Concept Card: Centre of Mass of a Rigid Body & Continuous Mass Distribution
1. Physical Framework & Integral Formulations:
For extended rigid bodies comprising a continuous distribution of matter, the discrete sum $\Sigma m_i \vec{r}_i$ transforms into a definite integral over infinitesimal mass elements $dm$:
$$\mathbf{\vec{r}_{\text{cm}} = \frac{\int \vec{r}\,dm}{\int dm} = \frac{1}{M}\int \vec{r}\,dm}$$
where $M = \int dm$ is the total mass of the body.
- Cartesian Coordinate Components:
$$x_{\text{cm}} = \frac{1}{M}\int x\,dm, \quad y_{\text{cm}} = \frac{1}{M}\int y\,dm, \quad z_{\text{cm}} = \frac{1}{M}\int z\,dm$$ - Differential Elements ($dm$) by Dimensionality:
- 1D Bodies (Thin wires, rods, rings): $dm = \lambda\,dx$ or $dm = \lambda R\,d\theta$, where $\lambda = \frac{dM}{dL}$ is linear mass density ($\text{kg/m}$).
- 2D Bodies (Laminae, thin plates, discs): $dm = \sigma\,dA = \sigma(2\pi r\,dr)$, where $\sigma = \frac{dM}{dA}$ is surface mass density ($\text{kg/m}^2$).
- 3D Bodies (Solid spheres, cones, cylinders): $dm = \rho\,dV$, where $\rho = \frac{dM}{dV}$ is volumetric mass density ($\text{kg/m}^3$).
2. The Role of Geometric & Mass Symmetry
- Plane of Symmetry: If a body exhibits reflectional symmetry across a geometric plane, its centre of mass MUST lie within that plane.
- Axis of Symmetry: If a body possesses an axis of rotational symmetry (or is the intersection of two orthogonal reflection planes), its centre of mass MUST lie on that axis.
- Inversion Center: For any homogeneous body possessing a geometric center of inversion (uniform spheres, circular rings, circular discs, rectangular blocks, cylinders), the centre of mass coincides exactly with its geometric centroid.
3. Standard Continuous Bodies Reference Table (High-Yield for JEE)
| Rigid Body Geometry | Mass Distribution | Reference Origin / Axis | Centre of Mass Location |
|---|---|---|---|
| Uniform Semicircular Ring | 1D wire, radius $R$ | Center of curvature, diameter along $x$-axis | $x_{\text{cm}} = 0, \quad y_{\text{cm}} = \frac{2R}{\pi} \approx 0.637 R$ |
| Uniform Semicircular Disc | 2D lamina, radius $R$ | Center of flat base, diameter along $x$-axis | $x_{\text{cm}} = 0, \quad y_{\text{cm}} = \frac{4R}{3\pi} \approx 0.424 R$ |
| Uniform Hollow Hemispherical Shell | 2D surface, radius $R$ | Center of circular flat base | $y_{\text{cm}} = \frac{R}{2} = 0.500 R$ |
| Uniform Solid Hemisphere | 3D solid, radius $R$ | Center of circular flat base | $y_{\text{cm}} = \frac{3R}{8} = 0.375 R$ |
| Uniform Hollow Right Circular Cone | 2D shell, height $H$ | Center of circular base | $y_{\text{cm}} = \frac{H}{3} \quad$ (from base) |
| Uniform Solid Right Circular Cone | 3D solid, height $H$ | Center of flat circular base | $y_{\text{cm}} = \frac{H}{4} \quad$ (from base) |
| Uniform Triangular Lamina | 2D plate, height $H$ | Base of triangle | $y_{\text{cm}} = \frac{H}{3} \quad$ (centroid of medians) |
4. The Negative Mass Method for Cavity Problems
When a portion of mass $m_{\text{cut}}$ is removed from an original complete body of mass $M_{\text{orig}}$, the remaining object is treated as the superposition of the original complete body ($M_{\text{orig}}, \vec{r}_{\text{orig}}$) and a fictitious negative mass ($-m_{\text{cut}}, \vec{r}_{\text{cut}}$):
$$\mathbf{\vec{r}_{\text{rem}} = \frac{M_{\text{orig}} \vec{r}_{\text{orig}} – m_{\text{cut}} \vec{r}_{\text{cut}}}{M_{\text{orig}} – m_{\text{cut}}}}$$
- For uniform planar laminae: $\vec{r}_{\text{rem}} = \frac{A_{\text{orig}} \vec{r}_{\text{orig}} – A_{\text{cut}} \vec{r}_{\text{cut}}}{A_{\text{orig}} – A_{\text{cut}}}$.
- For uniform 3D solids: $\vec{r}_{\text{rem}} = \frac{V_{\text{orig}} \vec{r}_{\text{orig}} – V_{\text{cut}} \vec{r}_{\text{cut}}}{V_{\text{orig}} – V_{\text{cut}}}$.
5. Common JEE Pitfalls & Traps
- Trap 1 (Cone Reference Datum): The CM of a solid cone is at $H/4$ measured from the base, but at $3H/4$ measured from the apex. Always check which datum the problem specifies.
- Trap 2 (Hollow vs. Solid Hemispheres): Do not confuse $R/2$ (hollow hemispherical shell) with $3R/8$ (solid hemisphere) or $4R/(3\pi)$ (semicircular disc).
- Trap 3 (Variable Density Integration): When density is non-uniform ($\lambda(x) = \lambda_0 x^n$), the CM cannot be found by geometric symmetry. Both $\int x \lambda(x)\,dx$ and $\int \lambda(x)\,dx$ must be evaluated independently.
Solved Examples
Example 1 (Direct First-Principles Derivations – Semicircular Ring & Disc):
(a) Derive from first principles the location of the centre of mass of a thin uniform semicircular wire of radius $R$.
(b) Using elemental semicircular rings, derive the centre of mass of a uniform semicircular thin disc of radius $R$.
Solution:
(a) Place the center of curvature at the origin $O(0, 0)$ with the diameter lying along the $x$-axis. By reflectional symmetry about the $y$-axis, $x_{\text{cm}} = 0$.
Consider an infinitesimal element $d\theta$ at angle $\theta$ from the positive $x$-axis ($0 \le \theta \le \pi$):
– Arc length: $dL = R\,d\theta$, mass: $dm = \lambda R\,d\theta$, coordinate: $y = R\sin\theta$.
Total mass: $M = \lambda \pi R$.
$$y_{\text{cm}} = \frac{1}{M}\int_0^\pi y\,dm = \frac{1}{\lambda \pi R}\int_0^\pi (R\sin\theta)(\lambda R\,d\theta) = \frac{R}{\pi}[-\cos\theta]_0^\pi = \frac{2R}{\pi}$$
Thus, $\vec{r}_{\text{cm}} = \left(0, \frac{2R}{\pi}\right)$.
(b) For a uniform semicircular disc of radius $R$ and surface density $\sigma$ ($x_{\text{cm}} = 0$ by symmetry):
Choose an elemental semicircular strip of radius $r$ and thickness $dr$ ($0 \le r \le R$):
– Area: $dA = \pi r\,dr$, mass: $dm = \sigma \pi r\,dr$.
– The centre of mass of this elemental ring is located at $y_{\text{ring}} = \frac{2r}{\pi}$.
Total mass: $M = \sigma \frac{\pi R^2}{2}$.
$$y_{\text{cm}} = \frac{1}{M}\int y_{\text{ring}}\,dm = \frac{1}{\sigma \frac{\pi R^2}{2}}\int_0^R \left(\frac{2r}{\pi}\right)(\sigma \pi r\,dr) = \frac{4}{\pi R^2}\left[\frac{r^3}{3}\right]_0^R = \frac{4R}{3\pi}$$
Thus, $\vec{r}_{\text{cm}} = \left(0, \frac{4R}{3\pi}\right)$.
Example 2 (Mathematical Manipulation – Non-Uniform Rod with Variable Density):
A thin straight rod of length $L$ lies along the $x$-axis between $x = 0$ and $x = L$. Its linear mass density varies as $\lambda(x) = \lambda_0 \left(1 + \frac{x}{L}\right)$, where $\lambda_0$ is a constant.
(a) Determine the total mass $M$ of the rod.
(b) Find the centre of mass $x_{\text{cm}}$ of the rod.
(c) Generalize for a rod with density $\lambda(x) = \lambda_0 \left(\frac{x}{L}\right)^n$, where $n \ge 0$.
Solution:
(a) Total mass $M$:
$$M = \int_0^L \lambda_0 \left(1 + \frac{x}{L}\right)dx = \lambda_0 \left[ x + \frac{x^2}{2L} \right]_0^L = \frac{3}{2}\lambda_0 L$$
(b) Centre of mass coordinate $x_{\text{cm}}$:
$$\int_0^L x \lambda(x)\,dx = \lambda_0 \int_0^L \left(x + \frac{x^2}{L}\right)dx = \lambda_0 \left[ \frac{x^2}{2} + \frac{x^3}{3L} \right]_0^L = \lambda_0 L^2 \left(\frac{1}{2} + \frac{1}{3}\right) = \frac{5}{6}\lambda_0 L^2$$
$$x_{\text{cm}} = \frac{\frac{5}{6}\lambda_0 L^2}{\frac{3}{2}\lambda_0 L} = \frac{5}{9}L \approx 0.556 L$$
(c) For generalized density $\lambda(x) = \lambda_0 \left(\frac{x}{L}\right)^n$:
$$M = \frac{\lambda_0 L}{n + 1}, \qquad \int_0^L x \lambda(x)\,dx = \frac{\lambda_0 L^2}{n + 2} \implies x_{\text{cm}} = \left(\frac{n + 1}{n + 2}\right)L$$
– For $n = 0$ (uniform): $x_{\text{cm}} = \frac{1}{2}L$.
– For $n = 1$ (linear): $x_{\text{cm}} = \frac{2}{3}L$.
– For $n = 2$ (quadratic): $x_{\text{cm}} = \frac{3}{4}L$.
Example 3 (Standard JEE Advanced Scenario – Circular Cavity in a Disc):
From a uniform circular disc of radius $R$ and mass $M$ centered at origin $O(0, 0)$, a circular hole of radius $r = R/2$ is cut out. The boundary of the hole touches the outer rim at $(R, 0)$ and passes through center $O(0, 0)$.
(a) Find the mass $m_{\text{cut}}$ of the excised portion.
(b) Find the centre of mass coordinates of the remaining body.
(c) What point mass $m”$ placed at the rim $(R, 0)$ will bring the combined CM back to origin $O$?
Solution:
(a) Original area $A_0 = \pi R^2$. Hole area $A_{\text{cut}} = \pi (R/2)^2 = \frac{1}{4}A_0$.
$$m_{\text{cut}} = \frac{1}{4}M, \qquad M_{\text{rem}} = \frac{3}{4}M$$
(b) Centers of mass: original disc at $(0, 0)$; hole center at $(R/2, 0)$.
$$x_{\text{rem}} = \frac{M(0) – m_{\text{cut}} x_{\text{cut}}}{M – m_{\text{cut}}} = \frac{-\left(\frac{M}{4}\right)\left(\frac{R}{2}\right)}{\frac{3M}{4}} = -\frac{R}{6}$$
The centre of mass shifts to $\left(-\frac{R}{6}, 0\right)$.
(c) Point mass $m”$ at $(R, 0)$ restoring CM to origin:
$$M_{\text{rem}} x_{\text{rem}} + m”(R) = 0 \implies \left(\frac{3M}{4}\right)\left(-\frac{R}{6}\right) + m” R = 0 \implies m” = \frac{M}{8}$$
Example 4 (Edge Case – Solid Hemisphere with Concentric Cavity & Limits):
A uniform solid hemisphere of radius $R_2$ has a concentric hemispherical cavity of radius $R_1$ ($R_1 < R_2$) carved out from its flat base.
(a) Determine the height $y_{\text{cm}}$ of the centre of mass above the flat base.
(b) Verify the limiting cases when $R_1 \to 0$ and $R_1 \to R_2$.
Solution:
(a) Modeled as solid hemisphere of radius $R_2$ ($V_2 = \frac{2}{3}\pi R_2^3, y_2 = \frac{3}{8}R_2$) minus cavity of radius $R_1$ ($V_1 = \frac{2}{3}\pi R_1^3, y_1 = \frac{3}{8}R_1$):
$$y_{\text{cm}} = \frac{V_2 y_2 – V_1 y_1}{V_2 – V_1} = \frac{\frac{3}{8}\left(R_2^4 – R_1^4\right)}{R_2^3 – R_1^3} = \mathbf{\frac{3}{8}\left(\frac{R_2^4 – R_1^4}{R_2^3 – R_1^3}\right)}$$
(b) Limits:
– **Limit $R_1 \to 0$ (Solid Hemisphere):** $y_{\text{cm}} = \frac{3}{8}\frac{R_2^4}{R_2^3} = \frac{3}{8}R_2$.
– **Limit $R_1 \to R_2$ (Thin Hemispherical Shell):**
Factoring: $\frac{R_2^4 – R_1^4}{R_2^3 – R_1^3} = \frac{(R_2 + R_1)(R_2^2 + R_1^2)}{R_2^2 + R_1 R_2 + R_1^2} \xrightarrow{R_1 \to R_2} \frac{(2R)(2R^2)}{3R^2} = \frac{4}{3}R$.
$$y_{\text{cm}} = \frac{3}{8}\left(\frac{4}{3}R\right) = \frac{1}{2}R$$
Matches the hollow hemispherical shell result identically.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
The centre of mass of a uniform solid hemisphere of radius $R$ lies on its axis of symmetry at a distance from its flat base equal to:
(A) $\frac{R}{2}$
(B) $\frac{3R}{8}$
(C) $\frac{4R}{3\pi}$
(D) $\frac{3R}{4}$
Problem 2 (JEE Main – Single Correct):
A uniform thin wire is bent into the shape of a semicircle of radius $R$. The distance of its centre of mass from the center of curvature is:
(A) $\frac{2R}{\pi}$
(B) $\frac{4R}{3\pi}$
(C) $\frac{R}{\pi}$
(D) $\frac{R}{2}$
Problem 3 (JEE Main – Single Correct):
The centre of mass of a uniform solid right circular cone of height $H$ and base radius $R$ lies on its central axis at a distance from its flat circular base equal to:
(A) $\frac{H}{2}$
(B) $\frac{H}{3}$
(C) $\frac{H}{4}$
(D) $\frac{3H}{8}$
Problem 4 (JEE Main – Single Correct):
A thin rod of length $L$ lying along the $x$-axis has a non-uniform linear mass density $\lambda(x) = k x$, where $x$ is measured from one end ($x = 0$) and $k$ is a constant. The position $x_{\text{cm}}$ of the centre of mass from $x = 0$ is:
(A) $\frac{L}{2}$
(B) $\frac{2L}{3}$
(C) $\frac{3L}{4}$
(D) $\frac{L}{3}$
Problem 5 (JEE Main – Single Correct):
From a uniform circular disc of radius $R$, a concentric circular disc of radius $R/2$ is removed. The centre of mass of the remaining annular ring:
(A) Shifts by $\frac{R}{4}$
(B) Shifts by $\frac{R}{6}$
(C) Remains at the original center
(D) Shifts by $\frac{R}{8}$
Problem 6 (JEE Advanced – One or More Correct):
For which of the following uniform rigid bodies does the centre of mass lie strictly outside the physical material of the body?
(A) A uniform thin circular ring
(B) A uniform thin semicircular wire
(C) A hollow hemispherical shell without a base
(D) A uniform solid cone
Problem 7 (JEE Advanced – One or More Correct):
For any rigid body with continuous mass distribution, which of the following statements is/are correct?
(A) If the mass density depends only on the perpendicular distance from an axis of geometric symmetry, the centre of mass must lie on that axis.
(B) The centre of mass and centre of gravity coincide in any uniform gravitational field.
(C) The first moment of mass $\int \vec{r}’\,dm$ evaluated about the centre of mass is identically zero.
(D) If symmetrical cavities of equal mass are carved out on opposite sides of the centre of mass, the location of the centre of mass remains unchanged.
Problem 8 (JEE Advanced – One or More Correct):
A circular disc of radius $R$ has a non-uniform surface mass density varying purely with radial distance as $\sigma(r) = \sigma_0 \left(1 – \frac{r}{R}\right)$, where $\sigma_0$ is a constant. Which of the following statements is/are correct?
(A) By rotational symmetry, the centre of mass lies at the geometric center $r = 0$.
(B) The total mass of the disc is $M = \frac{1}{3}\pi \sigma_0 R^2$.
(C) The average surface mass density over the disc area is $\bar{\sigma} = \frac{\sigma_0}{3}$.
(D) The centre of mass shifts toward the perimeter as $R$ increases.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
From a uniform circular disc of radius $R = 12.0\text{ cm}$, a smaller circular disc of radius $r = 4.0\text{ cm}$ tangent to the disc’s outer boundary is removed. Calculate the distance (in cm) by which the centre of mass of the remaining disc shifts from the original center.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A uniform solid right circular cone has a vertical height $H = 24.0\text{ cm}$. Calculate the distance (in cm) of its centre of mass from the apex (top vertex) of the cone.
Solutions & Explanations
Answer Key Summary:
1. (B) | 2. (A) | 3. (C) | 4. (B) | 5. (C) | 6. (A, B, C) | 7. (A, B, C, D) | 8. (A, B, C) | 9. 1 | 10. 18
Solution 1:
For a solid hemisphere of radius $R$, taking circular slices of radius $r(y) = \sqrt{R^2 – y^2}$ and volume $dV = \pi(R^2 – y^2)dy$:
$y_{\text{cm}} = \frac{\int_0^R y \pi(R^2 – y^2)dy}{\frac{2}{3}\pi R^3} = \frac{\frac{R^4}{4}}{\frac{2R^3}{3}} = \frac{3}{8}R$.
Correct Option: (B)
Solution 2:
For a semicircular wire of radius $R$: $dL = R\,d\theta$, $y = R\sin\theta$.
$y_{\text{cm}} = \frac{\int_0^\pi (R\sin\theta)(\lambda R\,d\theta)}{\lambda \pi R} = \frac{2R}{\pi}$.
Correct Option: (A)
Solution 3:
For a solid right circular cone of height $H$, choosing elemental discs at height $y$ yields $y_{\text{cm}} = \frac{H}{4}$ measured from the flat base.
Correct Option: (C)
Solution 4:
$x_{\text{cm}} = \frac{\int_0^L x(kx)dx}{\int_0^L (kx)dx} = \frac{\frac{L^3}{3}}{\frac{L^2}{2}} = \frac{2}{3}L$.
Correct Option: (B)
Solution 5:
Because the hole is concentric, circular symmetry is preserved. The CM remains at the original center (shift = 0).
Correct Option: (C)
Solution 6:
– (A) True: Center of circular ring is empty space.
– (B) True: $y = \frac{2R}{\pi} < R$ lies in the inner open region.
– (C) True: $y = \frac{R}{2}$ is inside the hollow interior.
– (D) False: A solid cone is completely filled, and $y = H/4$ lies inside the material.
Correct Options: (A, B, C)
Solution 7:
All four statements (A, B, C, D) are verified mathematical and dynamical properties of continuous rigid body mass distributions.
Correct Options: (A, B, C, D)
Solution 8:
– (A) True: $\sigma(r)$ depends only on $r$, so rotational symmetry places the CM at $r = 0$.
– (B) True: $M = \int_0^R \sigma_0(1 – r/R)2\pi r\,dr = \frac{\pi \sigma_0 R^2}{3}$.
– (C) True: $\bar{\sigma} = \frac{M}{\pi R^2} = \frac{\sigma_0}{3}$.
– (D) False: By rotational symmetry, the CM remains strictly at $r = 0$.
Correct Options: (A, B, C)
Solution 9:
$A_0 = \pi(12)^2 = 144\pi$, $A_{\text{cut}} = \pi(4)^2 = 16\pi = \frac{1}{9}A_0$.
Center of cavity: $x_{\text{cut}} = 12 – 4 = 8\text{ cm}$.
$x_{\text{rem}} = -\frac{(A_0/9)(8)}{A_0 – A_0/9} = -1.0\text{ cm}$. Shift magnitude is $1.0\text{ cm}$.
Correct Answer: 1
Solution 10:
From base: $y_{\text{base}} = \frac{H}{4} = \frac{24.0}{4} = 6.0\text{ cm}$.
From apex: $y_{\text{apex}} = H – y_{\text{base}} = 24.0 – 6.0 = 18.0\text{ cm}$.
Correct Answer: 18