Moment of Inertia and Radius of gyration Questions in English

(A) $1$. The mass of the ring is $m$ and its linear density is $\rho$. The circumference of the ring is $L = \frac{m}{\rho}$.
$2$. Since $L = 2 \pi R$,the radius of the ring is $R = \frac{m}{2 \pi \rho}$.
$3$. The moment of inertia of a ring about its diameter is $I_{diam} = \frac{1}{2} m R^2$.
$4$. According to the parallel axis theorem,the moment of inertia about a tangent parallel to the diameter is $I = I_{diam} + m R^2 = \frac{1}{2} m R^2 + m R^2 = \frac{3}{2} m R^2$.
$5$. Substituting $R = \frac{m}{2 \pi \rho}$ into the expression: $I = \frac{3}{2} m \left( \frac{m}{2 \pi \rho} \right)^2 = \frac{3}{2} m \left( \frac{m^2}{4 \pi^2 \rho^2} \right) = \frac{3 m^3}{8 \pi^2 \rho^2}$.

(A) The moment of inertia $I$ of a thin uniform rod of mass $M$ and length $L$ about an axis perpendicular to its length and passing through one end is given by $I = \frac{1}{3}ML^2$.
By definition,the radius of gyration $K$ is related to the moment of inertia by $I = MK^2$.
Equating the two expressions for $I$,we get $MK^2 = \frac{1}{3}ML^2$.
Canceling $M$ from both sides,we have $K^2 = \frac{L^2}{3}$.
Taking the square root of both sides,we get $K = \frac{L}{\sqrt{3}}$.
Therefore,the ratio $K: L = \frac{1}{\sqrt{3}}$ or $1: \sqrt{3}$.

(C) The moment of inertia of a circular ring of mass $m$ and radius $R$ about its diameter is given by $I = \frac{1}{2} mR^2$.
Given that the linear mass density is $\rho = \frac{m}{L}$,where $L$ is the circumference of the loop $(L = 2 \pi R)$.
Therefore,$\rho = \frac{m}{2 \pi R}$,which implies $R = \frac{m}{2 \pi \rho}$.
Substituting the value of $R$ into the formula for the moment of inertia:
$I = \frac{1}{2} m \left( \frac{m}{2 \pi \rho} \right)^2$
$I = \frac{1}{2} m \left( \frac{m^2}{4 \pi^2 \rho^2} \right)$
$I = \frac{m^3}{8 \pi^2 \rho^2}$.

(A) The moment of inertia $I$ of a solid sphere of mass $M$ and radius $R$ about its diameter is given by the formula $I = \frac{2}{5} MR^2$.
Since the mass $M$ is the same for both spheres,the moment of inertia is directly proportional to the square of the radius,i.e.,$I \propto R^2$.
Given the ratio of the radii is $\frac{R_1}{R_2} = \frac{2}{3}$.
Therefore,the ratio of the moments of inertia is $\frac{I_1}{I_2} = \left(\frac{R_1}{R_2}\right)^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9}$.
Thus,the ratio is $4:9$.

(A) The mass per unit length is given by $dm = \lambda|x|dx$.
First,calculate the moment of inertia about the axis $AA$ passing through the center of the rod:
$I_{AA} = \int x^2 dm = \int_{-L/2}^{L/2} x^2 (\lambda|x|) dx = 2\lambda \int_{0}^{L/2} x^3 dx = 2\lambda \left[ \frac{x^4}{4} \right]_0^{L/2} = \frac{\lambda}{2} \left( \frac{L}{2} \right)^4 = \frac{\lambda L^4}{32}$.
Given $\lambda = 16 \ kg/m^2$ and $L = 1 \ m$,$I_{AA} = \frac{16 \times 1^4}{32} = 0.5 \ kg \cdot m^2$.
Now,calculate the total mass $M$ of the rod:
$M = \int_{-L/2}^{L/2} \lambda|x| dx = 2\lambda \int_0^{L/2} x dx = 2\lambda \left[ \frac{x^2}{2} \right]_0^{L/2} = \lambda \left( \frac{L}{2} \right)^2 = \frac{\lambda L^2}{4}$.
With $\lambda = 16$ and $L = 1$,$M = \frac{16 \times 1^2}{4} = 4 \ kg$.
Using the parallel axis theorem to find the moment of inertia about axis $BB$ passing through one end:
$I_{BB} = I_{CM} + M h^2$,where $h = L/2 = 0.5 \ m$.
$I_{BB} = 0.5 + 4 \times (0.5)^2 = 0.5 + 4 \times 0.25 = 0.5 + 1 = 1.5 \ kg \cdot m^2$.

(D) Given,density of sphere,$\rho = A r^\alpha$ (where $r$ is the radial distance and $A$ and $\alpha$ are constants).
Consider an elemental spherical shell of radius $r$ and thickness $dr$.
Mass of elemental spherical shell,$dm = \text{Volume} \times \text{Density} = (4 \pi r^2) dr \cdot A r^\alpha = 4 \pi A r^{2+\alpha} dr$.
Mass of entire solid sphere,$M = 4 \pi A \int_0^R r^{2+\alpha} dr = 4 \pi A \left[ \frac{r^{3+\alpha}}{3+\alpha} \right]_0^R = \frac{4 \pi A}{3+\alpha} R^{3+\alpha}$.
Moment of inertia of the elemental spherical shell is $dI = \frac{2}{3} (dm) r^2 = \frac{2}{3} (4 \pi A r^{2+\alpha} dr) r^2 = \frac{8}{3} \pi A r^{4+\alpha} dr$.
Moment of inertia of the entire solid sphere,$I = \int_0^R dI = \frac{8}{3} \pi A \int_0^R r^{4+\alpha} dr = \frac{8}{3} \pi A \left[ \frac{r^{5+\alpha}}{5+\alpha} \right]_0^R = \frac{8 \pi A}{3(5+\alpha)} R^{5+\alpha}$.
Substituting $M$ into the expression for $I$,we get $I = \left( \frac{4 \pi A R^{3+\alpha}}{3+\alpha} \right) \cdot \frac{2}{3} \cdot \frac{3+\alpha}{5+\alpha} R^2 = M R^2 \left( \frac{2(3+\alpha)}{3(5+\alpha)} \right)$.
Given $I = \frac{6}{7} M R^2$,we equate: $\frac{2(3+\alpha)}{3(5+\alpha)} = \frac{6}{7}$.
$14(3+\alpha) = 18(5+\alpha) \Rightarrow 42 + 14\alpha = 90 + 18\alpha \Rightarrow -4\alpha = 48 \Rightarrow \alpha = -12$.

(C) The moment of inertia $(I)$ of a solid sphere about its diameter is given by the formula $I = \frac{2}{5}MR^2$,where $M$ is the mass and $R$ is the radius of the sphere.
Since the steel sphere has a higher density than the wooden sphere,for the same radius,the mass of the steel sphere is greater than that of the wooden sphere.
Therefore,the moment of inertia of the steel sphere is larger than that of the wooden sphere. Thus,Assertion $(A)$ is true.
However,the formula $I = \frac{2}{5}MR^2$ clearly shows that the moment of inertia depends on the mass $(M)$ of the body.
Therefore,Reason $(R)$ is false.
Hence,the correct option is $(C)$.

(C) The moment of inertia $I$ of a solid sphere of mass $M$ and radius $R$ about its diameter is given by $I = \frac{2}{5} M R^2$.
Since the mass $M$ is the product of volume $V$ and density $\rho$,we have $M = V \rho = \frac{4}{3} \pi R^3 \rho$.
Substituting this into the formula for $I$,we get $I = \frac{2}{5} (\frac{4}{3} \pi R^3 \rho) R^2 = \frac{8}{15} \pi R^5 \rho$.
Therefore,the ratio of the moments of inertia is $\frac{I_A}{I_B} = \frac{\frac{8}{15} \pi R^5 \rho_A}{\frac{8}{15} \pi R^5 \rho_B} = \frac{\rho_A}{\rho_B}$.

(A) The moment of inertia $(I)$ of a uniform circular disc is given by $I = \frac{1}{2} M R^2$.
Since the mass $(M)$ and thickness $(t)$ are constant,we express the radius $(R)$ in terms of density $(\rho)$:
$M = \pi R^2 t \rho \Rightarrow R^2 = \frac{M}{\pi t \rho}$.
Substituting this into the formula for $I$:
$I = \frac{1}{2} M \left( \frac{M}{\pi t \rho} \right) = \frac{M^2}{2 \pi t \rho}$.
Since $M$,$t$,and $\pi$ are constants,we have $I \propto \frac{1}{\rho}$.
Therefore,the disc made from the more dense material (higher $\rho$) will have a smaller rotational inertia.

(B) Given that the masses of the two solid spheres are equal,$m_A = m_B$.
Since mass $m = \text{Volume} \times \text{Density} = \frac{4}{3} \pi R^3 \rho$,we have:
$\frac{4}{3} \pi R_A^3 \rho_A = \frac{4}{3} \pi R_B^3 \rho_B$
$\Rightarrow \frac{R_A^3}{R_B^3} = \frac{\rho_B}{\rho_A} \Rightarrow \frac{R_A}{R_B} = \left(\frac{\rho_B}{\rho_A}\right)^{1/3}$.
The moment of inertia of a solid sphere about its diameter is $I = \frac{2}{5} m R^2$.
Since $m_A = m_B$,the ratio of moments of inertia is:
$\frac{I_B}{I_A} = \frac{\frac{2}{5} m_B R_B^2}{\frac{2}{5} m_A R_A^2} = \left(\frac{R_B}{R_A}\right)^2$.
Substituting the ratio of radii:
$\frac{I_B}{I_A} = \left[ \left( \frac{\rho_A}{\rho_B} \right)^{1/3} \right]^2 = \left( \frac{\rho_A}{\rho_B} \right)^{2/3}$.

(D) The moment of inertia of a solid sphere about its diameter is given by $I = \frac{2}{5} m R^2$.
Since mass $m$ remains constant,$I \propto R^2$.
The fractional change in moment of inertia is given by $\frac{\Delta I}{I} = 2 \frac{\Delta R}{R}$.
We know that the linear expansion is given by $\frac{\Delta R}{R} = \alpha \Delta t$,where $\alpha = 10^{-5} /^{\circ} C$ and $\Delta t = 200^{\circ} C$.
Substituting these values,we get $\frac{\Delta I}{I} = 2 \alpha \Delta t$.
To find the percentage increase,we multiply by $100$:
$\text{Percentage increase} = 2 \times 10^{-5} \times 200 \times 100 = 0.4 \%$.

(C) The moment of inertia of a disc about an axis passing through its centre and perpendicular to its plane is given by $I = \frac{1}{2} M R^2$.
Since the disc is melted and recasted into a solid sphere,the volume remains constant.
Volume of disc = Volume of sphere
$\pi R^2 \times \frac{R}{6} = \frac{4}{3} \pi R_1^3$
$\frac{R^3}{6} = \frac{4}{3} R_1^3$
$R_1^3 = \frac{R^3}{8} \implies R_1 = \frac{R}{2}$,where $R_1$ is the radius of the sphere.
The moment of inertia of a solid sphere about its diameter is $I' = \frac{2}{5} M R_1^2$.
Substituting $R_1 = \frac{R}{2}$ into the formula:
$I' = \frac{2}{5} M \left(\frac{R}{2}\right)^2 = \frac{2}{5} M \left(\frac{R^2}{4}\right) = \frac{1}{5} \left(\frac{1}{2} M R^2\right)$.
Since $I = \frac{1}{2} M R^2$,we get $I' = \frac{I}{5}$.

(B) Given: Total mass $M = 0.6 \ kg$,Length $L = 1 \ m = 100 \ cm$.
Mass per unit length $\lambda = \frac{0.6 \ kg}{100 \ cm} = 0.006 \ kg/cm$.
The axis is at $20 \ cm$. This divides the scale into two parts: $AB$ $(20 \ cm)$ and $BC$ $(80 \ cm)$.
Mass of part $AB$,$m_1 = \lambda \times 20 = 0.006 \times 20 = 0.12 \ kg$.
Mass of part $BC$,$m_2 = \lambda \times 80 = 0.006 \times 80 = 0.48 \ kg$.
The moment of inertia of a rod about an axis passing through one end is $I = \frac{1}{3} m l^2$.
For part $AB$ (axis at end $B$): $I_1 = \frac{1}{3} m_1 (l_1)^2 = \frac{1}{3} \times 0.12 \times (0.2)^2 = 0.04 \times 0.04 = 0.0016 \ kg-m^2$.
For part $BC$ (axis at end $B$): $I_2 = \frac{1}{3} m_2 (l_2)^2 = \frac{1}{3} \times 0.48 \times (0.8)^2 = 0.16 \times 0.64 = 0.1024 \ kg-m^2$.
Total moment of inertia $I = I_1 + I_2 = 0.0016 + 0.1024 = 0.104 \ kg-m^2$.

(D) The moment of inertia of a circular loop of mass $m$ and radius $R$ about an axis passing through its centre and perpendicular to its plane is $I = mR^2$.
Let the mass of loop $P$ be $m_P$ and its radius be $r_P = r$. The mass of loop $Q$ is $m_Q$ and its radius is $r_Q = nr$.
Since the wire is uniform,the mass is proportional to the circumference $(m \propto 2\pi R)$. Thus,$m_Q = n m_P$.
The moment of inertia of $P$ is $I_P = m_P r^2$.
The moment of inertia of $Q$ is $I_Q = m_Q (nr)^2 = (n m_P) (n^2 r^2) = n^3 m_P r^2$.
Given that $I_Q = 8 I_P$,we have $n^3 m_P r^2 = 8 m_P r^2$.
Therefore,$n^3 = 8$,which gives $n = 2$.

(D) Let the axis of rotation be the $AX$ line,which passes through vertex $A$ and is perpendicular to $AB$.
The moment of inertia $I$ of the system is the sum of the moments of inertia of individual particles about the axis $AX$.
$I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the $i$-th particle from the axis $AX$.
$1$. For the particle at vertex $A$: The distance $r_A = 0$,so $I_A = m(0)^2 = 0$.
$2$. For the particle at vertex $B$: The distance $r_B = a$,so $I_B = m(a)^2 = ma^2$.
$3$. For the particle at vertex $C$: The perpendicular distance from $AX$ is $r_C = a \cos 60^{\circ} = a \times \frac{1}{2} = \frac{a}{2}$.
So,$I_C = m(\frac{a}{2})^2 = \frac{ma^2}{4}$.
The total moment of inertia $I = I_A + I_B + I_C = 0 + ma^2 + \frac{ma^2}{4} = \frac{5ma^2}{4} \text{ } g-cm^2$.

(D) Let $M$ be the mass of both spheres. For the solid sphere: $M = \rho_{1} \frac{4}{3} \pi R^{3}$.
For the hollow sphere with inner radius $r$: $M = \rho_{2} \frac{4}{3} \pi (R^{3} - r^{3})$.
Equating the masses: $\rho_{1} R^{3} = \rho_{2} (R^{3} - r^{3})$.
This gives $\frac{\rho_{1}}{\rho_{2}} = 1 - \frac{r^{3}}{R^{3}}$,so $\frac{r^{3}}{R^{3}} = 1 - \frac{\rho_{1}}{\rho_{2}}$,which implies $\frac{r}{R} = (1 - \frac{\rho_{1}}{\rho_{2}})^{1/3}$.
The moment of inertia of the solid sphere is $I_{S} = \frac{2}{5} M R^{2}$.
The moment of inertia of the hollow sphere is $I_{H} = \frac{2}{5} M \frac{R^{5} - r^{5}}{R^{3} - r^{3}}$.
Using $M = \rho_{2} \frac{4}{3} \pi (R^{3} - r^{3})$,we get $I_{H} = \frac{2}{5} (\rho_{2} \frac{4}{3} \pi (R^{3} - r^{3})) \frac{R^{5} - r^{5}}{R^{3} - r^{3}} = \frac{2}{5} \rho_{2} \frac{4}{3} \pi (R^{5} - r^{5})$.
Since $M = \rho_{1} \frac{4}{3} \pi R^{3}$,we have $I_{S} = \frac{2}{5} (\rho_{1} \frac{4}{3} \pi R^{3}) R^{2} = \frac{2}{5} \rho_{1} \frac{4}{3} \pi R^{5}$.
Therefore,$\frac{I_{H}}{I_{S}} = \frac{\rho_{2} (R^{5} - r^{5})}{\rho_{1} R^{5}} = \frac{\rho_{2}}{\rho_{1}} [1 - (\frac{r}{R})^{5}]$.
Substituting $\frac{r}{R} = (1 - \frac{\rho_{1}}{\rho_{2}})^{1/3}$,we get $\frac{I_{H}}{I_{S}} = \frac{\rho_{2}}{\rho_{1}} [1 - (1 - \frac{\rho_{1}}{\rho_{2}})^{5/3}]$.

(A) Let the rectangular frame have corners $A, B, C,$ and $D$ such that side $AD = BC = a$ and side $AB = CD = b$. The masses $m$ are placed at each corner.
The axis of rotation passes along the side of length $b$ (let this be side $CD$).
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 of rotation.
$1$. The masses at $C$ and $D$ lie on the axis of rotation,so their perpendicular distances are $r_C = 0$ and $r_D = 0$. Thus,their contribution to the moment of inertia is $m(0)^2 + m(0)^2 = 0$.
$2$. The masses at $A$ and $B$ are at a perpendicular distance $a$ from the axis $CD$. Thus,their contribution is $m(a)^2 + m(a)^2 = 2ma^2$.
Therefore,the total moment of inertia of the system about the axis $CD$ is $I = 0 + 2ma^2 = 2ma^2$.

(A) The moment of inertia of a solid sphere about an axis passing through its centre is $I_{cm} = \frac{2}{5} mR^2$.
Using the parallel axis theorem,the moment of inertia about an axis at a distance $d$ from the centre is $I = I_{cm} + md^2$.
$I = \frac{2}{5} mR^2 + md^2$.
By definition,the radius of gyration $k$ is given by $I = mk^2$.
Therefore,$mk^2 = \frac{2}{5} mR^2 + md^2$.
$k^2 = \frac{2}{5} R^2 + d^2$.
Given $R = 10 \ cm$ and $d = 15 \ cm$:
$k^2 = \frac{2}{5} (10)^2 + (15)^2$.
$k^2 = \frac{2}{5} (100) + 225 = 40 + 225 = 265$.
Since $k = \sqrt{n}$,we have $k^2 = n$.
Thus,$n = 265$.

(A) Let $\rho$ be the density of the material of the cylinder.
The mass of the original cylinder is $M = \rho \cdot \pi R^2 L$.
The moment of inertia of the original cylinder about its axis is $I_1 = \frac{1}{2} M R^2 = \frac{1}{2} (\rho \pi R^2 L) R^2 = \frac{1}{2} \rho \pi R^4 L$.
The mass of the carved cylinder is $m = \rho \cdot \pi (R/3)^2 \cdot (L/2) = \rho \cdot \pi (R^2/9) \cdot (L/2) = \frac{\rho \pi R^2 L}{18} = \frac{M}{18}$.
The moment of inertia of the carved cylinder about its axis is $I_2 = \frac{1}{2} m (R/3)^2 = \frac{1}{2} (\frac{M}{18}) (\frac{R^2}{9}) = \frac{1}{324} M R^2$.
Now,the ratio $I_1/I_2$ is:
$\frac{I_1}{I_2} = \frac{\frac{1}{2} M R^2}{\frac{1}{324} M R^2} = \frac{1}{2} \times 324 = 162$.

(C) The moment of inertia of a circular disc about its central axis is given by $I = \frac{1}{2} M R^2$.
Since the discs are made of the same material,their density $\rho$ is the same.
The mass of a disc is $M = \text{Volume} \times \text{density} = (\pi R^2 T) \rho$.
Therefore,the moment of inertia is $I = \frac{1}{2} (\pi R^2 T \rho) R^2 = \frac{1}{2} \pi \rho R^4 T$.
Given that $I_1 = I_2$,we have:
$\frac{1}{2} \pi \rho R_1^4 T_1 = \frac{1}{2} \pi \rho R_2^4 T_2$
$R_1^4 T_1 = R_2^4 T_2$
$\frac{T_1}{T_2} = \frac{R_2^4}{R_1^4} = \left( \frac{R_2}{R_1} \right)^4$.
Given $\frac{R_1}{R_2} = 2$,so $\frac{R_2}{R_1} = \frac{1}{2}$.
$\frac{T_1}{T_2} = \left( \frac{1}{2} \right)^4 = \frac{1}{16}$.
Comparing this with $\frac{T_1}{T_2} = \frac{1}{\alpha}$,we get $\alpha = 16$.

(B) Let the vertical rod be $1$ and the horizontal rod be $2$.
For rod $1$,the moment of inertia about an axis passing through $P$ (its end) and perpendicular to its length is $I_1 = \frac{ML^2}{3}$.
For rod $2$,the moment of inertia about its center of mass is $I_{cm} = \frac{ML^2}{12}$. Using the parallel axis theorem,the moment of inertia about an axis passing through $P$ and perpendicular to the plane is $I_2 = I_{cm} + Md^2$,where $d = L$ is the distance from $P$ to the center of rod $2$.
$I_2 = \frac{ML^2}{12} + M(L)^2 = \frac{13ML^2}{12}$.
The total moment of inertia is $I = I_1 + I_2 = \frac{ML^2}{3} + \frac{13ML^2}{12} = \frac{4ML^2 + 13ML^2}{12} = \frac{17ML^2}{12}$.
Comparing this with $\frac{x}{12} ML^2$,we get $x = 17$.