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System of Particles and Rotational Motion - Motion of Centre of Mass

Grade 11CBSEPhysics

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

🔑Concepts

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The motion of the centre of mass of a system of particles is governed by the net external force acting on the system. It moves as if the entire mass of the system were concentrated at that point and all external forces were applied there.

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Internal forces between the particles of a system do not affect the motion of the centre of mass. According to Newton's Third Law, these forces occur in equal and opposite pairs and their sum is zero: ∑F⃗int=0\sum \vec{F}_{int} = 0.

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The total linear momentum of a system of particles is the product of the total mass of the system MM and the velocity of its centre of mass V⃗cm\vec{V}_{cm}, expressed as P⃗=MV⃗cm\vec{P} = M\vec{V}_{cm}.

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If the net external force F⃗ext\vec{F}_{ext} acting on a system is zero, the acceleration of the centre of mass A⃗cm\vec{A}_{cm} is zero. This implies that the velocity of the centre of mass V⃗cm\vec{V}_{cm} remains constant.

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In phenomena like radioactive decay or the explosion of a projectile in mid-air, the internal forces cause the fragments to fly apart, but the centre of mass continues to move along the same original trajectory as long as no additional external force is applied.

📐Formulae

V⃗cm=m1v⃗1+m2v⃗2+⋯+mnv⃗nM\vec{V}_{cm} = \frac{m_1\vec{v}_1 + m_2\vec{v}_2 + \dots + m_n\vec{v}_n}{M}

A⃗cm=m1a⃗1+m2a⃗2+⋯+mna⃗nM\vec{A}_{cm} = \frac{m_1\vec{a}_1 + m_2\vec{a}_2 + \dots + m_n\vec{a}_n}{M}

F⃗ext=MA⃗cm\vec{F}_{ext} = M\vec{A}_{cm}

P⃗=MV⃗cm=∑miv⃗i\vec{P} = M\vec{V}_{cm} = \sum m_i\vec{v}_i

dP⃗dt=F⃗ext\frac{d\vec{P}}{dt} = \vec{F}_{ext}

💡Examples

Problem 1:

Two particles of masses m1=2 kgm_1 = 2\text{ kg} and m2=3 kgm_2 = 3\text{ kg} are moving with velocities v⃗1=10i^ m/s\vec{v}_1 = 10\hat{i}\text{ m/s} and v⃗2=5j^ m/s\vec{v}_2 = 5\hat{j}\text{ m/s} respectively. Find the velocity of the centre of mass of the system.

Solution:

V⃗cm=m1v⃗1+m2v⃗2m1+m2\vec{V}_{cm} = \frac{m_1\vec{v}_1 + m_2\vec{v}_2}{m_1 + m_2} V⃗cm=2(10i^)+3(5j^)2+3\vec{V}_{cm} = \frac{2(10\hat{i}) + 3(5\hat{j})}{2 + 3} V⃗cm=20i^+15j^5\vec{V}_{cm} = \frac{20\hat{i} + 15\hat{j}}{5} V⃗cm=(4i^+3j^) m/s\vec{V}_{cm} = (4\hat{i} + 3\hat{j})\text{ m/s}

Explanation:

The velocity of the centre of mass is the weighted average of the velocities of individual particles, where the weights are their respective masses.

Problem 2:

A projectile of mass MM is fired and explodes in mid-air into two fragments of masses m1m_1 and m2m_2 while following a parabolic path. Describe the motion of the centre of mass after the explosion.

Solution:

The explosion is caused by internal forces. The only external force acting on the projectile (neglecting air resistance) is gravity (Mg⃗M\vec{g}). Since the external force remains the same before and after the explosion, the acceleration of the centre of mass remains A⃗cm=g⃗\vec{A}_{cm} = \vec{g}. Therefore, the centre of mass continues to follow the same parabolic path as the original projectile.

Explanation:

Internal forces cannot change the motion of the centre of mass of a system. Only the external force (gravity) determines the path of the CM.

Problem 3:

Calculate the total mass MM of a system consisting of three particles with masses 15 kg15\text{ kg}, 25 kg25\text{ kg}, and 10 kg10\text{ kg}.

Solution:

1525+1050\begin{array}{r} 15 \\ 25 \\ + 10 \\ \hline 50 \end{array} M=50 kgM = 50\text{ kg}

Explanation:

The total mass of the system is the simple arithmetic sum of the individual masses of the particles.