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System of Particles and Rotational Motion - Center of Mass

Grade 11CBSEPhysics

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

🔑Concepts

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The Center of Mass (CM) of a system is a point where the entire mass of the system is assumed to be concentrated for the study of its translational motion.

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For a system of nn particles with masses m1,m2,...,mnm_1, m_2, ..., m_n and position vectors r⃗1,r⃗2,...,r⃗n\vec{r}_1, \vec{r}_2, ..., \vec{r}_n, the position vector of the center of mass is R⃗cm=∑mir⃗iM\vec{R}_{cm} = \frac{\sum m_i \vec{r}_i}{M}, where M=∑miM = \sum m_i.

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If the origin of the coordinate system is located at the center of mass, then ∑mir⃗i=0\sum m_i \vec{r}_i = 0.

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For continuous mass distributions like a uniform rod, ring, or sphere, the CM is found using integration: R⃗cm=1M∫r⃗dm\vec{R}_{cm} = \frac{1}{M} \int \vec{r} dm.

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The center of mass of a body with a uniform density and a geometric shape (like a circle or cylinder) coincides with its geometric center (centroid).

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The velocity of the center of mass is given by V⃗cm=dR⃗cmdt=∑miv⃗iM\vec{V}_{cm} = \frac{d\vec{R}_{cm}}{dt} = \frac{\sum m_i \vec{v}_i}{M}.

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The total linear momentum of a system is equal to the product of the total mass and the velocity of its center of mass: P⃗=MV⃗cm\vec{P} = M\vec{V}_{cm}.

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Newton's Second Law for a system of particles states that F⃗ext=MA⃗cm\vec{F}_{ext} = M\vec{A}_{cm}, where F⃗ext\vec{F}_{ext} is the sum of all external forces acting on the system.

📐Formulae

R⃗cm=m1r⃗1+m2r⃗2+...+mnr⃗nm1+m2+...+mn\vec{R}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + ... + m_n\vec{r}_n}{m_1 + m_2 + ... + m_n}

Xcm=∑mixi∑mi,Ycm=∑miyi∑mi,Zcm=∑mizi∑miX_{cm} = \frac{\sum m_i x_i}{\sum m_i}, Y_{cm} = \frac{\sum m_i y_i}{\sum m_i}, Z_{cm} = \frac{\sum m_i z_i}{\sum m_i}

R⃗cm=1M∫r⃗dm\vec{R}_{cm} = \frac{1}{M} \int \vec{r} dm

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

A⃗cm=F⃗extM\vec{A}_{cm} = \frac{\vec{F}_{ext}}{M}

P⃗=MV⃗cm\vec{P} = M\vec{V}_{cm}

💡Examples

Problem 1:

Three particles of masses m1=1 kgm_1 = 1\,kg, m2=2 kgm_2 = 2\,kg, and m3=3 kgm_3 = 3\,kg are placed at the corners of an equilateral triangle of side 1 m1\,m. Find the coordinates of the center of mass.

Solution:

Let m1m_1 be at (0,0)(0, 0), m2m_2 at (1,0)(1, 0), and m3m_3 at (0.5,32)(0.5, \frac{\sqrt{3}}{2}). Xcm=1(0)+2(1)+3(0.5)1+2+3=2+1.56=3.56≈0.583 mX_{cm} = \frac{1(0) + 2(1) + 3(0.5)}{1+2+3} = \frac{2 + 1.5}{6} = \frac{3.5}{6} \approx 0.583\,m. Ycm=1(0)+2(0)+3(32)6=3312=34≈0.433 mY_{cm} = \frac{1(0) + 2(0) + 3(\frac{\sqrt{3}}{2})}{6} = \frac{3\sqrt{3}}{12} = \frac{\sqrt{3}}{4} \approx 0.433\,m.

Explanation:

We used the discrete particle formula for the xx and yy coordinates separately by setting a reference coordinate system at one of the vertices.

Problem 2:

A projectile is fired and explodes in mid-air into two fragments. What happens to the path of the center of mass of the fragments?

Solution:

The center of mass of the fragments will continue to follow the same parabolic path that the projectile would have followed if it had not exploded.

Explanation:

Since the explosion is caused by internal forces, the net external force acting on the system (which is gravity Mg⃗M\vec{g}) remains unchanged. Therefore, the acceleration of the center of mass A⃗cm\vec{A}_{cm} remains g⃗\vec{g}, and its trajectory remains parabolic.