Systems Of Particles And Rotational Motion | Class 11 Physics Notes
By ConceptScroll Team · Published on 17 July 2026 · 4 min read
Systems Of Particles And Rotational Motion – this guide gives you a concise, exam-ready overview of Systems Of Particles And Rotational Motion from Class 11 Physics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.
Centre of Mass
The center of mass (CM) of a system of particles is a crucial concept that simplifies the study of motion. It is defined as the point where the entire mass of the system can be considered to be concentrated for the purpose of analyzing translational motion. Mathematically, for a system of particles with masses m₁, m₂, ..., m_n located at position vectors r₁, r₂, ..., r_n, the position vector R of the center of mass is given by R = (Σ m_i r_i) / (Σ m_i). This formula represents the weighted average of the position vectors, weighted by the masses. The CM moves as if all external forces act on it, and internal forces cancel out due to Newton's third law. For continuous bodies, the summation is replaced by an integral over the mass distribution. The CM is independent of the coordinate system chosen and is a fixed property of the system's mass distribution. The velocity and acceleration of the CM are the mass-weighted averages of the velocities and accelerations of the particles. This concept is extensively used in mechanics to analyze the motion of complex systems by focusing on the CM motion.
📊 Diagram: Diagrams show two particles on a line with masses and positions marked, and the center of mass located between them. Another figure depicts velocity vectors of particles and the resulting velocity of the center of mass.
🧪 Activity: Activity: Determining the center of mass of a system of two or more particles using a simple physical model or graphically.
🔗 Connection: Understanding the center of mass leads to the study of linear momentum of a system and its conservation.
Frequently asked questions
6.1 Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body ?
For each of the given bodies with uniform mass density, the centre of mass (CM) is located at the geometric centre due to symmetry:
(i) Sphere: The CM is at the centre of the sphere.
(ii) Cylinder: The CM is at the midpoint of the axis of the cylinder, i.e., at the centre along its length and at the centre of the circular cross-section.
(iii) Ring: The CM is at the centre of the ring (the centre of the circle formed by the ring).
(iv) Cube: The CM is at the geometric centre of the cube.
Reg
6.2 In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10-10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.
Let the mass of hydrogen atom = m Mass of chlorine atom = 35.5 m Distance between nuclei = d = 1.27 × 10^{-10} m
Taking hydrogen at origin (x=0), chlorine at x=d,
Centre of mass (CM) position from hydrogen nucleus:
x_CM = (m × 0 + 35.5 m × d) / (m + 35.5 m) = (35.5 d) / 36.5
x_CM = (35.5 / 36.5) × 1.27 × 10^{-10} m ≈ 1.236 × 10^{-10} m
So, the CM lies approximately 1.236 Å from the hydrogen nucleus towards chlorine.
6.3 A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system ?
The speed of the centre of mass (CM) of the system (trolley + child) remains the same as the initial speed V of the trolley.
Explanation: Since there are no external horizontal forces acting on the system (assuming smooth floor and no friction), the velocity of the CM cannot change due to internal motions of the child on the trolley. The child running inside the trolley is an internal motion and does not affect the CM velocity.
Therefore, the speed of the CM of the system remains V.
6.4 Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b.
The area of the parallelogram formed by vectors a and b is |a × b|.
Since a triangle is half of a parallelogram, the area of the triangle formed by vectors a and b is:
Area = (1/2) |a × b|
Proof: The magnitude of the cross product |a × b| = |a||b| sin θ, where θ is the angle between a and b.
The area of the parallelogram with sides a and b is base × height = |a||b| sin θ = |a × b|.
Therefore, the area of the triangle = (1/2) × area of parallelogram = (1/2) |a × b|.
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