Concept Card: Moment of Inertia of Simple Bodies – Ring, Disc & Rod
1. Physical Framework & Integral Method:
For continuous, extended bodies with uniform mass distribution, the moment of inertia about a specified axis is evaluated using definite integration over infinitesimal mass elements $dm$:
$$\mathbf{I = \int r^2 dm}$$
where $r$ is the strictly perpendicular distance from the element $dm$ to the axis of rotation. The standard integration strategy requires selecting a symmetric mass element whose parts are all equidistant from the axis.
2. Thin Uniform Rod (Mass $M$, Length $L$)
Consider a uniform rod of length $L$ and total mass $M$ with linear mass density $\lambda = \frac{M}{L}$.
- Case A: Perpendicular Axis Passing Through the Centre of Mass ($I_{\text{cm}}$):
Take the origin at the midpoint of the rod ($-L/2 \le x \le L/2$). An element of length $dx$ at distance $x$ has mass $dm = \frac{M}{L}dx$.
$$I_{\text{cm}} = \int_{-L/2}^{L/2} x^2 \left(\frac{M}{L}dx\right) = \frac{M}{L} \left[\frac{x^3}{3}\right]_{-L/2}^{L/2} = \frac{M}{3L} \left[\frac{L^3}{8} – \left(-\frac{L^3}{8}\right)\right] = \mathbf{\frac{1}{12} M L^2}$$
Radius of gyration: $\mathbf{k_{\text{cm}} = \frac{L}{\sqrt{12}} = \frac{L}{2\sqrt{3}}}$. - Case B: Perpendicular Axis Passing Through One End ($I_{\text{end}}$):
Take the origin at the end of the rod ($0 \le x \le L$):
$$I_{\text{end}} = \int_0^L x^2 \left(\frac{M}{L}dx\right) = \frac{M}{L} \left[\frac{x^3}{3}\right]_0^L = \mathbf{\frac{1}{3} M L^2}$$
Radius of gyration: $\mathbf{k_{\text{end}} = \frac{L}{\sqrt{3}}}$.
Note: $I_{\text{end}} = 4 I_{\text{cm}}$, consistent with the Parallel Axis Theorem ($I_{\text{end}} = I_{\text{cm}} + M(L/2)^2 = \frac{1}{12}M L^2 + \frac{1}{4}M L^2 = \frac{1}{3}M L^2$). - Case C: Axis Inclined at Angle $\theta$ to the Rod:
If the axis of rotation passes through the center of mass but makes an angle $\theta$ with the rod, the perpendicular distance of element $dx$ from the axis is $r = x\sin\theta$.
$$\mathbf{I_{\text{inclined, cm}} = \frac{1}{12} M L^2 \sin^2\theta} \quad \text{and} \quad \mathbf{I_{\text{inclined, end}} = \frac{1}{3} M L^2 \sin^2\theta}$$
3. Thin Circular Ring / Hoop (Mass $M$, Radius $R$)
All mass of a thin circular hoop resides at a fixed radius $R$ from the center.
- Case A: Central Axis Perpendicular to the Plane of the Ring ($I_z$):
Every mass element $dm$ on the rim is at an identical perpendicular distance $r = R$ from the central axis.
$$I_z = \int R^2 dm = R^2 \int dm = \mathbf{M R^2}$$
Radius of gyration: $\mathbf{k_z = R}$. - Case B: Diametrical Axis in the Plane of the Ring ($I_d$):
By planar symmetry, all diametrical axes in the plane are identical ($I_x = I_y = I_d$). Applying the Perpendicular Axis Theorem ($I_z = I_x + I_y = 2 I_d$):
$$2 I_d = M R^2 \implies \mathbf{I_d = \frac{1}{2} M R^2}$$
Radius of gyration: $\mathbf{k_d = \frac{R}{\sqrt{2}}}$. - Case C: Circular Arc / Ring Sector:
For a circular arc of mass $M$, radius $R$, and subtended central angle $\theta$, all mass remains at distance $R$ from the center of curvature. Therefore, its moment of inertia about the perpendicular central axis remains strictly $\mathbf{I = M R^2}$, regardless of the angle $\theta$!
4. Uniform Circular Disc (Mass $M$, Radius $R$)
Mass is uniformly distributed across the surface area $A = \pi R^2$, with mass per unit area $\sigma = \frac{M}{\pi R^2}$.
- Case A: Central Axis Perpendicular to the Plane of the Disc ($I_z$):
Choose a concentric circular ring element of radius $r$ and thickness $dr$ ($0 \le r \le R$).
Area of element: $dA = 2\pi r\,dr$. Mass of element: $dm = \sigma dA = \left(\frac{M}{\pi R^2}\right)(2\pi r\,dr) = \frac{2M}{R^2} r\,dr$.
All mass of this elemental ring is at perpendicular distance $r$ from the central axis:
$$I_z = \int_0^R r^2 dm = \int_0^R r^2 \left(\frac{2M}{R^2} r\,dr\right) = \frac{2M}{R^2} \int_0^R r^3\,dr = \frac{2M}{R^2} \left[\frac{r^4}{4}\right]_0^R = \mathbf{\frac{1}{2} M R^2}$$
Radius of gyration: $\mathbf{k_z = \frac{R}{\sqrt{2}}}$. - Case B: Diametrical Axis in the Plane of the Disc ($I_d$):
By symmetry in the disc plane, $I_x = I_y = I_d$. From the Perpendicular Axis Theorem ($I_z = I_x + I_y = 2 I_d$):
$$2 I_d = \frac{1}{2} M R^2 \implies \mathbf{I_d = \frac{1}{4} M R^2}$$
Radius of gyration: $\mathbf{k_d = \frac{R}{2}}$. - Case C: Circular Disc Sector:
For a sector of a uniform disc of mass $M$, radius $R$, and subtended angle $\theta$, the moment of inertia about the central perpendicular axis through the apex is identically $\mathbf{I = \frac{1}{2} M R^2}$.
5. Summary Comparison & Hierarchy (Same Mass $M$ and Radius $R$)
$$\mathbf{I_{\text{ring, central}} (M R^2) > I_{\text{disc, central}} \left(\frac{1}{2}M R^2\right) = I_{\text{ring, diameter}} \left(\frac{1}{2}M R^2\right) > I_{\text{disc, diameter}} \left(\frac{1}{4}M R^2\right)}$$
6. Common JEE Pitfalls & Traps
- Trap 1 (Circular Arc Angle Dependency): Students often erroneously multiply by $\frac{\theta}{2\pi}$. If the mass of the arc is already specified as $M$, the moment of inertia about the center of curvature is simply $M R^2$, independent of $\theta$.
- Trap 2 (Inclined Rod Dimension): For a rod inclined at angle $\theta$ to the axis of rotation, the perpendicular distance from any point on the rod to the axis is $x\sin\theta$. Always remember the factor $\sin^2\theta$!
- Trap 3 (Ring vs. Disc Diametrical Axes): Remember that $I_{\text{ring, diameter}} = \frac{1}{2}M R^2$, which equals the perpendicular central moment of inertia of a disc ($I_{\text{disc, central}} = \frac{1}{2}M R^2$). Do not confuse the two!
Solved Examples
Example 1 (Direct Application – Inclined Rod Dynamics):
A uniform thin rod of mass $M = 2.0\text{ kg}$ and length $L = 1.2\text{ m}$ is mounted on a frictionless axle.
(a) Find its moment of inertia $I_c$ and radius of gyration $k_c$ about an axis passing through its midpoint inclined at an angle $\theta = 30^\circ$ to the rod.
(b) Find its moment of inertia $I_e$ and radius of gyration $k_e$ about a parallel axis passing through one end of the rod inclined at the same angle $\theta = 30^\circ$.
(c) Under an applied torque $\tau = 1.2\text{ N}\cdot\text{m}$ about the central inclined axis, calculate the angular acceleration $\alpha$.
Solution:
(a) For an axis passing through the center of mass inclined at angle $\theta$:
$$I_c = \frac{1}{12} M L^2 \sin^2\theta = \frac{1}{12} (2.0) (1.2)^2 \sin^2(30^\circ)$$
Since $(1.2)^2 = 1.44$ and $\sin(30^\circ) = 0.50 \implies \sin^2(30^\circ) = 0.25$:
$$I_c = \frac{1}{12} (2.0) (1.44) (0.25) = \frac{0.72}{12} = \mathbf{0.060\text{ kg}\cdot\text{m}^2}$$
Radius of gyration:
$$k_c = \sqrt{\frac{I_c}{M}} = \sqrt{\frac{0.060}{2.0}} = \sqrt{0.030} \approx \mathbf{0.1732\text{ m}}$$
(b) For the parallel axis passing through one end:
$$I_e = \frac{1}{3} M L^2 \sin^2\theta = 4 I_c = 4(0.060) = \mathbf{0.240\text{ kg}\cdot\text{m}^2}$$
Radius of gyration:
$$k_e = \sqrt{\frac{I_e}{M}} = \sqrt{\frac{0.240}{2.0}} = \sqrt{0.120} \approx \mathbf{0.3464\text{ m}}$$
(c) Angular acceleration about the central axis:
$$\alpha = \frac{\tau}{I_c} = \frac{1.2\text{ N}\cdot\text{m}}{0.060\text{ kg}\cdot\text{m}^2} = \mathbf{20.0\text{ rad/s}^2}$$
Example 2 (Slicing & Sectoring Invariance of Discs and Rings):
A uniform circular disc has an initial total mass $M_0 = 6.0\text{ kg}$ and radius $R = 0.50\text{ m}$. A sector of angle $\theta_0 = 60^\circ$ is cut out and removed.
(a) Determine the mass $M_{\text{rem}}$ of the remaining $300^\circ$ sector.
(b) Calculate the moment of inertia $I_{\text{rem}}$ of the remaining sector about the central perpendicular axis passing through the apex.
(c) Prove that if this remaining sector is rolled without deformation into a cone, its moment of inertia about the cone’s axis remains unchanged in form.
Solution:
(a) Mass is distributed uniformly across $360^\circ$. The remaining sector subtends an angle $\theta = 360^\circ – 60^\circ = 300^\circ$.
$$M_{\text{rem}} = M_0 \left(\frac{300^\circ}{360^\circ}\right) = M_0 \left(\frac{5}{6}\right) = (6.0)\left(\frac{5}{6}\right) = \mathbf{5.0\text{ kg}}$$
(b) For any disc sector of mass $M_{\text{rem}}$ and radius $R$, the surface density is $\sigma = \frac{M_{\text{rem}}}{\frac{1}{2}R^2 \theta}$.
An elemental sector arc at radius $r$ has mass $dm = \sigma (r\theta\,dr)$. Integrating from $0$ to $R$:
$$I_{\text{rem}} = \int_0^R r^2 dm = \int_0^R r^2 \sigma \theta r\,dr = \sigma \theta \left[\frac{r^4}{4}\right]_0^R = \frac{\sigma \theta R^4}{4} = \frac{1}{2} \left(\frac{1}{2}\sigma \theta R^2\right) R^2 = \mathbf{\frac{1}{2} M_{\text{rem}} R^2}$$
Numerical value:
$$I_{\text{rem}} = \frac{1}{2} (5.0) (0.50)^2 = \frac{1}{2} (5.0) (0.25) = \mathbf{0.625\text{ kg}\cdot\text{m}^2}$$
Fundamental Insight: Slicing out a sector reduces mass proportionally, but leaves the coefficient $\frac{1}{2}$ completely unchanged!
(c) When rolled into a cone, every mass element that was at distance $r$ along the slant surface is now at perpendicular distance $r’ = r\sin\alpha$ from the cone axis (where $\alpha$ is the semi-vertical angle). The distance $r’$ from the axis shrinks by $\sin\alpha$, modifying $I_{\text{cone}} = \frac{1}{2}M_{\text{rem}} (R\sin\alpha)^2 = \frac{1}{2}M_{\text{rem}} R_{\text{base}}^2$.
Example 3 (Composite Rigid System – Spoked Wheel):
A lightweight carriage wheel is modeled as a circular rim (ring) of mass $M_1 = 3.0\text{ kg}$ and radius $R = 0.50\text{ m}$, supported by four identical radial spokes. Each spoke is a uniform thin rod of mass $M_2 = 0.75\text{ kg}$ and length $R = 0.50\text{ m}$, extending from the central axle to the outer rim.
(a) Find the moment of inertia of the circular rim about the central axis perpendicular to the wheel plane.
(b) Find the moment of inertia of the four spokes about the same axis.
(c) Calculate the total moment of inertia $I_{\text{total}}$ of the wheel and its effective radius of gyration $k_{\text{total}}$.
Solution:
(a) The circular rim is a thin ring of mass $M_1$ and radius $R$ rotating about its central perpendicular axis:
$$I_{\text{rim}} = M_1 R^2 = (3.0\text{ kg})(0.50\text{ m})^2 = (3.0)(0.25) = \mathbf{0.75\text{ kg}\cdot\text{m}^2}$$
(b) Each radial spoke is a thin uniform rod of mass $M_2$ and length $R$ rotating about an axis perpendicular to the rod passing through one of its ends (the central axle):
$$I_{\text{spoke}} = \frac{1}{3} M_2 R^2 = \frac{1}{3} (0.75) (0.50)^2 = (0.25)(0.25) = 0.0625\text{ kg}\cdot\text{m}^2$$
For all four identical spokes combined:
$$I_{\text{spokes}} = 4 \times I_{\text{spoke}} = 4(0.0625) = \mathbf{0.25\text{ kg}\cdot\text{m}^2}$$
(c) By the superposition principle of rotational inertia:
$$I_{\text{total}} = I_{\text{rim}} + I_{\text{spokes}} = 0.75 + 0.25 = \mathbf{1.00\text{ kg}\cdot\text{m}^2}$$
Total mass of the wheel: $M_{\text{total}} = M_1 + 4 M_2 = 3.0 + 4(0.75) = 3.0 + 3.0 = 6.0\text{ kg}$.
Radius of gyration:
$$k_{\text{total}} = \sqrt{\frac{I_{\text{total}}}{M_{\text{total}}}} = \sqrt{\frac{1.00\text{ kg}\cdot\text{m}^2}{6.0\text{ kg}}} = \frac{1}{\sqrt{6}}\text{ m} \approx \mathbf{0.4082\text{ m}}$$
Example 4 (Edge Case – Non-Uniform Disc with Radial Density Gradient):
A circular disc of radius $R$ has a non-uniform surface mass density that increases linearly with radial distance from the center: $\sigma(r) = \sigma_0 \left(\frac{r}{R}\right)$, where $\sigma_0$ is a positive constant.
(a) Determine the total mass $M$ of the disc in terms of $\sigma_0$ and $R$.
(b) Derive the moment of inertia $I$ of the disc about the perpendicular axis passing through its center in terms of $M$ and $R$.
(c) Compare this result with that of a uniform disc and explain the physical reason for the difference.
Solution:
(a) Divide the disc into concentric elemental rings of radius $r$ and thickness $dr$ ($0 \le r \le R$).
Mass of the elemental ring: $dm = \sigma(r)\,(2\pi r\,dr) = \sigma_0 \left(\frac{r}{R}\right) (2\pi r\,dr) = \frac{2\pi \sigma_0}{R} r^2\,dr$.
Total mass $M$:
$$M = \int_0^R dm = \frac{2\pi \sigma_0}{R} \int_0^R r^2\,dr = \frac{2\pi \sigma_0}{R} \left[\frac{r^3}{3}\right]_0^R = \mathbf{\frac{2}{3} \pi \sigma_0 R^2}$$
(b) All mass of the elemental ring is at perpendicular distance $r$ from the central axis:
$$dI = r^2 dm = r^2 \left(\frac{2\pi \sigma_0}{R} r^2\,dr\right) = \frac{2\pi \sigma_0}{R} r^4\,dr$$
Integrating over the entire disc:
$$I = \int_0^R dI = \frac{2\pi \sigma_0}{R} \int_0^R r^4\,dr = \frac{2\pi \sigma_0}{R} \left[\frac{r^5}{5}\right]_0^R = \mathbf{\frac{2}{5} \pi \sigma_0 R^4}$$
To express $I$ in terms of total mass $M$, take the ratio $\frac{I}{M}$:
$$\frac{I}{M} = \frac{\frac{2}{5}\pi \sigma_0 R^4}{\frac{2}{3}\pi \sigma_0 R^2} = \frac{3}{5} R^2 \implies \mathbf{I = \frac{3}{5} M R^2}$$
Radius of gyration: $\mathbf{k = \sqrt{\frac{3}{5}} R \approx 0.7746 R}$.
(c) Physical Comparison: For a uniform disc, $I_{\text{uniform}} = \frac{1}{2} M R^2 = 0.50 M R^2$.
For this non-uniform disc, $I = \frac{3}{5} M R^2 = 0.60 M R^2$.
Because the density increases linearly outwards, a greater fraction of the total mass is distributed near the outer perimeter, resulting in a higher moment of inertia ($0.60 > 0.50$).
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
A uniform thin rod of length $L$ and mass $M$ is bent at its midpoint to form an ‘L’-shape with two mutually perpendicular arms of equal length $L/2$. The moment of inertia of this bent rod about an axis perpendicular to the plane of the ‘L’ passing through the vertex (the bend) is:
(A) $\frac{1}{6} M L^2$
(B) $\frac{1}{12} M L^2$
(C) $\frac{1}{24} M L^2$
(D) $\frac{1}{3} M L^2$
Problem 2 (JEE Main – Single Correct):
A thin circular hoop and a uniform circular disc have identical mass $M$ and identical radius $R$. The ratio of the moment of inertia of the hoop about its diameter to that of the disc about its central perpendicular axis is:
(A) $1 : 1$
(B) $2 : 1$
(C) $1 : 2$
(D) $4 : 1$
Problem 3 (JEE Main – Single Correct):
A uniform disc of mass $M$ and radius $R$ has a concentric circular hole of radius $R/3$ cut out. The moment of inertia of the remaining annular disc about its central perpendicular axis in terms of its mass $M$ and outer radius $R$ is:
(A) $\frac{4}{9} M R^2$
(B) $\frac{5}{9} M R^2$
(C) $\frac{1}{2} M R^2$
(D) $\frac{8}{9} M R^2$
Problem 4 (JEE Main – Single Correct):
A circular wire loop of radius $R$ has mass $M$. It is cut and straightened into a uniform thin rod of length $L = 2\pi R$. The moment of inertia of this rod about an axis perpendicular to it passing through its center is:
(A) $\frac{\pi^2}{3} M R^2$
(B) $\frac{\pi^2}{6} M R^2$
(C) $\frac{4\pi^2}{3} M R^2$
(D) $\frac{\pi^2}{12} M R^2$
Problem 5 (JEE Main – Single Correct):
A uniform circular disc of radius $R$ and mass $M$ is rotating about an axis passing through its diameter. Its radius of gyration is:
(A) $R/2$
(B) $R/\sqrt{2}$
(C) $R/4$
(D) $R/\sqrt{3}$
Problem 6 (JEE Advanced – One or More Correct):
A circular arc of radius $R$ subtends an angle $\theta$ at its center of curvature and has mass $M$:
(A) Its moment of inertia about an axis perpendicular to its plane passing through the center of curvature is $M R^2$.
(B) The radius of gyration about the center of curvature is $R$, completely independent of $\theta$.
(C) If the arc is folded without stretching into a smaller radius, its moment of inertia about the original center changes.
(D) Its moment of inertia about an axis passing through its own center of mass is less than $M R^2$.
Problem 7 (JEE Advanced – One or More Correct):
Consider a uniform thin rod of mass $M$ and length $L$ lying in the $xy$-plane along the $x$-axis from $x = -L/2$ to $x = +L/2$:
(A) Its moment of inertia about the $x$-axis is zero (assuming an ideal 1D line mass).
(B) Its moment of inertia about the $y$-axis is $\frac{1}{12}M L^2$.
(C) Its moment of inertia about the $z$-axis is $\frac{1}{12}M L^2$.
(D) The relation $I_z = I_x + I_y$ is strictly satisfied for this 1D rod.
Problem 8 (JEE Advanced – One or More Correct):
Two thin uniform rods, each of mass $m$ and length $L$, are joined end-to-end to form a straight rod of mass $2m$ and length $2L$:
(A) The moment of inertia of the combined rod about its midpoint perpendicular to its length is $\frac{2}{3}m L^2$.
(B) The moment of inertia of each individual rod about its own center of mass is $\frac{1}{12}m L^2$.
(C) The distance of the center of mass of each individual rod from the midpoint of the combined rod is $L/2$.
(D) Applying the parallel axis theorem to each half rod verifies that $2 \left[\frac{1}{12}m L^2 + m(L/2)^2\right] = \frac{2}{3}m L^2$.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A uniform thin rod of mass $M = 6.0\text{ kg}$ and length $L = 2.0\text{ m}$ is rotated about an axis perpendicular to the rod passing through one of its ends. Calculate its moment of inertia in $\text{kg}\cdot\text{m}^2$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A uniform circular disc has a radius $R = 40.0\text{ cm}$. Calculate the radius of gyration of the disc (in $\text{cm}$) about an axis passing through one of its diameters.
Solutions & Explanations
Answer Key Summary:
1. (B) | 2. (A) | 3. (B) | 4. (A) | 5. (A) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (A, B, C, D) | 9. 8 | 10. 20
Solution 1:
The bent rod consists of two arms, each of mass $m = M/2$ and length $l = L/2$.
The vertex is an endpoint for both arms.
Moment of inertia of each arm about the axis through the vertex:
$$I_{\text{arm}} = \frac{1}{3} m l^2 = \frac{1}{3} \left(\frac{M}{2}\right) \left(\frac{L}{2}\right)^2 = \frac{1}{3} \left(\frac{M}{2}\right) \left(\frac{L^2}{4}\right) = \frac{M L^2}{24}$$
Total moment of inertia of both arms:
$$I = 2 \times I_{\text{arm}} = 2 \left(\frac{M L^2}{24}\right) = \mathbf{\frac{1}{12} M L^2}$$
Correct Option: (B)
Solution 2:
– For the hoop about its diameter: $I_{\text{hoop}, d} = \frac{1}{2} M R^2$.
– For the disc about its central perpendicular axis: $I_{\text{disc}, z} = \frac{1}{2} M R^2$.
$$\text{Ratio} = \frac{\frac{1}{2}M R^2}{\frac{1}{2}M R^2} = 1 : 1$$
Correct Option: (A)
Solution 3:
For any uniform annular disc of mass $M$ with inner radius $R_1$ and outer radius $R_2$, the moment of inertia about the central perpendicular axis is:
$$I = \frac{1}{2} M (R_1^2 + R_2^2)$$
Here $R_1 = R/3$ and $R_2 = R$:
$$I = \frac{1}{2} M \left[\left(\frac{R}{3}\right)^2 + R^2\right] = \frac{1}{2} M \left(\frac{R^2}{9} + R^2\right) = \frac{1}{2} M \left(\frac{10}{9}R^2\right) = \mathbf{\frac{5}{9} M R^2}$$
Correct Option: (B)
Solution 4:
Straightened length $L = 2\pi R$.
For a thin rod of mass $M$ and length $L$ about its center:
$$I = \frac{1}{12} M L^2 = \frac{1}{12} M (2\pi R)^2 = \frac{1}{12} M (4\pi^2 R^2) = \mathbf{\frac{\pi^2}{3} M R^2}$$
Correct Option: (A)
Solution 5:
For a disc about its diameter, $I_d = \frac{1}{4} M R^2$.
Radius of gyration: $k_d = \sqrt{\frac{I_d}{M}} = \sqrt{\frac{\frac{1}{4}M R^2}{M}} = \mathbf{\frac{R}{2}}$.
Correct Option: (A)
Solution 6:
– (A) True: Every particle on the arc is at distance $R$ from the center of curvature $\implies I = \int R^2 dm = R^2 \int dm = M R^2$.
– (B) True: $k = \sqrt{I/M} = \sqrt{M R^2 / M} = R$, independent of $\theta$.
– (C) True: Changing the radial distance alters $R^2$.
– (D) True: By the parallel axis theorem, $I_{\text{center}} = I_{\text{cm}} + M d^2 \implies I_{\text{cm}} = M R^2 – M d^2 < M R^2$.
Correct Options: (A, B, C, D)
Solution 7:
– (A) True: For an idealized 1D rod along $x$, all mass elements have $y = 0, z = 0 \implies r_x = \sqrt{y^2 + z^2} = 0 \implies I_x = 0$.
– (B) True: $I_y = \int x^2 dm = \frac{1}{12}M L^2$.
– (C) True: $I_z = \int x^2 dm = \frac{1}{12}M L^2$.
– (D) True: $I_z = I_x + I_y \implies \frac{1}{12}M L^2 = 0 + \frac{1}{12}M L^2$.
Correct Options: (A, B, C, D)
Solution 8:
– Total mass $M_{\text{tot}} = 2m$, total length $L_{\text{tot}} = 2L$.
$$I_{\text{mid}} = \frac{1}{12} M_{\text{tot}} L_{\text{tot}}^2 = \frac{1}{12} (2m) (2L)^2 = \frac{1}{12} (2m) (4L^2) = \frac{8}{12} m L^2 = \frac{2}{3} m L^2$$
– Individual half rod: mass $m$, length $L$. $I_{\text{cm, half}} = \frac{1}{12}m L^2$.
– Center of mass of each half rod is at distance $d = L/2$ from the midpoint.
– Parallel axis theorem on each half: $I_{\text{half, mid}} = \frac{1}{12}m L^2 + m(L/2)^2 = \frac{1}{12}m L^2 + \frac{1}{4}m L^2 = \frac{1}{3}m L^2$.
– For both halves: $2 \times \frac{1}{3}m L^2 = \frac{2}{3}m L^2$.
All statements (A, B, C, D) are verified.
Correct Options: (A, B, C, D)
Solution 9:
$I_{\text{end}} = \frac{1}{3} M L^2 = \frac{1}{3} (6.0\text{ kg}) (2.0\text{ m})^2 = \frac{1}{3} (6.0) (4.0) = \mathbf{8.0\text{ kg}\cdot\text{m}^2}$.
Correct Answer: 8
Solution 10:
For a disc about its diameter: $k_d = \frac{R}{2}$.
Given $R = 40.0\text{ cm}$:
$$k_d = \frac{40.0\text{ cm}}{2} = \mathbf{20.0\text{ cm}}$$
Correct Answer: 20