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.
Motion of the Centre of Mass
The motion of the center of mass (CM) of a system of particles is governed by the net external force acting on the system. According to Newton's second law, the acceleration of the CM is given by A = F_ext / M, where F_ext is the net external force and M is the total mass. This means the CM moves as if all the mass of the system is concentrated at that point and all external forces act there. The internal forces, being equal and opposite, do not influence the CM motion. The velocity and acceleration of the CM can be found by differentiating its position vector with respect to time. This principle is widely used in analyzing the motion of complex systems such as rockets, vehicles, and celestial bodies. The chapter also discusses the motion of the CM in different coordinate systems and frames of reference. The concept simplifies the study of dynamics by reducing a system of particles to a single point mass moving under external forces.
📊 Diagram: Diagrams show a system of particles with external forces acting, the center of mass marked, and vectors representing velocity and acceleration of the CM.
🧪 Activity: No specific activity in this section.
🔗 Connection: This section leads to the study of angular velocity and rotational motion of rigid bodies.
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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