JEE Physics Centre of Mass Notes Set 04

Centre of Mass of Particles

For a collection of particles the centre of mass is the mass-weighted average position. It is the single point that behaves as if the whole mass were concentrated there when external forces act on the system.
\[ \vec R_{cm} = \frac{\sum m_i\vec r_i}{M},\qquad x_{cm}=\frac{\sum m_ix_i}{M},\ \ y_{cm}=\frac{\sum m_iy_i}{M},\ \ z_{cm}=\frac{\sum m_iz_i}{M} \]
For two particles the centre of mass lies on the line joining them, nearer the heavier one. The sum of the moments of all masses about the centre of mass is zero: \( \sum m_i\vec r_{i/cm}=\vec 0 \). The centre of mass may lie outside the body (a ring, a boomerang) and may have no mass at all at that point.

Continuous Bodies

For a continuous body replace the sum by an integral over mass elements: \( \vec R_{cm}=\frac1M\int\vec r\,dm \). Choose the origin and axes so that symmetry removes as many coordinates as possible; a body with a line or plane of symmetry has its centre of mass on that line or plane. For a rod the element is \( dm=\lambda\,dx \), where the density \( \lambda \) may depend on x.
Rod with \( \lambda=A+Bx \): \( x_{cm}=\frac{L(3A+2BL)}{3(2A+BL)} \). Uniform rod: L/2. Density proportional to x: 2L/3. Density proportional to x squared: 3L/4.

Standard Positions

Body with a Part Removed

Treat the removed portion as a body of negative mass placed at its own centre of mass, then use the normal formula:
\[ x_{cm}=\frac{Mx - m x'}{M-m},\qquad y_{cm}=\frac{My-my'}{M-m} \]
Here M and m are the masses of the complete body and of the removed part, and (x', y') is the centre of mass of the removed part. For uniform sheets, use areas in place of masses; for uniform solids, use volumes.
Example: a hole of radius R/2 touching the rim of a disc of radius R has area one quarter of the disc, so the centre of mass shifts by \( \frac{(1/4)(R/2)}{3/4}=\frac R6 \) away from the hole.

Motion of the Centre of Mass

Differentiating the position formula gives \( \vec v_{cm}=\frac{\sum m_i\vec v_i}{M} \) and \( \vec a_{cm}=\frac{\sum m_i\vec a_i}{M} \). The total linear momentum is \( \vec P=M\vec v_{cm} \), and Newton's second law for the whole system reads \( \vec F_{ext}=M\vec a_{cm} \). Internal forces never change the motion of the centre of mass.
If \( \vec F_{ext}=0 \), then \( \vec v_{cm} \) is constant. A system starting from rest keeps its centre of mass at the same place: a bomb exploding at rest, a person walking on a boat, two bodies attracting each other in free space.

Useful Results from Fixed Centre of Mass

Person of mass m walking a distance L along a plank of mass M on a smooth surface: the plank moves \( \frac{mL}{M+m} \) in the opposite direction.
Two bodies released from rest in free space: displacements are in inverse ratio of masses, \( s_1:s_2=m_2:m_1 \), and they meet at the original centre of mass.
Atwood machine with masses \( m_1>m_2 \): each mass has \( a=\frac{m_1-m_2}{m_1+m_2}g \), and the centre of mass accelerates downward at \( a_{cm}=\left(\frac{m_1-m_2}{m_1+m_2}\right)^2g \).
Fragments after an explosion from rest have momenta that add to zero, so the lighter fragment is the faster one.

Impulse and Momentum

The impulse of a force is the area under its force-time graph and equals the change of momentum: \( \int\vec F\,dt=\Delta\vec p \). This is Newton's second law in integrated form. A short, violent impact involves a large force over a small time; the same momentum change spread over a long time needs only a small force.