Concept Card: Moment of Inertia of Spheres – Solid & Hollow Spherical Shells
1. Physical Framework & Spherical Symmetry:
Unlike asymmetric or planar bodies, a uniform sphere possesses complete isotropic symmetry in three dimensions. Consequently:
- Every axis passing through the center of a uniform sphere is a principal diametrical axis. There is no distinction between “perpendicular” and “in-plane” axes; every central axis yields an identical moment of inertia:
$$I_x = I_y = I_z = I_{\text{cm}}$$ - Spherical Symmetry Theorem: For any spherically symmetric mass distribution centered at the origin:
$$I_x + I_y + I_z = \int (y^2 + z^2)dm + \int (x^2 + z^2)dm + \int (x^2 + y^2)dm = 2\int (x^2 + y^2 + z^2)dm = 2\int r^2 dm$$
Since $I_x = I_y = I_z = I_d$, we obtain the remarkably elegant relation:
$$\mathbf{3 I_d = 2\int r^2 dm \implies I_d = \frac{2}{3} \int r^2 dm}$$
2. Thin Spherical Shell / Hollow Sphere (Mass $M$, Radius $R$)
All mass $M$ of an infinitesimally thin spherical shell is located at a constant radial distance $r = R$ from the center.
- Diametrical Axis ($I_{\text{cm}}$):
Applying the spherical symmetry theorem directly, since $r = R$ for all mass elements:
$$I_{\text{cm}} = \frac{2}{3} \int R^2 dm = \frac{2}{3} R^2 \int dm = \mathbf{\frac{2}{3} M R^2}$$
Radius of gyration: $\mathbf{k = \sqrt{\frac{2}{3}} R \approx 0.8165 R}$. - Tangential Axis ($I_{\text{tangent}}$):
By the Parallel Axis Theorem, for an axis tangent to the outer spherical surface ($d = R$):
$$I_{\text{tangent}} = I_{\text{cm}} + M R^2 = \frac{2}{3} M R^2 + M R^2 = \mathbf{\frac{5}{3} M R^2}$$
Radius of gyration: $\mathbf{k_t = \sqrt{\frac{5}{3}} R \approx 1.291 R}$.
3. Uniform Solid Sphere (Mass $M$, Radius $R$)
Mass is distributed uniformly throughout the volume $V = \frac{4}{3}\pi R^3$ with constant volume density $\rho = \frac{M}{\frac{4}{3}\pi R^3} = \frac{3M}{4\pi R^3}$.
- Diametrical Axis ($I_{\text{cm}}$) – Concentric Shell Method:
Decompose the solid sphere into thin concentric spherical shells of radius $x$ and thickness $dx$ ($0 \le x \le R$).
Volume of elemental shell: $dV = 4\pi x^2\,dx$.
Mass of elemental shell: $dm = \rho\,dV = \left(\frac{3M}{4\pi R^3}\right)(4\pi x^2\,dx) = \frac{3M}{R^3} x^2\,dx$.
Each thin shell has a diametrical moment of inertia $dI = \frac{2}{3} x^2 dm$. Integrating over all shells:
$$I_{\text{cm}} = \int_0^R \frac{2}{3} x^2 \left(\frac{3M}{R^3} x^2\,dx\right) = \frac{2M}{R^3} \int_0^R x^4\,dx = \frac{2M}{R^3} \left[\frac{x^5}{5}\right]_0^R = \mathbf{\frac{2}{5} M R^2}$$
Radius of gyration: $\mathbf{k = \sqrt{\frac{2}{5}} R \approx 0.6325 R}$. - Tangential Axis ($I_{\text{tangent}}$):
Applying the Parallel Axis Theorem for a tangent line ($d = R$):
$$I_{\text{tangent}} = I_{\text{cm}} + M R^2 = \frac{2}{5} M R^2 + M R^2 = \mathbf{\frac{7}{5} M R^2}$$
Radius of gyration: $\mathbf{k_t = \sqrt{\frac{7}{5}} R \approx 1.183 R}$.
4. Thick Spherical Shell (Inner Radius $R_1$, Outer Radius $R_2$, Mass $M$)
For a hollow spherical body with finite wall thickness between radii $R_1$ and $R_2$:
$$\mathbf{I = \frac{2}{5} M \left(\frac{R_2^5 – R_1^5}{R_2^3 – R_1^3}\right)}$$
- Limiting Case 1 (Solid Sphere): As $R_1 \to 0$, $I \to \frac{2}{5} M R_2^2$.
- Limiting Case 2 (Thin Shell): As $R_1 \to R_2 = R$, by L’Hôpital’s rule $\frac{R_2^5 – R_1^5}{R_2^3 – R_1^3} \to \frac{5 R^4 dR}{3 R^2 dR} = \frac{5}{3} R^2$, yielding $I \to \frac{2}{5}M\left(\frac{5}{3}R^2\right) = \frac{2}{3} M R^2$.
5. Rolling Dynamics Comparison on an Inclined Plane
When a rigid body of mass $M$, radius $R$, and radius of gyration $k$ rolls without slipping down an incline of angle $\theta$:
$$\mathbf{a_{\text{rolling}} = \frac{g\sin\theta}{1 + \frac{k^2}{R^2}} = \frac{g\sin\theta}{1 + \frac{I_{\text{cm}}}{M R^2}}}$$
- Solid Sphere: $\frac{k^2}{R^2} = \frac{2}{5} = 0.40 \implies a_{\text{solid}} = \frac{g\sin\theta}{1 + 0.40} = \mathbf{\frac{5}{7} g\sin\theta \approx 0.714 g\sin\theta}$.
- Hollow Sphere: $\frac{k^2}{R^2} = \frac{2}{3} \approx 0.667 \implies a_{\text{hollow}} = \frac{g\sin\theta}{1 + 2/3} = \mathbf{\frac{3}{5} g\sin\theta = 0.600 g\sin\theta}$.
- Crucial JEE Takeaway: Because $\frac{k^2}{R^2}_{\text{solid}} < \frac{k^2}{R^2}_{\text{hollow}}$, the solid sphere has smaller rotational inertia, accelerates faster, and always wins the race down the incline ($t_{\text{solid}} < t_{\text{hollow}}$).
6. Common JEE Pitfalls & Traps
- Trap 1 (Coefficient Confusion): Never confuse $\frac{2}{5}$ with $\frac{2}{3}$. Physical check: In the hollow shell, all mass is located at the outermost boundary $R$, so its rotational resistance is strictly greater than that of the solid sphere: $\frac{2}{3} = 0.667 > \frac{2}{5} = 0.400$.
- Trap 2 (Tangent Axis Parallel Shift): Always remember to add $M R^2$ when shifting to a tangent line:
$$\text{Solid: } \frac{2}{5} + 1 = \frac{7}{5} = 1.4 M R^2 \quad | \quad \text{Hollow: } \frac{2}{3} + 1 = \frac{5}{3} \approx 1.67 M R^2$$ - Trap 3 (Energy Partitioning in Pure Rolling): In pure rolling ($v = \omega R$), the fraction of total kinetic energy stored in rotation is $\frac{K_{\text{rot}}}{K_{\text{tot}}} = \frac{k^2/R^2}{1 + k^2/R^2}$. For a solid sphere, this is $\frac{2/5}{7/5} = \frac{2}{7} \approx 28.6\%$; for a hollow sphere, it is $\frac{2/3}{5/3} = \frac{2}{5} = 40.0\%$.
Solved Examples
Example 1 (Direct Application – Diametrical vs. Tangential Dynamics):
A uniform solid sphere of mass $M = 5.0\text{ kg}$ and radius $R = 0.40\text{ m}$ is mounted on a frictionless axle.
(a) Calculate the moment of inertia $I_d$ and radius of gyration $k_d$ about a diametrical axis passing through its center.
(b) Calculate the moment of inertia $I_t$ and radius of gyration $k_t$ about a tangent line touching its surface.
(c) Determine the torque $\tau$ required to accelerate the sphere from rest to an angular speed $\omega = 30.0\text{ rad/s}$ in $t = 3.0\text{ s}$ about both axes.
Solution:
(a) For the diametrical axis:
$$I_d = \frac{2}{5} M R^2 = \frac{2}{5} (5.0\text{ kg}) (0.40\text{ m})^2 = \frac{2}{5} (5.0) (0.16) = \mathbf{0.320\text{ kg}\cdot\text{m}^2}$$
Radius of gyration:
$$k_d = \sqrt{\frac{I_d}{M}} = \sqrt{\frac{0.320}{5.0}} = \sqrt{0.064} = \sqrt{\frac{2}{5}} (0.40) \approx \mathbf{0.2530\text{ m}}$$
(b) For the tangential axis, by the Parallel Axis Theorem ($d = R$):
$$I_t = I_d + M R^2 = \frac{7}{5} M R^2 = \frac{7}{5} (5.0) (0.16) = 7(0.16) = \mathbf{1.120\text{ kg}\cdot\text{m}^2}$$
Radius of gyration:
$$k_t = \sqrt{\frac{I_t}{M}} = \sqrt{\frac{1.120}{5.0}} = \sqrt{0.224} = \sqrt{\frac{7}{5}} (0.40) \approx \mathbf{0.4733\text{ m}}$$
(c) Angular acceleration required:
$$\alpha = \frac{\omega – \omega_0}{t} = \frac{30.0 – 0}{3.0} = 10.0\text{ rad/s}^2$$
Torque about diametrical axis:
$$\tau_d = I_d \alpha = (0.320\text{ kg}\cdot\text{m}^2)(10.0\text{ rad/s}^2) = \mathbf{3.20\text{ N}\cdot\text{m}}$$
Torque about tangential axis:
$$\tau_t = I_t \alpha = (1.120\text{ kg}\cdot\text{m}^2)(10.0\text{ rad/s}^2) = \mathbf{11.20\text{ N}\cdot\text{m}}$$
Rotating about a tangent requires $3.5$ times more torque than about the center!
Example 2 (Rolling Race Down an Incline – Solid vs. Hollow Sphere):
A solid sphere and a thin hollow sphere of identical mass $M$ and identical outer radius $R$ are released simultaneously from rest at the top of a rough inclined plane of length $s = 7.0\text{ m}$ and inclination angle $\theta = 30^\circ$. Both roll without slipping. Take $g = 9.8\text{ m/s}^2$.
(a) Calculate the linear acceleration of each sphere down the plane.
(b) Find the ratio of the times taken by the spheres to reach the bottom ($t_{\text{hollow}} / t_{\text{solid}}$).
(c) Calculate the fraction of total kinetic energy that exists in rotational form for each sphere at the bottom.
Solution:
(a) In pure rolling down an incline, $a = \frac{g\sin\theta}{1 + I_{\text{cm}}/(M R^2)}$:
– For the solid sphere ($I_{\text{cm}} = \frac{2}{5}M R^2$):
$$a_{\text{solid}} = \frac{g\sin 30^\circ}{1 + 2/5} = \frac{5}{7} g \left(\frac{1}{2}\right) = \frac{5}{14} g = \frac{5}{14}(9.8) = \mathbf{3.50\text{ m/s}^2}$$
– For the hollow sphere ($I_{\text{cm}} = \frac{2}{3}M R^2$):
$$a_{\text{hollow}} = \frac{g\sin 30^\circ}{1 + 2/3} = \frac{3}{5} g \left(\frac{1}{2}\right) = \frac{3}{10} g = \frac{3}{10}(9.8) = \mathbf{2.94\text{ m/s}^2}$$
(b) Since $s = \frac{1}{2} a t^2 \implies t = \sqrt{\frac{2s}{a}}$, for the same distance $s$:
$$\frac{t_{\text{hollow}}}{t_{\text{solid}}} = \sqrt{\frac{a_{\text{solid}}}{a_{\text{hollow}}}} = \sqrt{\frac{5/14}{3/10}} = \sqrt{\frac{5}{14} \times \frac{10}{3}} = \sqrt{\frac{25}{21}} \approx \mathbf{1.091}$$
The solid sphere reaches the bottom approximately $9.1\%$ faster.
(c) Total kinetic energy: $K_{\text{tot}} = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2}M v^2 + \frac{1}{2}I_{\text{cm}}\omega^2 = \frac{1}{2}M v^2\left(1 + \frac{k^2}{R^2}\right)$.
$$\frac{K_{\text{rot}}}{K_{\text{tot}}} = \frac{\frac{k^2}{R^2}}{1 + \frac{k^2}{R^2}}$$
– For solid sphere: $\frac{2/5}{1 + 2/5} = \frac{2/5}{7/5} = \mathbf{\frac{2}{7} \approx 28.57\%}$.
– For hollow sphere: $\frac{2/3}{1 + 2/3} = \frac{2/3}{5/3} = \mathbf{\frac{2}{5} = 40.00\%}$.
Example 3 (Standard JEE Advanced Scenario – Excavated Cavity in a Solid Sphere):
A uniform solid sphere of radius $R$ and uniform mass density has an initial mass $M_0$. A spherical cavity of radius $r = R/2$ is scooped out such that it touches the outer surface and passes through the center $O$ of the original sphere.
(a) Determine the mass $m_{\text{cav}}$ of the excavated sphere and the remaining mass $M$ in terms of $M_0$.
(b) Calculate the moment of inertia $I_x$ of the remaining body about the $x$-axis (the diameter passing through the center of both the original sphere and the cavity).
(c) Calculate the moment of inertia $I_z$ of the remaining body about an axis through $O$ perpendicular to the line of centers.
Solution:
(a) Volume of original sphere $V_0 = \frac{4}{3}\pi R^3$. Volume of cavity $V_{\text{cav}} = \frac{4}{3}\pi (R/2)^3 = \frac{1}{8}V_0$.
Since density $\rho$ is uniform:
$$m_{\text{cav}} = \frac{1}{8} M_0 \implies M = M_0 – m_{\text{cav}} = \frac{7}{8} M_0 \implies M_0 = \frac{8}{7} M$$
(b) Let the $x$-axis connect the center of the sphere $O(0, 0, 0)$ to the center of the cavity $O'(R/2, 0, 0)$.
Since both centers lie directly on the $x$-axis, this axis is a diameter for both the original sphere and the cavity ($d = 0$ for both):
$$I_{0, x} = \frac{2}{5} M_0 R^2$$
$$I_{\text{cav}, x} = \frac{2}{5} m_{\text{cav}} r^2 = \frac{2}{5} \left(\frac{M_0}{8}\right) \left(\frac{R}{2}\right)^2 = \frac{2}{5} \left(\frac{M_0}{8}\right) \left(\frac{R^2}{4}\right) = \frac{1}{32} \left(\frac{2}{5} M_0 R^2\right)$$
By the superposition subtraction principle:
$$I_x = I_{0, x} – I_{\text{cav}, x} = \left(1 – \frac{1}{32}\right) \left(\frac{2}{5} M_0 R^2\right) = \frac{31}{32} \left(\frac{2}{5} M_0 R^2\right) = \mathbf{\frac{31}{80} M_0 R^2}$$
In terms of remaining mass $M = \frac{7}{8} M_0 \implies M_0 = \frac{8}{7}M$:
$$I_x = \frac{31}{80} \left(\frac{8}{7}M\right) R^2 = \mathbf{\frac{31}{70} M R^2}$$
(c) For the $z$-axis through $O$ perpendicular to the line of centers:
The original sphere has $I_{0, z} = \frac{2}{5} M_0 R^2$.
The center of the cavity $O’$ is at perpendicular distance $d = R/2$ from the $z$-axis. Applying the Parallel Axis Theorem for the cavity:
$$I_{\text{cav}, z} = I_{\text{cav}, \text{cm}} + m_{\text{cav}} d^2 = \frac{2}{5} m_{\text{cav}} r^2 + m_{\text{cav}} \left(\frac{R}{2}\right)^2 = m_{\text{cav}} \left[\frac{2}{5}\left(\frac{R^2}{4}\right) + \frac{R^2}{4}\right] = m_{\text{cav}} \left(\frac{R^2}{4}\right) \left(\frac{7}{5}\right) = \frac{7}{20} m_{\text{cav}} R^2$$
Substitute $m_{\text{cav}} = \frac{1}{8}M_0$:
$$I_{\text{cav}, z} = \frac{7}{20} \left(\frac{M_0}{8}\right) R^2 = \frac{7}{160} M_0 R^2$$
Subtracting from the original sphere ($I_{0, z} = \frac{2}{5}M_0 R^2 = \frac{64}{160} M_0 R^2$):
$$I_z = I_{0, z} – I_{\text{cav}, z} = \frac{64 – 7}{160} M_0 R^2 = \mathbf{\frac{57}{160} M_0 R^2}$$
In terms of remaining mass $M$ ($M_0 = \frac{8}{7}M$):
$$I_z = \frac{57}{160} \left(\frac{8}{7}M\right) R^2 = \mathbf{\frac{57}{140} M R^2}$$
Example 4 (Edge Case – Spherically Symmetric Radial Density Gradient):
A planet-sized solid sphere of radius $R$ has a non-uniform density that increases linearly from center to surface according to $\rho(r) = \rho_0 \left(\frac{r}{R}\right)$, where $\rho_0$ is a positive constant.
(a) Determine the total mass $M$ of the sphere in terms of $\rho_0$ and $R$.
(b) Derive the moment of inertia $I$ about any diameter in terms of $M$ and $R$.
(c) Find the radius of gyration $k$ and compare the result with a uniform solid sphere.
Solution:
(a) Decompose into thin concentric shells of radius $r$ and thickness $dr$ ($0 \le r \le R$).
Mass element: $dm = \rho(r) (4\pi r^2\,dr) = \rho_0 \left(\frac{r}{R}\right) (4\pi r^2\,dr) = \frac{4\pi \rho_0}{R} r^3\,dr$.
Total mass $M$:
$$M = \int_0^R dm = \frac{4\pi \rho_0}{R} \int_0^R r^3\,dr = \frac{4\pi \rho_0}{R} \left[\frac{r^4}{4}\right]_0^R = \mathbf{\pi \rho_0 R^3}$$
(b) Each elemental spherical shell has moment of inertia $dI = \frac{2}{3} r^2 dm$ about any diameter:
$$dI = \frac{2}{3} r^2 \left(\frac{4\pi \rho_0}{R} r^3\,dr\right) = \frac{8\pi \rho_0}{3R} r^5\,dr$$
Integrating over the entire sphere:
$$I = \int_0^R dI = \frac{8\pi \rho_0}{3R} \int_0^R r^5\,dr = \frac{8\pi \rho_0}{3R} \left[\frac{r^6}{6}\right]_0^R = \frac{8\pi \rho_0 R^5}{18} = \mathbf{\frac{4}{9} \pi \rho_0 R^5}$$
Substitute $\pi \rho_0 R^3 = M$:
$$I = \frac{4}{9} \left(\pi \rho_0 R^3\right) R^2 = \mathbf{\frac{4}{9} M R^2}$$
(c) Radius of gyration:
$$k = \sqrt{\frac{I}{M}} = \sqrt{\frac{4}{9}R^2} = \mathbf{\frac{2}{3} R \approx 0.667 R}$$
Comparative Takeaway: For a uniform solid sphere, $I = \frac{2}{5}M R^2 = 0.400 M R^2$. Here, $I = \frac{4}{9}M R^2 \approx 0.444 M R^2$. Because density increases toward the surface, more mass is located at larger radial distances, yielding a higher moment of inertia.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
The ratio of the moment of inertia of a thin hollow spherical shell to that of a uniform solid sphere of the same mass $M$ and radius $R$ about their respective diameters is:
(A) $3 : 5$
(B) $5 : 3$
(C) $2 : 5$
(D) $5 : 2$
Problem 2 (JEE Main – Single Correct):
A uniform solid sphere of mass $M$ and radius $R$ rotates about a tangent to its surface. The radius of gyration of the sphere about this tangential axis is:
(A) $\sqrt{\frac{2}{5}} R$
(B) $\sqrt{\frac{7}{5}} R$
(C) $\sqrt{\frac{5}{3}} R$
(D) $\frac{7}{5} R$
Problem 3 (JEE Main – Single Correct):
A solid sphere rolls down an inclined plane without slipping. The percentage of its total kinetic energy that is associated with rotational motion is:
(A) $20.0\%$
(B) $28.6\%$
(C) $40.0\%$
(D) $50.0\%$
Problem 4 (JEE Main – Single Correct):
Two solid spheres are made of the same uniform material. The radius of the second sphere is double the radius of the first sphere ($R_2 = 2 R_1$). The ratio of their moments of inertia about their diameters ($I_2 / I_1$) is:
(A) $4$
(B) $8$
(C) $16$
(D) $32$
Problem 5 (JEE Main – Single Correct):
A thin spherical shell of mass $M$ and radius $R$ is rolling without slipping on a horizontal surface with linear speed $v$. Its total kinetic energy is:
(A) $\frac{1}{2} M v^2$
(B) $\frac{7}{10} M v^2$
(C) $\frac{5}{6} M v^2$
(D) $M v^2$
Problem 6 (JEE Advanced – One or More Correct):
A solid sphere ($S$) and a thin hollow spherical shell ($H$) of identical mass $M$ and outer radius $R$ are placed on a rough horizontal surface and given identical forward linear impulses:
(A) The solid sphere has a smaller radius of gyration than the hollow sphere.
(B) For pure rolling motion with the same linear velocity $v$, the hollow sphere possesses greater total kinetic energy.
(C) Under the action of the same tangential braking torque $\tau$, the solid sphere exhibits greater angular deceleration.
(D) When released from rest on an inclined plane, the solid sphere experiences a smaller frictional force than the hollow sphere during pure rolling.
Problem 7 (JEE Advanced – One or More Correct):
For any spherically symmetric mass distribution centered at the origin:
(A) All diametrical axes passing through the center have identical moments of inertia.
(B) The moment of inertia about the center of mass is strictly minimum compared to any other parallel axis.
(C) The relation $I_x + I_y + I_z = 2\int r^2 dm$ holds true.
(D) The perpendicular axis theorem $I_z = I_x + I_y$ is strictly applicable to a 3D solid sphere.
Problem 8 (JEE Advanced – One or More Correct):
A uniform solid sphere of mass $M$ and radius $R$ has a concentric spherical cavity of radius $R/2$ excavated:
(A) The mass of the material removed is $M/8$ of the original solid sphere.
(B) The moment of inertia of the remaining shell about its diameter is $\frac{31}{70} M_{\text{rem}} R^2$.
(C) The radius of gyration of the remaining body about its diameter is $\sqrt{\frac{31}{70}} R$.
(D) The excavated sphere had a moment of inertia of $\frac{1}{32}$ of the original sphere about the same diameter.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A thin hollow spherical shell of mass $M = 6.0\text{ kg}$ and radius $R = 2.0\text{ m}$ rotates about an axis tangent to its surface. Calculate its moment of inertia about this tangential axis in $\text{kg}\cdot\text{m}^2$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A uniform solid sphere of mass $M = 2.5\text{ kg}$ has a moment of inertia of $I_d = 0.36\text{ kg}\cdot\text{m}^2$ about its diameter. Calculate the radius of the sphere (in $\text{cm}$).
Solutions & Explanations
Answer Key Summary:
1. (B) | 2. (B) | 3. (B) | 4. (D) | 5. (C) | 6. (A, B, C, D) | 7. (A, B, C) | 8. (A, B, C, D) | 9. 40 | 10. 60
Solution 1:
$I_{\text{hollow}} = \frac{2}{3} M R^2$ and $I_{\text{solid}} = \frac{2}{5} M R^2$.
$$\frac{I_{\text{hollow}}}{I_{\text{solid}}} = \frac{2/3}{2/5} = \frac{5}{3}$$
Correct Option: (B)
Solution 2:
$I_{\text{tangent}} = I_{\text{cm}} + M R^2 = \frac{2}{5} M R^2 + M R^2 = \frac{7}{5} M R^2$.
$$k_t = \sqrt{\frac{I_{\text{tangent}}}{M}} = \sqrt{\frac{7}{5}} R$$
Correct Option: (B)
Solution 3:
$K_{\text{rot}} = \frac{1}{2} I \omega^2 = \frac{1}{2}\left(\frac{2}{5}M R^2\right)\left(\frac{v}{R}\right)^2 = \frac{1}{5} M v^2$.
$K_{\text{trans}} = \frac{1}{2} M v^2$.
$K_{\text{tot}} = \frac{1}{2} M v^2 + \frac{1}{5} M v^2 = \frac{7}{10} M v^2$.
$$\frac{K_{\text{rot}}}{K_{\text{tot}}} = \frac{1/5}{7/10} = \frac{2}{7} \approx 28.57\%$$
Correct Option: (B)
Solution 4:
$M = \rho \left(\frac{4}{3}\pi R^3\right) \propto R^3$.
$$I = \frac{2}{5} M R^2 \propto (R^3)(R^2) = R^5$$
$$\frac{I_2}{I_1} = \left(\frac{R_2}{R_1}\right)^5 = 2^5 = 32$$
Correct Option: (D)
Solution 5:
For a thin spherical shell, $I = \frac{2}{3} M R^2$. In pure rolling, $\omega = v/R$.
$$K_{\text{tot}} = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2 = \frac{1}{2} M v^2 + \frac{1}{2}\left(\frac{2}{3}M R^2\right)\left(\frac{v^2}{R^2}\right) = \frac{1}{2} M v^2 + \frac{1}{3} M v^2 = \frac{5}{6} M v^2$$
Correct Option: (C)
Solution 6:
– (A) True: $k_S = \sqrt{2/5}R \approx 0.632 R < k_H = \sqrt{2/3}R \approx 0.816 R$.
– (B) True: $K_{\text{tot}} = \frac{1}{2}M v^2(1 + k^2/R^2)$. Since $k_H > k_S$, $K_{H} > K_S$.
– (C) True: $\alpha = \tau/I$. Since $I_S < I_H$, $\alpha_S > \alpha_H$.
– (D) True: On an incline, friction required for pure rolling is $f = \frac{M g\sin\theta}{1 + M R^2/I_{\text{cm}}}$. Since $I_S < I_H$, $f_{\text{solid}} < f_{\text{hollow}}$.
Correct Options: (A, B, C, D)
Solution 7:
– (A) True: Spherical symmetry implies all central axes are identical.
– (B) True: By Parallel Axis Theorem $I = I_{\text{cm}} + M d^2$, $I_{\text{cm}}$ is the absolute minimum.
– (C) True: Sum of perpendicular distances squared equals $2\int r^2 dm$.
– (D) False: The perpendicular axis theorem ($I_z = I_x + I_y$) is valid strictly for 2D planar laminas, NOT for 3D solid bodies!
Correct Options: (A, B, C)
Solution 8:
– (A) True: Volume of cavity is $(1/2)^3 = 1/8$ of original sphere, so mass removed is $M_0/8$.
– (B) True: $I = \frac{31}{70} M_{\text{rem}} R^2$.
– (C) True: $k = \sqrt{I / M_{\text{rem}}} = \sqrt{\frac{31}{70}} R$.
– (D) True: $I_{\text{cav}} = \frac{2}{5} (M_0/8) (R/2)^2 = \frac{1}{32} \left(\frac{2}{5} M_0 R^2\right)$.
Correct Options: (A, B, C, D)
Solution 9:
For a thin spherical shell about a tangent:
$$I_{\text{tangent}} = \frac{5}{3} M R^2 = \frac{5}{3} (6.0\text{ kg}) (2.0\text{ m})^2 = \frac{5}{3} (6.0) (4.0) = 5(2.0)(4.0) = \mathbf{40.0\text{ kg}\cdot\text{m}^2}$$
Correct Answer: 40
Solution 10:
For a solid sphere about its diameter:
$$I_d = \frac{2}{5} M R^2 \implies 0.36 = \frac{2}{5} (2.5) R^2 = 1.0 R^2 \implies R^2 = 0.36\text{ m}^2 \implies R = 0.60\text{ m}$$
In centimeters: $R = 0.60 \times 100 = \mathbf{60\text{ cm}}$.
Correct Answer: 60