Centre of Mass — Revision Notes

⚡ 30-Second Revision

Discrete Particles: — $\vec{R}_{CM} = \frac{\sum m_i \vec{r}_i}{\sum m_i}$ \n- Continuous Bodies: $\vec{R}_{CM} = \frac{\int \vec{r} \,dm}{\int dm}$ \n- Velocity of CoM: $\vec{V}_{CM} = \frac{\sum m_i \vec{v}_i}{\sum m_i} = \frac{\vec{P}_{sys}}{M}$ \n- Acceleration of CoM: $\vec{A}_{CM} = \frac{\vec{F}_{ext, net}}{M}$ \n- CoM of Uniform Rod (length L): At L/2 from an end. \n- CoM of Uniform Semicircular Arc (radius R): $2R/\pi$ from center along axis of symmetry. \n- CoM of Uniform Semicircular Plate (radius R): $4R/(3\pi)$ from center along axis of symmetry. \n- CoM of Uniform Solid Hemisphere (radius R): $3R/8$ from center along axis of symmetry. \n- CoM of Uniform Hollow Hemisphere (radius R): $R/2$ from center along axis of symmetry. \n- Conservation: If $\vec{F}_{ext, net} = 0$, then $\vec{V}_{CM}$ is constant (momentum conserved). If system is initially at rest, $\vec{V}_{CM} = \vec{0}$ always.

2-Minute Revision

The Centre of Mass (CoM) is a crucial point representing the average position of mass in a system. For discrete particles, calculate it as a weighted average: $X_{CM} = \frac{\sum m_i x_i}{\sum m_i}$.

For continuous bodies, use integration or recall standard formulas for uniform shapes. The most powerful aspect is that the translational motion of the CoM is governed solely by external forces: $\vec{F}_{ext, net} = M \vec{A}_{CM}$.

Internal forces (like those in an explosion) do not affect the CoM's motion. If the net external force is zero, the CoM's velocity remains constant, leading to the conservation of linear momentum. Remember that the CoM can be outside the physical body (e.

g., a ring) and is distinct from the Centre of Gravity, though they coincide in a uniform gravitational field. For composite bodies or those with holes, use the 'negative mass' technique. Always choose a convenient origin to simplify calculations.

5-Minute Revision

The Centre of Mass (CoM) is a conceptual point that simplifies the dynamics of multi-particle systems and rigid bodies. It's the unique point where the entire mass of the system can be considered concentrated for translational motion.

Mathematically, for discrete particles, its position vector is $\vec{R}_{CM} = \frac{\sum m_i \vec{r}_i}{\sum m_i}$. For continuous bodies, this becomes an integral $\vec{R}_{CM} = \frac{\int \vec{r} \,dm}{\int dm}$.

\n\nKey takeaways for NEET: \n1. Calculation: Be proficient in calculating CoM for discrete particles (e.g., particles at vertices of a triangle/square). For continuous bodies, memorize standard formulas (e.

g., uniform rod at $L/2$, semicircular arc at $2R/\pi$, solid hemisphere at $3R/8$). For composite bodies or those with holes, use the 'negative mass' method: treat the hole as a negative mass at its own CoM and apply the discrete particle formula.

\n2. Motion of CoM: The most significant principle is Newton's Second Law for a system: $\vec{F}_{ext, net} = M \vec{A}_{CM}$. This means only external forces can change the translational motion of the CoM.

Internal forces (like those in an explosion) have no effect on the CoM's path. If $\vec{F}_{ext, net} = 0$, then $\vec{V}_{CM}$ is constant (conservation of linear momentum). This is crucial for problems involving explosions (CoM continues its path) or a person walking on a frictionless boat (CoM remains stationary if initially at rest).

\n3. Conceptual Understanding: Understand that CoM is a mathematical point and can lie outside the physical body (e.g., a ring). Differentiate CoM from Centre of Gravity (CoG); they coincide only in a uniform gravitational field.

CoM is fundamental for understanding collisions, rotational motion, and stability.\n\nWorked Example: A 2 kg mass is at (1,0) m and a 3 kg mass is at (0,2) m. Find their CoM.\n$X_{CM} = \frac{(2\,\text{kg})(1\,\text{m}) + (3\,\text{kg})(0\,\text{m})}{2\,\text{kg} + 3\,\text{kg}} = \frac{2}{5} = 0.

4\,\text{m}$.\n$Y_{CM} = \frac{(2\,\text{kg})(0\,\text{m}) + (3\,\text{kg})(2\,\text{m})}{2\,\text{kg} + 3\,\text{kg}} = \frac{6}{5} = 1.2\,\text{m}$.\nSo, CoM is at$(0.4, 1.2)\,\text{m}$.

Prelims Revision Notes

The Centre of Mass (CoM) is a critical concept for NEET, simplifying the analysis of multi-particle systems and rigid bodies. It's the point where the entire mass of a system can be considered concentrated for translational motion.

\n\n1. Definition and Formulas: \n* Discrete Particles: For $n$ particles with masses $m_i$ at position vectors $\vec{r}_i$: \n $\vec{R}_{CM} = \frac{\sum_{i=1}^{n} m_i\vec{r}_i}{\sum_{i=1}^{n} m_i}$.

\n In Cartesian coordinates: $X_{CM} = \frac{\sum m_i x_i}{M}$, $Y_{CM} = \frac{\sum m_i y_i}{M}$, $Z_{CM} = \frac{\sum m_i z_i}{M}$. \n* Continuous Bodies: For a body with mass element $dm$ at $\vec{r}$: \n $\vec{R}_{CM} = \frac{\int \vec{r} \,dm}{\int dm}$.

\n For uniform bodies, $dm = \lambda \,dx$ (linear), $dm = \sigma \,dA$ (surface), $dm = \rho \,dV$ (volume). \n\n2. CoM of Standard Uniform Bodies (Memorize!): \n* Rod of length L: At L/2 from either end.

\n* Ring/Disc (circular): At its geometric center. \n* Sphere/Spherical Shell: At its geometric center. \n* Semicircular Arc (radius R): $2R/\pi$ from the center along the axis of symmetry.

\n* Semicircular Plate (radius R): $4R/(3\pi)$ from the center along the axis of symmetry. \n* Solid Hemisphere (radius R): $3R/8$ from the center of the flat base along the axis of symmetry. \n* Hollow Hemisphere (radius R): $R/2$ from the center of the flat base along the axis of symmetry.

\n* Solid Cone (height H): $H/4$ from the base along the axis of symmetry. \n\n3. Motion of CoM: \n* Velocity of CoM: $\vec{V}_{CM} = \frac{\sum m_i \vec{v}_i}{M} = \frac{\vec{P}_{sys}}{M}$.

\n* Acceleration of CoM: $\vec{A}_{CM} = \frac{\sum m_i \vec{a}_i}{M} = \frac{\vec{F}_{ext, net}}{M}$. \n* Key Principle: Only external forces affect the translational motion of the CoM. Internal forces cancel out and do not change $\vec{V}_{CM}$.

\n* Conservation of Momentum: If $\vec{F}_{ext, net} = 0$, then $\vec{A}_{CM} = 0$, implying $\vec{V}_{CM}$ is constant. If the system is initially at rest, $\vec{V}_{CM}$ remains zero. \n\n4. CoM of Composite Bodies / Bodies with Holes: \n* Treat each component as a particle located at its own CoM.

\n* For a body with a hole, use the 'negative mass' concept: $X_{CM} = \frac{M_{full} X_{full} - M_{hole} X_{hole}}{M_{full} - M_{hole}}$. \n\n5. Common Misconceptions: \n* CoM is not always inside the body (e.

g., ring). \n* CoM is not always at the geometric center (unless uniform and symmetric). \n* CoM $\neq$ CoG (unless uniform gravitational field).

Vyyuha Quick Recall

For CoM of common shapes, remember: \nRod: L/2 (from end) \nSemi-circular Arc: 2R/pi (from center) \nSemi-circular Plate: 4R/3pi (from center) \nHollow Hemisphere: R/2 (from base) \nSolid Hemisphere: 3R/8 (from base) \nCone: H/4 (from base) \n\nMnemonic: 'Rods Are Simple, Plates Are Heavy, Hemispheres Have Half, Solids Are Three-Eighths, Cones Are Quarters.' (Relates to the complexity/density and fraction of R or H)