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

(A) Given: Mass $M = 1 \, kg$, Length $L = 5 \, m$.
The moment of inertia of a thin rod about an axis passing through its center of mass and perpendicular to its length is $I_{CM} = \frac{ML^2}{12}$.
$I_{CM} = \frac{1 \times 5^2}{12} = \frac{25}{12} \approx 2.083 \, kg \cdot m^2$.
The axis of rotation is at a distance $x = 1 \, m$ from one end. The center of mass is at the midpoint, which is $L/2 = 2.5 \, m$ from either end.
The distance $d$ between the center of mass and the new axis is $d = |2.5 \, m - 1 \, m| = 1.5 \, m$.
Using the parallel axis theorem, $I = I_{CM} + Md^2$.
$I = 2.083 + (1 \times 1.5^2) = 2.083 + 2.25 = 4.333 \, kg \cdot m^2$.
Thus, the moment of inertia is $4.33 \, kg \cdot m^2$.

(C) Given, mass of solid cylinder, $M = 10 \,kg$.
Radius, $R = 40 \,cm = 0.4 \,m$.
Length, $L = 9R = 9 \times 0.4 = 3.6 \,m$.
The moment of inertia of a solid cylinder about an axis passing through its centre of mass and perpendicular to its axis is given by:
$I_{COM} = M \left( \frac{L^2}{12} + \frac{R^2}{4} \right)$
Substituting the values:
$I_{COM} = 10 \left( \frac{(3.6)^2}{12} + \frac{(0.4)^2}{4} \right)$
$I_{COM} = 10 \left( \frac{12.96}{12} + \frac{0.16}{4} \right)$
$I_{COM} = 10 (1.08 + 0.04) = 10 \times 1.12 = 11.2 \,kg-m^2$.
According to the parallel axis theorem, the moment of inertia $I'$ about an axis passing through its edge and perpendicular to its axis is:
$I' = I_{COM} + M \left( \frac{L}{2} \right)^2$
$I' = 11.2 + 10 \left( \frac{3.6}{2} \right)^2$
$I' = 11.2 + 10 (1.8)^2$
$I' = 11.2 + 10 (3.24) = 11.2 + 32.4 = 43.6 \,kg-m^2$.

(B) The moment of inertia of a solid cylinder of mass $M$, length $L$, and radius $R$ about an axis passing through its centre of mass and perpendicular to its longitudinal axis is given by $I_{CM} = M(\frac{L^2}{12} + \frac{R^2}{4})$.
Given $L = 2R$, we substitute this into the formula:
$I_1 = M(\frac{(2R)^2}{12} + \frac{R^2}{4}) = M(\frac{4R^2}{12} + \frac{R^2}{4}) = M(\frac{R^2}{3} + \frac{R^2}{4}) = M(\frac{7R^2}{12})$.
The moment of inertia about an axis passing through one end and perpendicular to the longitudinal axis is given by the parallel axis theorem: $I = I_{CM} + M d^2$, where $d = \frac{L}{2} = R$.
$I_2 = I_1 + M R^2 = M(\frac{7R^2}{12}) + M R^2 = M(\frac{7R^2 + 12R^2}{12}) = M(\frac{19R^2}{12})$.
Thus, $I_2 - I_1 = M R^2$.

(D) The length of the wire is $l$. Let $R$ be the radius of the circular loop. Then, $2 \pi R = l$, which gives $R = \frac{l}{2 \pi}$.
The mass of the wire is $m = \rho l$.
The moment of inertia of a circular loop about its diameter is $I_{diam} = \frac{1}{2} m R^2$.
According to the parallel axis theorem, the moment of inertia about a tangent $AB$ is $I = I_{cm} + m R^2$, where $I_{cm} = I_{diam} = \frac{1}{2} m R^2$.
Therefore, $I = \frac{1}{2} m R^2 + m R^2 = \frac{3}{2} m R^2$.
Substituting $m = \rho l$ and $R = \frac{l}{2 \pi}$:
$I = \frac{3}{2} (\rho l) \left( \frac{l}{2 \pi} \right)^2 = \frac{3}{2} \rho l \left( \frac{l^2}{4 \pi^2} \right) = \frac{3 \rho l^3}{8 \pi^2}$.

(D) $1$. The moment of inertia of a disc about its diameter $(I_d)$ is given by $I_d = \frac{1}{4} M R^2$.
$2$. According to the parallel axis theorem, the moment of inertia about any axis parallel to the diameter is $I = I_{cm} + M h^2$, where $h$ is the distance between the axes.
$3$. For a tangent parallel to the diameter, the distance $h = R$.
$4$. Therefore, $I = \frac{1}{4} M R^2 + M R^2 = \frac{5}{4} M R^2$.

(D) The moment of inertia of a circular disc about an axis passing through its centre and perpendicular to its plane is given by $I = \frac{1}{2} M R^2$.
From this, we have $M R^2 = 2 I$.
The moment of inertia of the disc about its diameter is $I_d = \frac{1}{4} M R^2$.
Using the parallel axis theorem, the moment of inertia about an axis parallel to the diameter and touching the edge of the rim is $I' = I_d + M R^2$.
Substituting the values, $I' = \frac{1}{4} M R^2 + M R^2 = \frac{5}{4} M R^2$.
Since $M R^2 = 2 I$, we get $I' = \frac{5}{4} (2 I) = \frac{5}{2} I$.

(B) Let the mass of the square lamina be $m$. The moment of inertia of the square lamina about an axis passing through its center of mass and parallel to the $X$-axis is $I_{cm,x} = \frac{ma^2}{12}$.
Using the parallel axis theorem, the moment of inertia about an axis parallel to the $X$-axis passing through $(0, 2a)$ is:
$I_1 = I_{cm,x} + m(2a)^2 = \frac{ma^2}{12} + 4ma^2 = \frac{49ma^2}{12}$.
The moment of inertia of the square lamina about an axis passing through its center of mass and perpendicular to the $xy$-plane is $I_{cm,z} = I_{cm,x} + I_{cm,y} = \frac{ma^2}{12} + \frac{ma^2}{12} = \frac{ma^2}{6}$.
Using the parallel axis theorem, the moment of inertia about an axis passing through $(d, 0)$ and perpendicular to the $xy$-plane is:
$I_2 = I_{cm,z} + md^2 = \frac{ma^2}{6} + md^2$.
Equating $I_1$ and $I_2$:
$\frac{49ma^2}{12} = \frac{ma^2}{6} + md^2$.
$\frac{49a^2}{12} - \frac{2a^2}{12} = d^2$.
$d^2 = \frac{47a^2}{12}$.
$d = \sqrt{\frac{47}{12}} a$.

(D) The moment of inertia of a solid cylinder about an axis passing through its center of mass and perpendicular to its longitudinal axis is given by:
$I_{XY} = M \left( \frac{l^2}{12} + \frac{R^2}{4} \right)$
Using the parallel axis theorem, the moment of inertia $I_{AB}$ about an axis passing through its edge and perpendicular to the longitudinal axis is:
$I_{AB} = I_{XY} + M \left( \frac{l}{2} \right)^2$
Substituting the expression for $I_{XY}$:
$I_{AB} = M \left( \frac{l^2}{12} + \frac{R^2}{4} \right) + M \frac{l^2}{4}$
$I_{AB} = M \left( \frac{l^2}{3} + \frac{R^2}{4} \right)$
Given that the length $l = 6R$, we substitute this into the equation:
$I_{AB} = M \left[ \frac{(6R)^2}{3} + \frac{R^2}{4} \right]$
$I_{AB} = M \left[ \frac{36R^2}{3} + \frac{R^2}{4} \right]$
$I_{AB} = M \left[ 12R^2 + \frac{R^2}{4} \right]$
$I_{AB} = M \left( \frac{48R^2 + R^2}{4} \right) = \frac{49 M R^2}{4}$

(B) Let $\sigma$ be the surface mass density of the disc. The mass of the original disc is $M = \sigma \pi r^2$.
Each cut-out disc has a radius $r' = r/4$. Its mass $m$ is given by $m = \sigma \pi (r/4)^2 = \sigma \pi r^2 / 16 = M/16$.
The moment of inertia of the original disc about axis $A$ is $I_0 = \frac{1}{2} Mr^2$.
The moment of inertia of one cut-out disc about its own central axis is $I_{cm} = \frac{1}{2} m (r/4)^2 = \frac{1}{2} m (r^2/16) = \frac{mr^2}{32}$.
Using the parallel axis theorem, the moment of inertia of one cut-out disc about axis $A$ is $I_{cut} = I_{cm} + md^2$, where $d = 3r/4$.
$I_{cut} = \frac{mr^2}{32} + m(3r/4)^2 = \frac{mr^2}{32} + \frac{9mr^2}{16} = \frac{mr^2 + 18mr^2}{32} = \frac{19mr^2}{32}$.
The moment of inertia of the remaining part is $I_{rem} = I_0 - 2 \times I_{cut}$.
$I_{rem} = \frac{1}{2} Mr^2 - 2 \times \left( \frac{19mr^2}{32} \right) = \frac{1}{2} Mr^2 - \frac{19mr^2}{16}$.
Substituting $m = M/16$, we get $I_{rem} = \frac{1}{2} Mr^2 - \frac{19(M/16)r^2}{16} = \frac{1}{2} Mr^2 - \frac{19}{256} Mr^2$.
$I_{rem} = \frac{128}{256} Mr^2 - \frac{19}{256} Mr^2 = \frac{109}{256} Mr^2$.
Comparing this with $\frac{x}{256} Mr^2$, we find $x = 109$.

(D) Let the mass of each cylinder be $M' = M/4$.
Consider the two cylinders perpendicular to the axis (labeled $I_1$ in the diagram). The axis passes through their centers perpendicular to their length. The moment of inertia of one such cylinder is $I_1 = \frac{M'R^2}{4} + \frac{M'L^2}{12}$.
Consider the two cylinders parallel to the axis (labeled $I_2$ in the diagram). The axis is at a distance $L/2$ from their centers. The moment of inertia of one such cylinder about its own center (axis along length) is $\frac{M'R^2}{2}$. By the parallel axis theorem, $I_2 = \frac{M'R^2}{2} + M'(L/2)^2 = \frac{M'R^2}{2} + \frac{M'L^2}{4}$.
The total moment of inertia is $I_{net} = 2I_1 + 2I_2 = 2(\frac{M'R^2}{4} + \frac{M'L^2}{12}) + 2(\frac{M'R^2}{2} + \frac{M'L^2}{4})$.
$I_{net} = \frac{M'R^2}{2} + \frac{M'L^2}{6} + M'R^2 + \frac{M'L^2}{2} = \frac{3}{2}M'R^2 + \frac{2}{3}M'L^2$.
Substituting $M' = M/4$: $I_{net} = \frac{3}{2}(M/4)R^2 + \frac{2}{3}(M/4)L^2 = \frac{3}{8}MR^2 + \frac{1}{6}ML^2$.

(D) The moment of inertia of a solid sphere about its tangent is given by the parallel axis theorem: $I = I_{cm} + MR^2 = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$.
For the system of two spheres, the total moment of inertia about the common tangent passing through the point of contact is the sum of the moments of inertia of each sphere about that same axis.
$I_{total} = I_1 + I_2 = \frac{7}{5}m_1R_1^2 + \frac{7}{5}m_2R_2^2 = \frac{7}{5}[m_1R_1^2 + m_2R_2^2]$.
Given: $m_1 = 5 \ kg$, $R_1 = 0.1 \ m$; $m_2 = 10 \ kg$, $R_2 = 0.2 \ m$.
$I = \frac{7}{5} [5 \times (0.1)^2 + 10 \times (0.2)^2]$.
$I = \frac{7}{5} [5 \times 0.01 + 10 \times 0.04] = \frac{7}{5} [0.05 + 0.4] = \frac{7}{5} [0.45]$.
$I = 7 \times 0.09 = 0.63 \ kg \cdot m^{2}$.

(15) $1$. Moment of inertia of the rod about axis $AB$ (passing through one end and perpendicular to the length):
$I_{\text{rod}} = \frac{1}{3}ML^2 = \frac{1}{3} \times 40 \times (3)^2 = \frac{1}{3} \times 40 \times 9 = 120 \text{ kg m}^2$.
$2$. Moment of inertia of the solid sphere about an axis parallel to $AB$ at a distance $d = 3 \text{ m}$ from its center:
Using the parallel axis theorem, $I_{\text{sphere}} = I_{\text{cm}} + md^2 = \frac{2}{5}mR^2 + md^2$.
Given $m = 10 \text{ kg}$ and $d = 3 \text{ m}$:
$I_{\text{sphere}} = \frac{2}{5} \times 10 \times R^2 + 10 \times (3)^2 = 4R^2 + 90$.
$3$. Equating the two moments of inertia:
$120 = 4R^2 + 90$
$4R^2 = 30$
$R^2 = \frac{30}{4} = 7.5$.
$4$. Given $R = \sqrt{\frac{\alpha}{2}}$, so $R^2 = \frac{\alpha}{2}$.
Equating the values of $R^2$:
$\frac{\alpha}{2} = 7.5$
$\alpha = 15$.

(C) Total mass $M = m \times L$.
The radius $R$ is found from $L = 2\pi R$, so $R = \frac{L}{2\pi}$.
The moment of inertia of a ring about its diameter is $I_{diam} = \frac{1}{2}MR^2$.
Using the parallel axis theorem, the moment of inertia about the tangent $yy'$ is $I = I_{cm} + MR^2$, where $I_{cm} = I_{diam} = \frac{1}{2}MR^2$.
Thus, $I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.
Substituting $M = mL$ and $R = \frac{L}{2\pi}$:
$I = \frac{3}{2} (mL) \left(\frac{L}{2\pi}\right)^2 = \frac{3}{2} mL \left(\frac{L^2}{4\pi^2}\right) = \frac{3mL^3}{8\pi^2}$.
Therefore, the correct option is $C$.

(B) $1$. The moment of inertia of a disc about its diameter is $I_d = MR^2/4$.
$2$. The moment of inertia of a disc about an axis passing through its center and normal to the disc (perpendicular axis theorem) is $I_{cm} = I_d + I_d = MR^2/4 + MR^2/4 = MR^2/2$.
$3$. To find the moment of inertia about an axis passing through a point on its edge and normal to the disc, we use the parallel axis theorem: $I = I_{cm} + Md^2$, where $d = R$ is the distance between the center and the edge.
$4$. Substituting the values: $I = MR^2/2 + M(R)^2 = MR^2/2 + MR^2 = (3/2)MR^2$.
$5$. Comparing this with the given expression $(x/2)MR^2$, we get $x/2 = 3/2$, which implies $x = 3$.

(C) 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 parallel to the one passing through the centre of mass at a distance $d$ is given by $I = I_{cm} + Md^2$.
Here, the distance $d = R/3$.
Substituting the values, we get:
$I = \frac{2}{5}MR^2 + M(R/3)^2$
$I = \frac{2}{5}MR^2 + M(R^2/9)$
$I = MR^2(\frac{2}{5} + \frac{1}{9})$
$I = MR^2(\frac{18 + 5}{45})$
$I = \frac{23}{45}MR^2$.

(B) $1$. Moment of inertia of a disc of mass $M_d = M/2$ and radius $R$ about a tangent in its plane:
Using the parallel axis theorem, $I_{tangent} = I_{cm} + M_d R^2$.
For a disc, $I_{cm} = (1/4) M_d R^2$.
So, $I_{disc} = (1/4) (M/2) R^2 + (M/2) R^2 = (1/8) M R^2 + (4/8) M R^2 = (5/8) M R^2$.
$2$. Moment of inertia of a solid sphere of mass $M$ and radius $R$ about its tangent:
Using the parallel axis theorem, $I_{tangent} = I_{cm} + M R^2$.
For a solid sphere, $I_{cm} = (2/5) M R^2$.
So, $I_{sphere} = (2/5) M R^2 + M R^2 = (7/5) M R^2$.
$3$. Ratio of the moment of inertia of the disc to the sphere:
Ratio $= I_{disc} / I_{sphere} = [(5/8) M R^2] / [(7/5) M R^2] = (5/8) * (5/7) = 25/56$.
Thus, the ratio is $25 : 56$.

(C) The moment of inertia of a solid sphere about its diameter is $I_{cm} = \frac{2}{5} mr^2$.
In the given system, the axis of rotation $AA'$ passes through the centers of all five spheres.
Since the axis of rotation passes through the center of each sphere, the distance $d$ between the axis of rotation and the center of each sphere is $0$.
Using the parallel axis theorem, $I = I_{cm} + md^2$, for each sphere, the moment of inertia about the axis $AA'$ is simply $I_{cm} = \frac{2}{5} mr^2$.
Since there are five such spheres, the total moment of inertia of the system is $I_{total} = 5 \times I_{cm} = 5 \times (\frac{2}{5} mr^2) = 2 mr^2$.
Wait, re-evaluating the image: The central column has $3$ spheres, and there are $2$ spheres on the sides. All $5$ spheres have their centers on the axis $AA'$.
Therefore, the total moment of inertia is $5 \times \frac{2}{5} mr^2 = 2 mr^2$. None of the options match $2 mr^2$. Let's re-examine the geometry. If the side spheres are at a distance $d=2r$ from the axis, $I = \frac{2}{5}mr^2 + m(2r)^2 = 4.4mr^2$. If the side spheres are touching the central axis, their centers are at distance $r$ from the axis. Then $I_{side} = \frac{2}{5}mr^2 + mr^2 = 1.4mr^2$. Total $I = 3 \times (0.4mr^2) + 2 \times (1.4mr^2) = 1.2mr^2 + 2.8mr^2 = 4mr^2$. This matches option $(C)$.

(C) The moment of inertia of a solid sphere about its diameter is given by $I_{cm} = 2/5 MR^2$.
According to the parallel axis theorem, the moment of inertia $I$ about an axis parallel to the diameter at a distance $d = R/3$ from the centre is given by $I = I_{cm} + Md^2$.
Substituting the values, we get $I = 2/5 MR^2 + M(R/3)^2$.
$I = 2/5 MR^2 + M(R^2/9)$.
Taking the common denominator as $45$, we get $I = (18/45) MR^2 + (5/45) MR^2$.
$I = 23/45 MR^2$.

(A) The moment of inertia of a hollow sphere about an axis passing through its center is $I_{cm} = \frac{2}{3}MR^2$.
By the parallel axis theorem, the moment of inertia $I$ about an axis at a distance $h$ from the center is given by $I = I_{cm} + Mh^2$.
Substituting the value of $I_{cm}$, we get $I = \frac{2}{3}MR^2 + Mh^2$.
The radius of gyration $K$ is defined by the relation $I = MK^2$.
Given that $K = R$, we have $I = MR^2$.
Equating the two expressions for $I$:
$MR^2 = \frac{2}{3}MR^2 + Mh^2$
$R^2 = \frac{2}{3}R^2 + h^2$
$h^2 = R^2 - \frac{2}{3}R^2$
$h^2 = \frac{1}{3}R^2$
$h = \frac{R}{\sqrt{3}}$
Thus, the correct option is $A$.

(A) The system consists of one central ring and six surrounding rings.
$1$. For the central ring, the moment of inertia about the axis passing through its centre and normal to its plane is $I_{central} = MR^2$.
$2$. For each of the six surrounding rings, the distance between the centre of the central ring and the centre of the surrounding ring is $d = 2R$.
$3$. Using the parallel axis theorem, the moment of inertia of one surrounding ring about the axis passing through the centre of the central ring is $I_{surrounding} = I_{CM} + Md^2 = MR^2 + M(2R)^2 = MR^2 + 4MR^2 = 5MR^2$.
$4$. Since there are six such surrounding rings, their total moment of inertia is $6 \times 5MR^2 = 30MR^2$.
$5$. The total moment of inertia of the system is $I_{total} = I_{central} + 30MR^2 = MR^2 + 30MR^2 = 31MR^2$.

(C) The moment of inertia of a solid sphere of mass $M_s$ and radius $R$ about its diameter is $I_{cm} = \frac{2}{5} M_s R^2$.
Using the parallel axis theorem, the moment of inertia about its tangent is $I_{sphere} = I_{cm} + M_s R^2 = \frac{2}{5} M_s R^2 + M_s R^2 = \frac{7}{5} M_s R^2$.
Given $M_s = 5 \text{ kg}$, $I_{sphere} = \frac{7}{5} \times 5 \times R^2 = 7 R^2$.
The moment of inertia of a disc of mass $M_d$ and radius $R$ about its diameter is $I_{cm} = \frac{1}{4} M_d R^2$.
Using the parallel axis theorem, the moment of inertia about a tangent in its plane is $I_{disc} = I_{cm} + M_d R^2 = \frac{1}{4} M_d R^2 + M_d R^2 = \frac{5}{4} M_d R^2$.
Given $M_d = 4 \text{ kg}$, $I_{disc} = \frac{5}{4} \times 4 \times R^2 = 5 R^2$.
The ratio is $\frac{I_{sphere}}{I_{disc}} = \frac{7 R^2}{5 R^2} = \frac{7}{5}$.
Thus, $x = 7$ and $y = 5$.