Concept Card: Combined Parallel & Perpendicular Axis Theorems – Composite Systems & Cavities
1. Problem-Solving Synthesis & Strategy:
In high-yield JEE Advanced problems, determining the moment of inertia about arbitrary or non-standard axes frequently requires the sequential application of two fundamental theorems alongside the principle of superposition:
- Perpendicular Axis Theorem ($I_z = I_x + I_y$): Strictly valid only for two-dimensional planar laminas lying in the $xy$-plane ($z = 0$). Used to interconvert between central perpendicular axes and planar diametrical/centroidal axes.
- Parallel Axis Theorem ($I = I_{\text{cm}} + M d^2$): Universally valid for any 1D, 2D, or 3D rigid body. Used to translate from a known centre of mass axis to any parallel axis displaced by perpendicular distance $d$.
- Superposition & Subtraction Principle:
- Composite Systems: $I_{\text{total}} = \sum_{i} I_i$ (evaluate all parts about the same common axis).
- Cavity / Cutout Problems: $I_{\text{remaining}} = I_{\text{original}} – I_{\text{cavity}}$ (both evaluated about the same target axis).
2. Standard Two-Step Shift Workflows for Planar Laminas
A. Tangential In-Plane Axis of a Circular Disc:
To find the moment of inertia of a uniform circular disc ($M, R$) about a tangent lying in the plane of the disc:
- Step 1 (Perpendicular Axis Theorem): The perpendicular central axis has $I_z = \frac{1}{2} M R^2$. By planar symmetry, $I_x = I_y = I_d$.
$$I_z = 2 I_d \implies I_d = \frac{1}{4} M R^2$$ - Step 2 (Parallel Axis Theorem): The in-plane tangent is parallel to a central diameter at a perpendicular distance $d = R$:
$$I_{\text{tangent, in-plane}} = I_d + M R^2 = \frac{1}{4} M R^2 + M R^2 = \mathbf{\frac{5}{4} M R^2}$$
Radius of gyration: $\mathbf{k = \frac{\sqrt{5}}{2} R \approx 1.118 R}$.
B. Tangential In-Plane Axis of a Thin Circular Hoop:
- Step 1: $I_z = M R^2 \implies I_d = \frac{1}{2} I_z = \frac{1}{2} M R^2$.
- Step 2: $I_{\text{tangent, in-plane}} = I_d + M R^2 = \frac{1}{2} M R^2 + M R^2 = \mathbf{\frac{3}{2} M R^2}$.
3. Standard Transition Matrix for Classical Planar & 3D Bodies
| Body & Dimensions | Central Perpendicular ($I_z$) | Central In-Plane / Diametrical ($I_{\text{cm}}$) | Tangential In-Plane Axis ($I_{\text{tangent}}$) | Tangential Perpendicular ($I_{t, \perp}$) |
|---|---|---|---|---|
| Circular Ring ($M, R$) | $M R^2$ | $\frac{1}{2} M R^2$ | $\frac{3}{2} M R^2$ | $2 M R^2$ |
| Circular Disc ($M, R$) | $\frac{1}{2} M R^2$ | $\frac{1}{4} M R^2$ | $\frac{5}{4} M R^2$ | $\frac{3}{2} M R^2$ |
| Square Plate ($M, a$) | $\frac{1}{6} M a^2$ | $\frac{1}{12} M a^2$ (any central line) | Edge axis: $\frac{1}{3} M a^2$ | Corner perpendicular: $\frac{2}{3} M a^2$ |
| Solid Cylinder ($M, R, L$) | Along cylinder axis: $\frac{1}{2} M R^2$ | Transverse CM: $\frac{1}{4}M R^2 + \frac{1}{12}M L^2$ | — | End flat face transverse: $\frac{1}{4}M R^2 + \frac{1}{3}M L^2$ |
4. Cavity & Cutout Protocol (Off-Center Cavities)
For an object with an off-center cutout (e.g., a hole of radius $r$ in a disc of radius $R$ at distance $d$ from center):
- Find original mass $M_0$ and removed cavity mass $m_{\text{cav}}$ using area/volume density: $m_{\text{cav}} = M_0 \left(\frac{A_{\text{cav}}}{A_0}\right)$.
- Calculate $I_{\text{orig}}$ of the complete un-punctured body about the target axis.
- Calculate $I_{\text{cav, cm}}$ about the cavity’s own parallel center of mass axis.
- Apply the Parallel Axis Theorem to shift the cavity’s inertia to the target axis: $I_{\text{cav, target}} = I_{\text{cav, cm}} + m_{\text{cav}} d^2$.
- Subtract: $\mathbf{I_{\text{remaining}} = I_{\text{orig}} – I_{\text{cav, target}}}$.
5. Common JEE Pitfalls & Traps
- Trap 1 (The 3D Cylindrical Extension Blunder): Never attempt to calculate the transverse moment of inertia of a 3D cylinder by writing $I_{\text{transverse}} = I_{\text{axial}} + I_{\text{something}}$. The perpendicular axis theorem fails for 3D bodies! Instead, slice the cylinder into discs and integrate via the parallel axis theorem: $I = \int dI_{\text{disc}} + \int x^2 dm = \frac{1}{4}M R^2 + \frac{1}{12}M L^2$.
- Trap 2 (Non-CM Intermediate Shifts): In multi-body assemblies, never shift directly between two non-CM parallel axes. Always route through the component’s centre of mass!
- Trap 3 (Cavity Mass Subtraction Mistake): Forgetting that $m_{\text{cav}}$ is a fraction of the total un-punctured mass, or writing $I_{\text{rem}} = \frac{1}{2} M_{\text{rem}} R^2 – \frac{1}{2} m_{\text{cav}} r^2$ without parallel-shifting the cavity.
Solved Examples
Example 1 (Direct Two-Step Application – Disc Tangential Shifts):
A uniform circular disc of mass $M = 4.0\text{ kg}$ and radius $R = 0.50\text{ m}$ lies in the horizontal $xy$-plane centered at origin $O$.
(a) Using the perpendicular axis theorem, determine the moment of inertia $I_d$ about a central diameter.
(b) Using the parallel axis theorem, determine the moment of inertia $I_{t, \parallel}$ about a tangent line lying in the plane of the disc.
(c) Determine the moment of inertia $I_{t, \perp}$ about a tangent line perpendicular to the plane of the disc.
(d) If the disc rotates at angular velocity $\omega = 10.0\text{ rad/s}$ about the in-plane tangent, calculate its rotational kinetic energy.
Solution:
(a) For the disc, the perpendicular central axis ($z$-axis) has:
$$I_z = \frac{1}{2} M R^2 = \frac{1}{2} (4.0\text{ kg})(0.50\text{ m})^2 = 0.50\text{ kg}\cdot\text{m}^2$$
By planar symmetry, any two orthogonal diameters in the $xy$-plane have identical moments of inertia: $I_x = I_y = I_d$.
By the Perpendicular Axis Theorem:
$$I_z = I_x + I_y = 2 I_d \implies I_d = \frac{1}{4} M R^2 = \frac{1}{4} (4.0)(0.25) = \mathbf{0.250\text{ kg}\cdot\text{m}^2}$$
(b) The tangent line lying in the plane of the disc is parallel to the diameter at a perpendicular distance $d = R = 0.50\text{ m}$.
By the Parallel Axis Theorem:
$$I_{t, \parallel} = I_d + M R^2 = \frac{1}{4} M R^2 + M R^2 = \mathbf{\frac{5}{4} M R^2}$$
Numerical value:
$$I_{t, \parallel} = \frac{5}{4} (4.0\text{ kg})(0.50\text{ m})^2 = \frac{5}{4}(4.0)(0.25) = \mathbf{1.250\text{ kg}\cdot\text{m}^2}$$
(c) The tangent line perpendicular to the disc plane is parallel to the central $z$-axis at a perpendicular distance $d = R = 0.50\text{ m}$.
By the Parallel Axis Theorem:
$$I_{t, \perp} = I_z + M R^2 = \frac{1}{2} M R^2 + M R^2 = \mathbf{\frac{3}{2} M R^2}$$
Numerical value:
$$I_{t, \perp} = \frac{3}{2} (4.0\text{ kg})(0.50\text{ m})^2 = \frac{3}{2}(4.0)(0.25) = \mathbf{1.500\text{ kg}\cdot\text{m}^2}$$
(d) Rotational kinetic energy about the in-plane tangent:
$$K_{\text{rot}} = \frac{1}{2} I_{t, \parallel} \omega^2 = \frac{1}{2} (1.250\text{ kg}\cdot\text{m}^2)(10.0\text{ rad/s})^2 = \frac{1}{2}(1.250)(100.0) = \mathbf{62.50\text{ J}}$$
Example 2 (Solid Cylinder Transverse Axis – Slicing & Parallel Shift Integration):
A uniform solid cylinder of mass $M = 6.0\text{ kg}$, radius $R = 0.40\text{ m}$, and length $L = 1.0\text{ m}$ has its longitudinal symmetry axis along the $z$-axis.
(a) Derive the moment of inertia $I_{\text{trans}}$ of the cylinder about a transverse axis passing through its centre of mass perpendicular to its length.
(b) Calculate the moment of inertia $I_{\text{end}}$ about a parallel transverse axis passing through the center of one of its circular end faces.
(c) Compute the numerical values of $I_{\text{trans}}$ and $I_{\text{end}}$.
Solution:
(a) Divide the cylinder into thin circular disc slices of thickness $dz$ and radius $R$ at coordinate $z$ from the center ($-L/2 \le z \le L/2$).
Mass of each disc slice: $dm = \frac{M}{L}dz$.
About its own diametrical axis, each disc slice has moment of inertia: $dI_{\text{cm}} = \frac{1}{4} dm R^2 = \frac{1}{4} \left(\frac{M}{L}dz\right) R^2$.
By the Parallel Axis Theorem, the moment of inertia of this slice about the cylinder’s central transverse axis is:
$$dI = dI_{\text{cm}} + z^2 dm = \frac{1}{4} R^2 \left(\frac{M}{L}dz\right) + z^2 \left(\frac{M}{L}dz\right)$$
Integrating over the entire length from $-L/2$ to $+L/2$:
$$I_{\text{trans}} = \frac{M R^2}{4L} \int_{-L/2}^{L/2} dz + \frac{M}{L} \int_{-L/2}^{L/2} z^2 dz = \frac{M R^2}{4L}(L) + \frac{M}{L}\left[\frac{z^3}{3}\right]_{-L/2}^{L/2} = \mathbf{\frac{1}{4} M R^2 + \frac{1}{12} M L^2}$$
(b) The transverse axis through the end face is parallel to the central transverse axis at distance $d = L/2$.
Applying the Parallel Axis Theorem:
$$I_{\text{end}} = I_{\text{trans}} + M(L/2)^2 = \left(\frac{1}{4} M R^2 + \frac{1}{12} M L^2\right) + \frac{1}{4} M L^2 = \mathbf{\frac{1}{4} M R^2 + \frac{1}{3} M L^2}$$
(c) Numerical evaluations ($M = 6.0, R = 0.40, L = 1.0$):
$$I_{\text{trans}} = \frac{1}{4} (6.0)(0.40)^2 + \frac{1}{12} (6.0)(1.0)^2 = \frac{1}{4}(6.0)(0.16) + \frac{1}{12}(6.0)(1.0) = 0.240 + 0.500 = \mathbf{0.740\text{ kg}\cdot\text{m}^2}$$
$$I_{\text{end}} = 0.240 + \frac{1}{3} (6.0)(1.0)^2 = 0.240 + 2.000 = \mathbf{2.240\text{ kg}\cdot\text{m}^2}$$
Example 3 (Standard JEE Advanced Scenario – Disc with Off-Center Circular Hole):
A uniform thin circular disc of radius $R = 0.60\text{ m}$ has an initial mass $M_0 = 7.20\text{ kg}$. A circular hole of radius $r = R/2 = 0.30\text{ m}$ is punched out such that the hole touches both the outer circumference and the center $O$ of the original disc.
(a) Determine the mass $m_h$ of the removed hole and the remaining mass $M_{\text{rem}}$.
(b) Calculate the moment of inertia $I_z$ of the remaining disc about the central perpendicular axis through $O$.
(c) Calculate the moment of inertia $I_x$ about the diameter passing through the centers of both the disc and the hole.
(d) Calculate the moment of inertia $I_y$ about the diameter perpendicular to the line of centers passing through $O$, and verify that $I_z = I_x + I_y$.
Solution:
(a) Area of original disc: $A_0 = \pi R^2$. Area of hole: $A_h = \pi r^2 = \pi (R/2)^2 = \frac{1}{4}\pi R^2$.
Since the disc is uniform, mass is proportional to area:
$$m_h = \frac{1}{4} M_0 = \frac{1}{4} (7.20\text{ kg}) = 1.80\text{ kg}$$
Remaining mass: $M_{\text{rem}} = M_0 – m_h = 7.20 – 1.80 = \mathbf{5.40\text{ kg}}$, so $M_0 = \frac{4}{3} M_{\text{rem}}$.
(b) Center of the hole $O’$ is at distance $d = R/2 = 0.30\text{ m}$ along the $+x$ axis.
– Complete disc about central $z$-axis: $I_{0, z} = \frac{1}{2} M_0 R^2 = \frac{1}{2}(7.20)(0.36) = 1.296\text{ kg}\cdot\text{m}^2$.
– Hole about its own CM: $I_{h, \text{cm}, z} = \frac{1}{2} m_h r^2 = \frac{1}{2}(1.80)(0.09) = 0.081\text{ kg}\cdot\text{m}^2$.
– Shift hole to $O$ ($d = 0.30\text{ m}$): $I_{h, O, z} = 0.081 + (1.80)(0.30)^2 = 0.081 + 0.162 = 0.243\text{ kg}\cdot\text{m}^2$.
$$I_z = I_{0, z} – I_{h, O, z} = 1.296 – 0.243 = \mathbf{1.053\text{ kg}\cdot\text{m}^2} = \mathbf{\frac{13}{24} M_{\text{rem}} R^2}$$
(c) For the $x$-axis (line of centers):
Both the center of the disc and the center of the hole lie on the $x$-axis ($d = 0$ for both):
– Complete disc: $I_{0, x} = \frac{1}{4} M_0 R^2 = \frac{1}{4}(7.20)(0.36) = 0.648\text{ kg}\cdot\text{m}^2$.
– Hole about $x$-axis: $I_{h, x} = \frac{1}{4} m_h r^2 = \frac{1}{4}(1.80)(0.09) = 0.0405\text{ kg}\cdot\text{m}^2$.
$$I_x = I_{0, x} – I_{h, x} = 0.648 – 0.0405 = \mathbf{0.6075\text{ kg}\cdot\text{m}^2} = \mathbf{\frac{5}{16} M_{\text{rem}} R^2}$$
(d) For the $y$-axis (perpendicular to line of centers through $O$):
The hole center $O’$ is at perpendicular distance $d = R/2 = 0.30\text{ m}$ from the $y$-axis.
– Complete disc: $I_{0, y} = \frac{1}{4} M_0 R^2 = 0.648\text{ kg}\cdot\text{m}^2$.
– Hole about $y$-axis: $I_{h, y} = I_{h, \text{cm}, y} + m_h d^2 = 0.0405 + (1.80)(0.30)^2 = 0.0405 + 0.162 = 0.2025\text{ kg}\cdot\text{m}^2$.
$$I_y = I_{0, y} – I_{h, y} = 0.648 – 0.2025 = \mathbf{0.4455\text{ kg}\cdot\text{m}^2} = \mathbf{\frac{11}{48} M_{\text{rem}} R^2}$$
Verification via Perpendicular Axis Theorem:
$$I_x + I_y = 0.6075 + 0.4455 = 1.053\text{ kg}\cdot\text{m}^2 = I_z$$
$$\frac{5}{16} M_{\text{rem}} R^2 + \frac{11}{48} M_{\text{rem}} R^2 = \left(\frac{15 + 11}{48}\right) M_{\text{rem}} R^2 = \frac{26}{48} M_{\text{rem}} R^2 = \frac{13}{24} M_{\text{rem}} R^2 = I_z$$
Verified!
Example 4 (Edge Case – Square Frame of Four Thin Rods):
Four identical uniform thin rods, each of mass $m = 1.0\text{ kg}$ and length $L = 0.80\text{ m}$, are joined to form a rigid square frame $ABCD$ lying in the $xy$-plane.
(a) Find the moment of inertia $I_z$ about an axis perpendicular to the frame passing through its center $O$.
(b) Find the moment of inertia $I_x$ about an in-plane axis passing through $O$ parallel to side $AB$.
(c) Find the moment of inertia $I_{\text{side}}$ about an axis lying along side $AB$.
(d) Find the moment of inertia $I_{\text{diag}}$ about a diagonal of the square frame.
Solution:
(a) The centre of mass of each rod is at distance $d = L/2$ from center $O$.
For each rod, moment of inertia about its own CM perpendicular to plane is $\frac{1}{12} m L^2$.
By the Parallel Axis Theorem, for one rod about the center $O$:
$$I_1 = \frac{1}{12} m L^2 + m(L/2)^2 = \frac{1}{12} m L^2 + \frac{1}{4} m L^2 = \frac{1}{3} m L^2$$
For all 4 identical rods combined:
$$I_z = 4 \times \left(\frac{1}{3} m L^2\right) = \mathbf{\frac{4}{3} m L^2}$$
Numerical value: $I_z = \frac{4}{3}(1.0)(0.80)^2 = \frac{4}{3}(0.64) = \mathbf{0.8533\text{ kg}\cdot\text{m}^2}$.
(b) For an in-plane axis through $O$ parallel to side $AB$ (let this be the $x$-axis):
– The two sides parallel to the axis (top and bottom) have all their mass at perpendicular distance $y = L/2$:
$$I_{\text{parallel rods}} = 2 \times \left[ m (L/2)^2 \right] = 2 \left(\frac{1}{4} m L^2\right) = \frac{1}{2} m L^2$$
– The two sides perpendicular to the axis (left and right) have the axis passing through their midpoints:
$$I_{\text{perp rods}} = 2 \times \left(\frac{1}{12} m L^2\right) = \frac{1}{6} m L^2$$
Total moment of inertia about $x$-axis:
$$I_x = \frac{1}{2} m L^2 + \frac{1}{6} m L^2 = \mathbf{\frac{2}{3} m L^2}$$
Notice that $I_x = \frac{2}{3} m L^2$ and $I_y = \frac{2}{3} m L^2$, so $I_x + I_y = \frac{4}{3} m L^2 = I_z$, satisfying the Perpendicular Axis Theorem.
(c) For an axis along side $AB$:
– Rod $AB$ lies on the axis: $I_{AB} = 0$.
– Opposite rod $CD$ is at distance $L$: $I_{CD} = m L^2$.
– The two perpendicular rods ($AD$ and $BC$) rotate about one of their ends: $2 \times \left(\frac{1}{3} m L^2\right) = \frac{2}{3} m L^2$.
$$I_{\text{side}} = 0 + m L^2 + \frac{2}{3} m L^2 = \mathbf{\frac{5}{3} m L^2}$$
Equivalently by Parallel Axis Theorem from the parallel central axis: $I_{\text{side}} = I_x + (4m)(L/2)^2 = \frac{2}{3} m L^2 + 4m(L^2/4) = \frac{2}{3} m L^2 + m L^2 = \frac{5}{3} m L^2$. Perfect!
(d) By symmetry and the Perpendicular Axis Theorem, for any two orthogonal in-plane axes through $O$ (including the two diagonals $D_1$ and $D_2$):
$$I_{D_1} + I_{D_2} = I_z = \frac{4}{3} m L^2 \implies I_{\text{diag}} = \frac{1}{2} I_z = \mathbf{\frac{2}{3} m L^2}$$
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
A uniform circular disc of mass $M$ and radius $R$ rotates about an axis tangent to its perimeter and lying in the plane of the disc. Its moment of inertia about this axis is:
(A) $\frac{1}{4} M R^2$
(B) $\frac{1}{2} M R^2$
(C) $\frac{5}{4} M R^2$
(D) $\frac{3}{2} M R^2$
Problem 2 (JEE Main – Single Correct):
A thin circular hoop of mass $M$ and radius $R$ rotates about a tangential axis perpendicular to its plane. Its moment of inertia is:
(A) $M R^2$
(B) $\frac{3}{2} M R^2$
(C) $2 M R^2$
(D) $3 M R^2$
Problem 3 (JEE Main – Single Correct):
A uniform square plate of mass $M$ and side $a$ is rotated about an axis passing through one of its corners perpendicular to its plane. Its moment of inertia is:
(A) $\frac{1}{6} M a^2$
(B) $\frac{1}{3} M a^2$
(C) $\frac{2}{3} M a^2$
(D) $\frac{5}{6} M a^2$
Problem 4 (JEE Main – Single Correct):
The moment of inertia of a uniform solid cylinder of mass $M$, radius $R$, and length $L$ about a transverse axis passing through its centre of mass perpendicular to its length is:
(A) $\frac{1}{2} M R^2 + \frac{1}{12} M L^2$
(B) $\frac{1}{4} M R^2 + \frac{1}{12} M L^2$
(C) $\frac{1}{4} M R^2 + \frac{1}{3} M L^2$
(D) $\frac{1}{2} M R^2 + \frac{1}{3} M L^2$
Problem 5 (JEE Main – Single Correct):
A uniform circular disc of mass $M$ and radius $R$ has a concentric circular hole of radius $R/2$ bored out. The moment of inertia of the remaining annular disc about an axis passing through its diameter in its plane is:
(A) $\frac{5}{16} M R^2$
(B) $\frac{5}{32} M R^2$
(C) $\frac{5}{8} M R^2$
(D) $\frac{1}{4} M R^2$
Problem 6 (JEE Advanced – One or More Correct):
Which of the following statements regarding combined axis theorem applications is/are correct?
(A) The perpendicular axis theorem can be used to find the diametrical moment of inertia of a flat disc, and the parallel axis theorem can then be used to find the in-plane tangential moment of inertia.
(B) For a 3D solid cylinder, the transverse moment of inertia cannot be determined from the perpendicular axis theorem.
(C) In any cavity problem, the moment of inertia of the removed mass must be shifted to the target axis before subtraction.
(D) For any uniform regular hexagonal plate, the moment of inertia about every central in-plane axis is identical.
Problem 7 (JEE Advanced – One or More Correct):
A square frame is constructed from 4 identical thin rods, each of mass $m$ and length $L$:
(A) Its moment of inertia about the central perpendicular axis is $\frac{4}{3} m L^2$.
(B) Its moment of inertia about an in-plane axis bisecting opposite sides is $\frac{2}{3} m L^2$.
(C) Its moment of inertia about an in-plane diagonal is $\frac{2}{3} m L^2$.
(D) Its moment of inertia about an axis along one of the sides is $\frac{5}{3} m L^2$.
Problem 8 (JEE Advanced – One or More Correct):
A thin circular disc of mass $M$ and radius $R$ is compared with a thin circular ring of the same mass $M$ and radius $R$:
(A) The ratio of their in-plane tangential moments of inertia ($I_{\text{tangent, ring}} / I_{\text{tangent, disc}}$) is $6 : 5$.
(B) The ratio of their perpendicular tangential moments of inertia is $4 : 3$.
(C) Their central diametrical moments of inertia are in the ratio $2 : 1$.
(D) Their perpendicular central moments of inertia are in the ratio $2 : 1$.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A uniform circular disc of mass $M = 8.0\text{ kg}$ and radius $R = 1.0\text{ m}$ is rotated about a tangent line lying in the plane of the disc. Calculate its moment of inertia in $\text{kg}\cdot\text{m}^2$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A uniform solid cylinder of mass $M = 12.0\text{ kg}$, radius $R = 1.0\text{ m}$, and length $L = 2.0\text{ m}$ is rotated about a transverse axis passing through its centre of mass perpendicular to its length. Calculate its moment of inertia in $\text{kg}\cdot\text{m}^2$.
Solutions & Explanations
Answer Key Summary:
1. (C) | 2. (C) | 3. (C) | 4. (B) | 5. (A) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (A, B, C, D) | 9. 10 | 10. 7
Solution 1:
$I_d = \frac{1}{4} M R^2$. By the Parallel Axis Theorem: $I = I_d + M R^2 = \frac{1}{4} M R^2 + M R^2 = \frac{5}{4} M R^2$.
Correct Option: (C)
Solution 2:
$I_z = M R^2$. Shifting to a tangent perpendicular to the plane: $I = I_z + M R^2 = M R^2 + M R^2 = 2 M R^2$.
Correct Option: (C)
Solution 3:
$I_{\text{cm}, z} = \frac{1}{6} M a^2$. The distance from center to corner is $d = \frac{a}{\sqrt{2}}$.
By Parallel Axis Theorem: $I = I_{\text{cm}, z} + M d^2 = \frac{1}{6} M a^2 + M \left(\frac{a}{\sqrt{2}}\right)^2 = \frac{1}{6} M a^2 + \frac{1}{2} M a^2 = \frac{4}{6} M a^2 = \frac{2}{3} M a^2$.
Correct Option: (C)
Solution 4:
By slicing the solid cylinder into elemental discs and integrating via the parallel axis theorem: $I_{\text{trans}} = \frac{1}{4} M R^2 + \frac{1}{12} M L^2$.
Correct Option: (B)
Solution 5:
For an annular disc of mass $M$ with inner radius $R_1 = R/2$ and outer radius $R_2 = R$, the central perpendicular moment of inertia is $I_z = \frac{1}{2} M (R_1^2 + R_2^2) = \frac{1}{2} M \left(\frac{R^2}{4} + R^2\right) = \frac{5}{8} M R^2$.
By the Perpendicular Axis Theorem, for any in-plane diameter: $I_d = \frac{1}{2} I_z = \frac{1}{2} \left(\frac{5}{8} M R^2\right) = \frac{5}{16} M R^2$.
Correct Option: (A)
Solution 6:
– (A) True: Standard two-step procedure.
– (B) True: Perpendicular axis theorem is restricted strictly to 2D laminas.
– (C) True: In cavity problems, moments of inertia add/subtract algebraically only when computed about the identical reference axis.
– (D) True: Hexagonal planar symmetry forces all central in-plane axes to have $I = \frac{1}{2}I_z$.
Correct Options: (A, B, C, D)
Solution 7:
All four options (A, B, C, D) are verified in Example 4.
– $I_z = 4 \times \left(\frac{1}{12}m L^2 + m(L/2)^2\right) = \frac{4}{3}m L^2$.
– $I_x = I_{\text{diag}} = \frac{2}{3}m L^2$.
– $I_{\text{side}} = \frac{5}{3}m L^2$.
Correct Options: (A, B, C, D)
Solution 8:
– (A) True: $\frac{I_{t, \parallel, \text{ring}}}{I_{t, \parallel, \text{disc}}} = \frac{\frac{3}{2}M R^2}{\frac{5}{4}M R^2} = \frac{6}{5}$.
– (B) True: $\frac{I_{t, \perp, \text{ring}}}{I_{t, \perp, \text{disc}}} = \frac{2 M R^2}{\frac{3}{2} M R^2} = \frac{4}{3}$.
– (C) True: $\frac{I_{d, \text{ring}}}{I_{d, \text{disc}}} = \frac{\frac{1}{2}M R^2}{\frac{1}{4}M R^2} = 2 : 1$.
– (D) True: $\frac{I_{z, \text{ring}}}{I_{z, \text{disc}}} = \frac{M R^2}{\frac{1}{2}M R^2} = 2 : 1$.
All statements (A, B, C, D) are correct.
Correct Options: (A, B, C, D)
Solution 9:
$I = \frac{5}{4} M R^2 = \frac{5}{4} (8.0\text{ kg})(1.0\text{ m})^2 = \mathbf{10.0\text{ kg}\cdot\text{m}^2}$.
Correct Answer: 10
Solution 10:
$$I_{\text{trans}} = \frac{1}{4} M R^2 + \frac{1}{12} M L^2 = \frac{1}{4} (12.0\text{ kg})(1.0\text{ m})^2 + \frac{1}{12} (12.0\text{ kg})(2.0\text{ m})^2$$
$$I_{\text{trans}} = 3.0(1.0) + 1.0(4.0) = 3.0 + 4.0 = \mathbf{7.0\text{ kg}\cdot\text{m}^2}$$
Correct Answer: 7