JEE Physics Centre of Mass Notes Set 01

Centre of Mass of a System of Particles

The centre of mass is the single point at which the whole mass of a system can be taken to be concentrated when we study its translational motion. For particles of mass \( m_i \) at positions \( \vec r_i \),
\[ \vec r_{cm}=\frac{\sum m_i\vec r_i}{\sum m_i},\qquad x_{cm}=\frac{\sum m_ix_i}{\sum m_i} \]
and the same form holds for the y and z coordinates. It need not lie inside the body (a ring has it at its empty centre). If the origin is placed at the centre of mass, then \( \sum m_i\vec r_i=0 \), so the coordinates must take both signs. For two particles the centre of mass lies on the line joining them, nearer to the heavier one, at distances inversely proportional to the masses.
Illustration: masses 1 kg and 2 kg at x = 0 and x = 3 m give \( x_{cm}=\frac{0+6}3=2 \) m.

Continuous Bodies

For a continuous body replace the sums by integrals: \( \vec r_{cm}=\frac{\int\vec r\,dm}{\int dm} \), where \( dm=\lambda\,dx \) for a rod, \( \sigma\,dA \) for a plate and \( \rho\,dV \) for a solid. For a rod with density \( \lambda=kx^n \) measured from one end, \( x_{cm}=\frac{n+1}{n+2}L \): the centre sits at L/2 for n = 0, at 2L/3 for n = 1 and at 3L/4 for n = 2.
Standard results, with distances measured from the centre of curvature or the base:
1. Semicircular wire of radius R: \( \frac{2R}\pi \) from the centre.
2. Semicircular plate: \( \frac{4R}{3\pi} \).
3. Hollow hemispherical shell: \( \frac R2 \); solid hemisphere: \( \frac{3R}8 \).
4. Solid cone of height h: \( \frac h4 \) from the base; hollow cone: \( \frac h3 \).
5. Triangular plate of height h: \( \frac h3 \) from the base (the centroid).

Bodies with a Cut-Out

Treat the hole as a body of negative mass. If a uniform plate of area \( A_1 \) with centre at \( x_1 \) has a part of area \( A_2 \) with centre at \( x_2 \) removed, then \( x_{cm}=\frac{A_1x_1-A_2x_2}{A_1-A_2} \). The centre of mass always shifts away from the hole.
Illustration: a disc of radius a with a hole of radius a/2 touching the rim has its centre of mass \( \frac a6 \) from the disc centre, away from the hole. A disc of radius R with a hole of radius R/4 centred at 3R/4 gives a shift of \( \frac R{20} \).

Motion of the Centre of Mass

Differentiating the position formula gives \( \vec v_{cm}=\frac{\sum m_i\vec v_i}{M} \) and \( \vec a_{cm}=\frac{\sum m_i\vec a_i}{M} \). Since internal forces occur in equal and opposite pairs, they cancel in the sum, and \( \vec F_{ext}=M\vec a_{cm} \). The centre of mass therefore behaves like a particle of mass M acted on by the total external force only.
Consequences: an exploding projectile's fragments have a centre of mass that continues on the original parabola; a system with no external force has a centre of mass at rest or moving uniformly; and two balls in flight have \( a_{cm}=g \). The total linear momentum of a system is \( M\vec v_{cm} \).
When the centre of mass is initially at rest and no horizontal external force acts, the displacements of the parts satisfy \( \sum m_i\Delta x_i=0 \). This gives the man-and-plank or man-and-boat result: a man of mass m walking a distance d relative to a plank of mass M shifts the plank by \( \frac{m}{M+m}d \) in the opposite direction.

Conservation of Linear Momentum

If the resultant external force on a system is zero, or only non-impulsive forces (gravity, spring force) act during a short interaction, the total momentum is conserved: \( \sum m_i\vec v_i \) before equals that after. Typical uses are explosions from rest (fragment momenta add to zero), recoil of a gun, and a bullet embedded in a block.
Kinetic energy is not conserved in these processes. In an explosion it increases, because stored internal energy is released; in a sticking collision it decreases. The explosion energy of two fragments from rest, with momentum p each, is \( \frac{p^2}{2m_1}+\frac{p^2}{2m_2} \).
Kinetic energy and momentum are connected by \( K=\frac{p^2}{2m} \), so for fixed mass a 20% increase in momentum gives a 44% increase in kinetic energy, and doubling the momentum quadruples the energy.

Impulse

The impulse of a force is \( \vec J=\int\vec F\,dt=\Delta\vec p \), equal to the area under the F-t graph. The average force is \( \frac{\Delta p}{\Delta t} \). A ball reversing its velocity changes momentum by \( 2mv \), so a 0.15 kg ball at 20 m/s hit back at the same speed in 0.01 s feels an average force of 600 N. For a wall impact only the component normal to the wall changes.