \ kg$ mass (at $70 \ cm$) from the $100 \ cm$ mark is $r_2 = |100 \ cm - 70 \ cm| = 30 \ cm = 0.3 \ m$.
Therefore,the total moment of inertia is:
$I = m_1 r_1^2 + m_2 r_2^2$
$I = 1 \ kg \times (0.8 \ m)^2 + 4 \ kg \times (0.3 \ m)^2$
$I = 1 \times 0.64 + 4 \times 0.09$
$I = 0.64 + 0.36 = 1 \ kg \ m^2$.

(A) The moment of inertia $I$ of a system of particles about an axis is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the $i$-th particle from the axis.
For the $x$-axis,the perpendicular distance $r$ of a point $(x, y, z)$ is given by $r = \sqrt{y^2 + z^2}$.
$1$. For mass $m_1 = 1 \ kg$ at $(0, 0, 0)$: $r_1 = \sqrt{0^2 + 0^2} = 0$. So,$I_1 = 1 \times 0^2 = 0 \ kg-m^2$.
$2$. For mass $m_2 = 2 \ kg$ at $(2, 0, 0)$: $r_2 = \sqrt{0^2 + 0^2} = 0$. So,$I_2 = 2 \times 0^2 = 0 \ kg-m^2$.
$3$. For mass $m_3 = 3 \ kg$ at $(0, 3, 0)$: $r_3 = \sqrt{3^2 + 0^2} = 3 \ m$. So,$I_3 = 3 \times (3)^2 = 27 \ kg-m^2$.
$4$. For mass $m_4 = 4 \ kg$ at $(-2, -2, 0)$: $r_4 = \sqrt{(-2)^2 + 0^2} = 2 \ m$. So,$I_4 = 4 \times (-2)^2 = 4 \times 4 = 16 \ kg-m^2$.
The total moment of inertia about the $x$-axis is $I = I_1 + I_2 + I_3 + I_4 = 0 + 0 + 27 + 16 = 43 \ kg-m^2$.

(C) The moment of inertia of a single rod $AB$ about an axis passing through its center $P$ and perpendicular to its length is $I_C = \frac{1}{12}ML^2$.
Using the parallel axis theorem,the moment of inertia of rod $AB$ about an axis passing through the center $O$ of the square is:
$I = I_C + Md^2 = \frac{1}{12}ML^2 + M\left(\frac{L}{2}\right)^2 = \frac{1}{12}ML^2 + \frac{1}{4}ML^2 = \frac{1+3}{12}ML^2 = \frac{4}{12}ML^2 = \frac{1}{3}ML^2$.
Since there are four such rods,the total moment of inertia $I'$ about the axis passing through the center $O$ and perpendicular to the plane is:
$I' = 4 \times I = 4 \times \left(\frac{1}{3}ML^2\right) = \frac{4}{3}ML^2$.

(A) The moment of inertia $I$ of a system of particles about an axis is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the $i$-th particle from the axis.
The axis of rotation is the altitude passing through $m_1$.
$1$. For mass $m_1$: The particle lies on the axis,so its perpendicular distance $r_1 = 0$.
$2$. For mass $m_2$: The perpendicular distance from the axis is $r_2 = a \cos(60^\circ) = \frac{a}{2}$.
$3$. For mass $m_3$: The perpendicular distance from the axis is $r_3 = a \cos(60^\circ) = \frac{a}{2}$.
Now,calculating the total moment of inertia:
$I = m_1(0)^2 + m_2 \left( \frac{a}{2} \right)^2 + m_3 \left( \frac{a}{2} \right)^2$
$I = 0 + m_2 \frac{a^2}{4} + m_3 \frac{a^2}{4}$
$I = (m_2 + m_3) \frac{a^2}{4}$

(B) The moment of inertia $I$ of a system of particles is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the $i$-th particle from the axis of rotation.
For the $x$-axis,the distance $r_x = \sqrt{y^2 + z^2}$.
$I_x = 5(0^2+0^2) + 2(0^2+0^2) + 3(3^2+0^2) + 4((-2)^2+0^2) = 0 + 0 + 27 + 16 = 43$ units.
For the $y$-axis,the distance $r_y = \sqrt{x^2 + z^2}$.
$I_y = 5(0^2+0^2) + 2(2^2+0^2) + 3(0^2+0^2) + 4((-2)^2+0^2) = 0 + 8 + 0 + 16 = 24$ units.
For the $z$-axis,the distance $r_z = \sqrt{x^2 + y^2}$.
$I_z = 5(0^2+0^2) + 2(2^2+0^2) + 3(0^2+3^2) + 4((-2)^2+(-2)^2) = 0 + 8 + 27 + 4(4+4) = 35 + 32 = 67$ units.
Thus,the moments of inertia are $43, 24, 67$ units.

(D) The rod is bent at the midpoint $O$ at an angle of $90^{\circ}$. This divides the rod into two segments,$OA$ and $OB$,each of length $l = L/2$ and mass $m = M/2$.
The axis of rotation passes through point $O$ and is perpendicular to the plane of the bent rod. For a rod of length $l$ and mass $m$ rotating about an axis passing through one of its ends,the moment of inertia is $I = \frac{1}{3} ml^2$.
For segment $OA$,the moment of inertia about the axis through $O$ is:
$I_A = \frac{1}{3} \left( \frac{M}{2} \right) \left( \frac{L}{2} \right)^2 = \frac{1}{3} \left( \frac{M}{2} \right) \left( \frac{L^2}{4} \right) = \frac{ML^2}{24}$.
Similarly,for segment $OB$,the moment of inertia about the axis through $O$ is:
$I_B = \frac{1}{3} \left( \frac{M}{2} \right) \left( \frac{L}{2} \right)^2 = \frac{ML^2}{24}$.
The total moment of inertia of the system is the sum of the moments of inertia of the two segments:
$I = I_A + I_B = \frac{ML^2}{24} + \frac{ML^2}{24} = \frac{2ML^2}{24} = \frac{ML^2}{12}$.

(C) From the figure,the perpendicular distances of the three particles from the axis of rotation $AX$ are:
$r_A = 0$
$r_B = l$
$r_C = l \sin 30^\circ = \frac{l}{2}$
Therefore,the moment of inertia $I$ of the system about the axis $AX$ is given by:
$I = m r_A^2 + m r_B^2 + m r_C^2$
$I = m(0)^2 + m(l)^2 + m\left(\frac{l}{2}\right)^2$
$I = 0 + ml^2 + \frac{ml^2}{4}$
$I = \frac{5}{4}ml^2$

(D) The mass of disc $X$ is $M_X = (\pi R^2 t) \rho$,where $\rho$ is the density of iron.
The mass of disc $Y$ is $M_Y = (\pi (4R)^2 (t/4)) \rho = (4 \pi R^2 t) \rho = 4 M_X$.
The moment of inertia of a circular disc about its central axis is $I = \frac{1}{2} M R^2$.
For disc $X$: $I_X = \frac{1}{2} M_X R^2$.
For disc $Y$: $I_Y = \frac{1}{2} M_Y (4R)^2 = \frac{1}{2} (4 M_X) (16 R^2) = 64 (\frac{1}{2} M_X R^2) = 64 I_X$.
Therefore,$I_Y = 64 I_X$.

(B) For a circular disc of mass $M$ and radius $R$,the moment of inertia about an axis in its plane and tangent to it is $I_1 = \frac{5}{4}MR^2$.
Since $I = MK^2$,the radius of gyration $K_1 = \sqrt{\frac{I_1}{M}} = \sqrt{\frac{5}{4}}R = \frac{\sqrt{5}}{2}R$.
For a circular ring of mass $M$ and radius $R$,the moment of inertia about an axis in its plane and tangent to it is $I_2 = \frac{3}{2}MR^2$.
The radius of gyration $K_2 = \sqrt{\frac{I_2}{M}} = \sqrt{\frac{3}{2}}R$.
The ratio of the radii of gyration is $\frac{K_1}{K_2} = \frac{\sqrt{5}/2}{\sqrt{3/2}} = \sqrt{\frac{5}{4} \times \frac{2}{3}} = \sqrt{\frac{5}{6}} = \frac{\sqrt{5}}{\sqrt{6}}$.

(A) Let $r_1$ and $r_2$ be the distances of masses $m_1$ and $m_2$ from the center of mass $(CM)$ respectively.
By the definition of the center of mass,$m_1 r_1 = m_2 r_2$ and $r_1 + r_2 = r$.
Solving these equations,we get $r_1 = \frac{m_2 r}{m_1 + m_2}$ and $r_2 = \frac{m_1 r}{m_1 + m_2}$.
The moment of inertia $I$ about an axis passing through the center of mass is given by $I = m_1 r_1^2 + m_2 r_2^2$.
Substituting the values of $r_1$ and $r_2$:
$I = m_1 \left( \frac{m_2 r}{m_1 + m_2} \right)^2 + m_2 \left( \frac{m_1 r}{m_1 + m_2} \right)^2$
$I = m_1 \frac{m_2^2 r^2}{(m_1 + m_2)^2} + m_2 \frac{m_1^2 r^2}{(m_1 + m_2)^2}$
$I = \frac{m_1 m_2 r^2 (m_2 + m_1)}{(m_1 + m_2)^2}$
$I = \left( \frac{m_1 m_2}{m_1 + m_2} \right) r^2$.

(A) The moment of inertia $I$ of a system of particles about an axis is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the $i$-th particle from the axis of rotation.
In this case,the axis of rotation is the $XX'$ axis (the $x$-axis).
The perpendicular distance of a point $(x, y)$ from the $x$-axis is $|y|$.
$1$. For the $4 \ kg$ mass at $(0, 3)$,the distance $r_1 = |3| = 3 \ m$.
$I_1 = 4 \times (3)^2 = 4 \times 9 = 36 \ kg \cdot m^2$.
$2$. For the $2 \ kg$ mass at $(0, -2)$,the distance $r_2 = |-2| = 2 \ m$.
$I_2 = 2 \times (2)^2 = 2 \times 4 = 8 \ kg \cdot m^2$.
$3$. For the $3 \ kg$ mass at $(0, -4)$,the distance $r_3 = |-4| = 4 \ m$.
$I_3 = 3 \times (4)^2 = 3 \times 16 = 48 \ kg \cdot m^2$.
The total moment of inertia is $I = I_1 + I_2 + I_3 = 36 + 8 + 48 = 92 \ kg \cdot m^2$.

(B) The axis of rotation $XX'$ is at $100 \ cm$ $(1 \ m)$.
The position of the $4 \ kg$ mass is at $70 \ cm$ $(0.7 \ m)$. The distance from the axis is $r_1 = 100 \ cm - 70 \ cm = 30 \ cm = 0.3 \ m$.
The position of the $1 \ kg$ mass is at $20 \ cm$ $(0.2 \ m)$. The distance from the axis is $r_2 = 100 \ cm - 20 \ cm = 80 \ cm = 0.8 \ m$.
The moment of inertia $I$ is given by $I = \sum m_i r_i^2$.
$I = (4 \ kg) \times (0.3 \ m)^2 + (1 \ kg) \times (0.8 \ m)^2$.
$I = 4 \times 0.09 + 1 \times 0.64$.
$I = 0.36 + 0.64 = 1 \ kg \cdot m^2$.

(D) Let the four spheres be placed at corners $A, B, C,$ and $D$ of a square of side $a$. We want to find the moment of inertia about the side $AD$.
For spheres at $A$ and $D$,the axis $AD$ passes through their centers. The moment of inertia of a solid sphere about its diameter is $I_{cm} = \frac{2}{5}MR^2$.
Thus,$I_A = I_D = \frac{2}{5}MR^2$.
For spheres at $B$ and $C$,the distance of their centers from the axis $AD$ is $a$. Using the parallel axis theorem,$I = I_{cm} + Md^2$,we get:
$I_B = I_C = \frac{2}{5}MR^2 + Ma^2$.
The total moment of inertia of the system about axis $AD$ is:
$I_{AD} = I_A + I_D + I_B + I_C$
$I_{AD} = \frac{2}{5}MR^2 + \frac{2}{5}MR^2 + (\frac{2}{5}MR^2 + Ma^2) + (\frac{2}{5}MR^2 + Ma^2)$
$I_{AD} = 4(\frac{2}{5}MR^2) + 2Ma^2$
$I_{AD} = \frac{8}{5}MR^2 + 2Ma^2 = 2[\frac{4}{5}MR^2 + Ma^2]$.

(A) Given: Radius $R = 10 \ cm$,Thickness $x = 5 \ mm = 0.5 \ cm$,Density $\rho = 8 \ g/cc$.
The mass $M$ of the disc is given by: $M = \text{Volume} \times \text{Density} = (\pi R^2 x) \rho$.
Substituting the values: $M = \pi \times (10)^2 \times 0.5 \times 8 = \pi \times 100 \times 4 = 400\pi \ g$.
Using $\pi \approx 3.14$,$M = 400 \times 3.14 = 1256 \ g$.
The moment of inertia $I$ of a circular disc about an axis passing through its center and perpendicular to its plane is $I = \frac{1}{2} M R^2$.
$I = \frac{1}{2} \times (400\pi) \times (10)^2 = 200\pi \times 100 = 20000\pi \ g \ cm^2$.
$I = 20000 \times 3.14 = 62800 \ g \ cm^2 = 6.28 \times 10^4 \ g \ cm^2$.

(A) The moment of inertia of a disc about its central axis is $I = \frac{1}{2}MR^2$.
Since the mass $M$ and thickness $t$ are the same for both discs,we have $M = \rho V = \rho (\pi R^2 t)$.
Since $M$ and $t$ are constant,$\rho R^2 = \text{constant}$,which implies $R^2 \propto \frac{1}{\rho}$.
Therefore,the ratio of the radii squared is $\frac{R_1^2}{R_2^2} = \frac{d_2}{d_1}$.
The ratio of the moments of inertia is $\frac{I_1}{I_2} = \frac{\frac{1}{2} M R_1^2}{\frac{1}{2} M R_2^2} = \frac{R_1^2}{R_2^2}$.
Substituting the density relationship,we get $\frac{I_1}{I_2} = \frac{d_2}{d_1} = \frac{8.9}{7.2}$.

(D) The rod is bent at its midpoint into two halves,each of length $l = L/2$ and mass $m = M/2$.
Each half acts as a thin rod of length $L/2$ and mass $M/2$ rotating about an axis passing through one of its ends.
The moment of inertia of a thin rod of mass $m$ and length $l$ about an axis passing through one end and perpendicular to its length is given by $I = \frac{1}{3}ml^2$.
Here,$m = M/2$ and $l = L/2$.
So,the moment of inertia of one half about the bending point $O$ is $I_{half} = \frac{1}{3} \times (M/2) \times (L/2)^2 = \frac{1}{3} \times \frac{M}{2} \times \frac{L^2}{4} = \frac{ML^2}{24}$.
Since the two halves are identical and rotate about the same axis passing through $O$,the total moment of inertia is $I_{total} = I_{half} + I_{half} = 2 \times \frac{ML^2}{24} = \frac{ML^2}{12}$.

(A) The moment of inertia $(I)$ of a circular disc about its central axis is $I_{disc} = \frac{1}{2}MR^2$. The radius of gyration $(k)$ is defined by $I = Mk^2$. Thus,$k_{disc} = \sqrt{\frac{I}{M}} = \sqrt{\frac{R^2}{2}} = \frac{R}{\sqrt{2}}$.
The moment of inertia $(I)$ of a circular ring about its central axis is $I_{ring} = MR^2$. Thus,$k_{ring} = \sqrt{\frac{I}{M}} = \sqrt{R^2} = R$.
The ratio of the radius of gyration of the disc to that of the ring is $\frac{k_{disc}}{k_{ring}} = \frac{R/\sqrt{2}}{R} = \frac{1}{\sqrt{2}}$.
Therefore,the ratio is $1 : \sqrt{2}$.

(C) Let the distances of masses $m_1$ and $m_2$ from the center of mass be $l_1$ and $l_2$ respectively.
Given that the total length of the rod is $l$,so $l_1 + l_2 = l$.
By the definition of the center of mass,$m_1 l_1 = m_2 l_2$.
Substituting $l_2 = l - l_1$,we get $m_1 l_1 = m_2 (l - l_1)$,which simplifies to $l_1 = \frac{m_2 l}{m_1 + m_2}$.
Similarly,$l_2 = \frac{m_1 l}{m_1 + m_2}$.
The moment of inertia $I$ about an axis passing through the center of mass is given by $I = m_1 l_1^2 + m_2 l_2^2$.
Substituting the values of $l_1$ and $l_2$:
$I = m_1 \left( \frac{m_2 l}{m_1 + m_2} \right)^2 + m_2 \left( \frac{m_1 l}{m_1 + m_2} \right)^2$
$I = \frac{m_1 m_2^2 l^2 + m_2 m_1^2 l^2}{(m_1 + m_2)^2}$
$I = \frac{m_1 m_2 l^2 (m_2 + m_1)}{(m_1 + m_2)^2}$
$I = \frac{m_1 m_2}{m_1 + m_2} l^2$.

(B) Since the mass of the disc is negligible,the moment of inertia of the system is determined solely by the five particles attached to the rim.
Each particle has a mass $m = 2 \ kg$ and is at a distance $r = 0.1 \ m$ from the axis of rotation.
The moment of inertia $I$ for a system of particles is given by $I = \sum m_i r_i^2$.
Since all $5$ particles are at the same distance $r$ from the center,$I = 5 \times m \times r^2$.
Substituting the values: $I = 5 \times 2 \times (0.1)^2 = 10 \times 0.01 = 0.1 \ kg \cdot m^2$.

(B) The moment of inertia $I$ of a ring about its central axis is given by $I = MR^2$.
Since the rings are made of the same material with the same thickness $t$,the mass $M$ can be expressed as $M = \text{Volume} \times \text{Density} = (2\pi R \cdot A) \cdot \rho$,where $A$ is the cross-sectional area and $\rho$ is the density. Given the thickness is the same,$M \propto R$.
Thus,$I = MR^2 \propto (R)R^2 = R^3$ is incorrect for a ring; for a ring,$M = \text{Area} \times \text{Circumference} \times \rho = (t^2) \times (2\pi R) \times \rho$. So $M \propto R$.
Therefore,$I = MR^2 \propto R \cdot R^2 = R^3$.
Wait,if the thickness $t$ is the cross-sectional radius,$M = \pi t^2 (2\pi R) \rho \propto R$.
Then $\frac{I_1}{I_2} = \frac{R_1^3}{R_2^3} = (\frac{0.2}{0.6})^3 = (\frac{1}{3})^3 = \frac{1}{27}$.
However,if the rings are thin hoops where $M$ is proportional to the circumference $(M \propto R)$,then $I = MR^2 \propto R^3$. If the question implies the rings have the same cross-sectional area,$M \propto R$,then $I \propto R^3$. If the question implies the rings have the same mass,$I \propto R^2$. Given the standard interpretation of such problems where $M \propto R$,the ratio is $1:27$.

(B) The moment of inertia of a ring of mass $M$ and radius $R$ about an axis passing through its center and perpendicular to its plane is $I_{cm} = MR^2$.
According to the perpendicular axis theorem,$I_z = I_x + I_y$.
Since the ring is symmetric,the moment of inertia about any diameter is the same,so $I_x = I_y = I_d$.
Thus,$I_z = 2I_d$,which implies $MR^2 = 2I_d$.
Therefore,the moment of inertia about a diameter is $I_d = \frac{1}{2}MR^2$.

(A) complete circular ring of mass $M$ and radius $R$ has a moment of inertia $I = MR^2$ about an axis passing through its center and perpendicular to its plane.
Since a semicircular ring is just half of a full ring,we can consider it as a system of particles where each particle is at a distance $R$ from the center.
The moment of inertia is defined as $I = \sum m_i r_i^2$.
For a semicircular ring,every mass element $dm$ is at a distance $R$ from the center.
Therefore,$I = \int r^2 dm = \int R^2 dm = R^2 \int dm = MR^2$.
Thus,the moment of inertia of a semicircular ring about an axis passing through its center and perpendicular to its plane is $MR^2$.

(B) The moment of inertia $I$ of a body about an axis is given by $I = \int r^2 dm$. To maximize the moment of inertia for a fixed total mass, the mass should be distributed as far as possible from the axis of rotation. Since the density of iron $(\rho \approx 7870 \ kg/m^3)$ is significantly higher than that of aluminum $(\rho \approx 2700 \ kg/m^3)$, placing the denser material (iron) at the outer rim increases the moment of inertia compared to placing it at the center. Therefore, the inner part should be aluminum and the outer part should be iron.

(D) The moment of inertia $(I)$ of a rigid body is defined as $I = \sum m_i r_i^2$,where $m_i$ is the mass of the $i$-th particle and $r_i$ is its perpendicular distance from the axis of rotation.
$1$. It depends on the mass of the body $(m)$.
$2$. It depends on the distribution of mass relative to the axis of rotation $(r_i)$.
$3$. It depends on the position and orientation of the axis of rotation.
Since all the given factors influence the moment of inertia,the correct option is $(D)$.

(A) The moment of inertia $(I)$ of a rigid body is defined as $I = \sum m_i r_i^2$,where $m_i$ is the mass of the $i$-th particle and $r_i$ is its perpendicular distance from the axis of rotation.
From this definition,it is clear that the moment of inertia depends on the mass of the object,the distribution of mass relative to the axis,and the position/orientation of the axis of rotation.
It does not depend on the state of motion of the object,such as its angular velocity $(ω)$ or angular acceleration $(α)$.
Therefore,the correct option is $A$.

(B) The moment of inertia of a thin rod of mass $M$ and length $l$ about an axis passing through its center of mass and perpendicular to its length is given by $I_{cm} = Ml^2/12$.
Using the parallel axis theorem,$I = I_{cm} + Md^2$,where $d$ is the distance between the center of mass and the axis of rotation.
For an axis passing through one end,the distance $d = l/2$.
Substituting these values,we get $I = Ml^2/12 + M(l/2)^2$.
$I = Ml^2/12 + Ml^2/4$.
$I = (Ml^2 + 3Ml^2)/12 = 4Ml^2/12 = Ml^2/3$.
Therefore,the correct option is $B$.

(D) The moment of inertia $I$ of a body is given by the formula $I = MK^2$,where $M$ is the mass and $K$ is the radius of gyration.
Given: $M = 10 \ kg$ and $I = 160 \ kg \ m^2$.
Substituting the values into the formula:
$160 = 10 \times K^2$
$K^2 = \frac{160}{10} = 16$
$K = \sqrt{16} = 4 \ m$.
Therefore,the radius of gyration is $4 \ m$.