JEE Physics System Of Particles and Rotational Motion Notes Set 01

Centre of Mass of a Particle System

The centre of mass is the single point where the whole mass of a system can be taken to be concentrated when we study its translation. For particles of masses \( m_i \) at positions \( \vec r_i \): \( \vec R_{cm} = \dfrac{\sum m_i\vec r_i}{\sum m_i} \), i.e. \( x_{cm} = \dfrac{\sum m_ix_i}{M} \) and similarly for y and z. For two particles it lies on the line joining them and divides it in the inverse ratio of the masses: \( m_1r_1 = m_2r_2 \), so it is closer to the heavier one. With the origin at the centre of mass, \( \sum m_i\vec r_i = 0 \).

Continuous Bodies

For a continuous body replace the sum by an integral: \( x_{cm} = \dfrac{1}{M}\int x\,dm \). For a rod with linear density \( \lambda(x) \), \( dm = \lambda\,dx \). If \( \lambda = A + Bx \), then \( x_{cm} = \dfrac{L(3A + 2BL)}{3(2A + BL)} \). Symmetric uniform bodies have their centre of mass at the geometric centre (sphere, ring, rectangle, cylinder). The centre of mass may lie inside or outside the body, may be at a point with no mass, and does not depend on the choice of axes.

Standard Results to Remember

Composite and Cut-Out Bodies

For a body made of parts, treat each part as a point mass at its own centre of mass. For a body with a cavity, use the superposition idea: the cavity is a negative mass. If a part of mass m with centre of mass at x' is removed from a body of mass M with centre x, the rest has \( x_{cm} = \dfrac{Mx - mx'}{M - m} \). For a disc of radius R with a hole of radius r whose centre is at distance d, the shift of the centre of mass is \( \dfrac{r^2d}{R^2 - r^2} \) away from the hole.

Worked Examples

Worked Example 1. Masses 2 kg, 3 kg and 5 kg are at (0, 0), (4, 0) and (0, 3) m. Find the centre of mass.
Answer:
\( x_{cm} = \dfrac{2(0) + 3(4) + 5(0)}{10} = 1.2 \) m.
\( y_{cm} = \dfrac{2(0) + 3(0) + 5(3)}{10} = 1.5 \) m, so the centre of mass is at (1.2, 1.5) m.

Worked Example 2. A uniform disc of radius 20 cm has a hole of radius 10 cm centred 10 cm from the disc's centre. Find the centre of mass of the remaining part.
Answer:
Shift \( = \dfrac{r^2d}{R^2 - r^2} = \dfrac{100\times 10}{400 - 100} = 3.33 \) cm.
It lies 3.33 cm from the centre, on the side opposite to the hole.

Worked Example 3. A rod of length 2 m has linear density \( \lambda = 3x \) kg/m. Find its centre of mass from the x = 0 end.
Answer:
\( x_{cm} = \dfrac{\int_0^2 3x^2dx}{\int_0^2 3x\,dx} = \dfrac{8}{6} \).
\( x_{cm} = \dfrac{4}{3} \) m, which is two-thirds of the length.