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Motion of System of Particles and Rigid Body - Centre of Mass

Grade 11ICSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Centre of Mass (CM) of a system of particles is a specific point where the entire mass of the system can be considered to be concentrated for describing its translational motion.

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For a system of nn particles, the position vector R⃗cm\vec{R}_{cm} is the weighted average of the position vectors of the individual particles.

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The location of the CM depends on the distribution of mass. For a rigid body with uniform density and a symmetrical shape (like a sphere, cylinder, or cube), the CM coincides with its geometric centre.

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Internal forces between particles of a system do not affect the motion of the Centre of Mass. Only external forces F⃗ext\vec{F}_{ext} can change the state of motion of the CM.

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If the net external force acting on a system is zero (F⃗ext=0\vec{F}_{ext} = 0), the velocity of the centre of mass V⃗cm\vec{V}_{cm} remains constant.

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In a uniform gravitational field, the Centre of Mass coincides with the Centre of Gravity.

📐Formulae

R⃗cm=∑i=1nmir⃗iM=m1r⃗1+m2r⃗2+⋯+mnr⃗nm1+m2+⋯+mn\vec{R}_{cm} = \frac{\sum_{i=1}^{n} m_i \vec{r}_i}{M} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + \dots + m_n\vec{r}_n}{m_1 + m_2 + \dots + m_n}

Xcm=m1x1+m2x2+⋯+mnxnm1+m2+⋯+mnX_{cm} = \frac{m_1x_1 + m_2x_2 + \dots + m_nx_n}{m_1 + m_2 + \dots + m_n}

V⃗cm=∑miv⃗iM\vec{V}_{cm} = \frac{\sum m_i \vec{v}_i}{M}

A⃗cm=∑mia⃗iM=F⃗extM\vec{A}_{cm} = \frac{\sum m_i \vec{a}_i}{M} = \frac{\vec{F}_{ext}}{M}

R⃗cm=1M∫r⃗ dm (for continuous mass distribution)\vec{R}_{cm} = \frac{1}{M} \int \vec{r} \, dm \text{ (for continuous mass distribution)}

💡Examples

Problem 1:

Two particles of mass 2 kg2\text{ kg} and 4 kg4\text{ kg} are located on the x-axis at x=2 mx = 2\text{ m} and x=8 mx = 8\text{ m} respectively. Find the position of the centre of mass of the system.

Solution:

Given m1=2 kgm_1 = 2\text{ kg}, x1=2 mx_1 = 2\text{ m}, m2=4 kgm_2 = 4\text{ kg}, x2=8 mx_2 = 8\text{ m}. Using the formula: Xcm=m1x1+m2x2m1+m2X_{cm} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2} Xcm=(2×2)+(4×8)2+4=4+326=366=6 mX_{cm} = \frac{(2 \times 2) + (4 \times 8)}{2 + 4} = \frac{4 + 32}{6} = \frac{36}{6} = 6\text{ m}

Explanation:

The centre of mass lies on the line joining the two particles, closer to the heavier mass (4 kg4\text{ kg}).

Problem 2:

A projectile is fired and explodes in mid-air into several fragments. Describe the motion of the centre of mass of these fragments, assuming air resistance is negligible.

Solution:

The explosion is caused by internal forces. Since no new external force (other than gravity Mg⃗M\vec{g}) acts on the system during the explosion, the acceleration of the centre of mass remains a⃗cm=g⃗\vec{a}_{cm} = \vec{g}.

Explanation:

The centre of mass will continue to follow the original parabolic trajectory that the projectile would have followed if it had not exploded.

Centre of Mass - Revision Notes & Key Formulas | ICSE Class 11 Physics