JEE Physics Centre of Mass Notes Set 05

Centre of Mass of Particles

1. For particles of masses \( m_i \) at positions \( \vec r_i \), the centre of mass is \( \vec r_{cm} = \frac{\sum m_i\vec r_i}{\sum m_i} \). In components \( x_{cm} = \frac{\sum m_ix_i}{M} \), with similar expressions for y and z.
2. For two particles the centre of mass lies on the line joining them, closer to the heavier one: \( m_1 r_1 = m_2 r_2 \), where \( r_1, r_2 \) are the distances from the centre of mass.
3. The position of the centre of mass depends on the masses and the geometry, not on the choice of origin. The sum of mass-weighted position vectors about the centre of mass is zero: \( \sum m_i\vec r_i' = 0 \).
4. The centre of mass need not lie inside the body (a ring, a hollow sphere, a boomerang). If the body has a point, line or plane of symmetry with uniform density, the centre of mass lies on it.
5. The centre of mass and the centre of gravity coincide when g is uniform over the body.

Centre of Mass of Continuous Bodies

1. Replace the sum by an integral: \( x_{cm} = \frac{\int x\,dm}{\int dm} \). For a uniform rod of length L along x, take \( dm = \lambda\,dx \); for a variable density \( \lambda = \lambda(x) \), use that function.
2. Standard results for uniform bodies:
Rod of length L: at the middle.
Semicircular ring of radius R: \( \frac{2R}{\pi} \) from the centre along the axis of symmetry.
Semicircular disc: \( \frac{4R}{3\pi} \).
Hemispherical shell: \( \frac R2 \). Solid hemisphere: \( \frac{3R}{8} \).
Triangular plate: at the centroid, one-third of the height from the base.
Solid cone or pyramid of height h: \( \frac h4 \) from the base. Hollow cone (lateral surface only): \( \frac h3 \) from the base.
Circular arc of radius R and half-angle \( \alpha \): \( \frac{R\sin\alpha}{\alpha} \) from the centre. A sector of a disc: \( \frac{2R\sin\alpha}{3\alpha} \).
3. If the linear density of a rod varies as \( \lambda = kx \) from x = 0 to L, then \( x_{cm} = \frac{\int kx^2dx}{\int kx\,dx} = \frac{2L}{3} \).
4. For a body with several parts, treat each part as a point mass at its own centre of mass and use the discrete formula.

Combination of Structures and Cavity Problems

1. A composite body is a set of simple bodies. Find the mass and centre of mass of each and combine them: \( x_{cm} = \frac{m_1x_1 + m_2x_2 + \dots}{m_1+m_2+\dots} \).
2. For a body with a cavity, treat the cavity as a body of negative mass at its own centre: \( x_{cm} = \frac{M x_0 - m x_c}{M - m} \), where M and m are the masses the full body and the removed piece would have had.
3. A uniform disc of radius R with a hole of radius \( R/2 \) centred at distance \( R/2 \) from the centre has the hole's mass equal to one quarter of the full mass. The centre of mass of the remaining part is at \( \frac{M/4\cdot R/2}{3M/4} = \frac R6 \) from the centre, on the side away from the hole.
4. For a sphere of radius R with a spherical cavity of radius \( R/2 \) at distance \( R/2 \) from the centre, the cavity has mass \( M/8 \) and the centre of mass shifts by \( \frac R{14} \) away from the cavity.
5. Always check by writing the answer as a shift from the centre of the full body: the shift is opposite to the removed piece.

Motion of the Centre of Mass

1. \( \vec v_{cm} = \frac{\sum m_i\vec v_i}{M} \) and \( \vec a_{cm} = \frac{\sum m_i\vec a_i}{M} \). The total momentum is \( \vec P = M\vec v_{cm} \).
2. Newton's second law for a system: \( \vec F_{ext} = M\vec a_{cm} \). Internal forces cancel in pairs and cannot change the motion of the centre of mass.
3. If \( \vec F_{ext} = 0 \), the centre of mass moves with constant velocity and the total momentum is conserved. If it starts at rest, it stays at rest, even though the parts may move.
4. A shell exploding in flight: the centre of mass of the fragments continues along the original parabola until a fragment lands.
5. A man walking on a boat on still water with no horizontal external force: the centre of mass of the man and boat stays fixed. If the man moves a distance \( \ell \) relative to the boat, the boat moves \( \frac{m}{M+m}\ell \) in the opposite direction, with \( m \) the mass of the man and M the mass of the boat.
6. A system with a spring: if the external horizontal force is zero, the centre of mass moves uniformly, while the two blocks oscillate about it. In the frame of the centre of mass the two blocks have equal and opposite momenta.
7. The kinetic energy of a system splits as \( K = \frac12Mv_{cm}^2 + K_{rel} \), where \( K_{rel} \) is the kinetic energy relative to the centre of mass. For two bodies \( K_{rel} = \frac12\mu v_{rel}^2 \) with the reduced mass \( \mu = \frac{m_1m_2}{m_1+m_2} \).

Spring-Block System

1. Two blocks m1 and m2 joined by a spring of constant k, on a smooth surface, with one given a speed. The centre of mass moves at constant \( v_{cm} \). The spring is at its maximum extension or compression when the relative velocity is zero, that is both blocks move with \( v_{cm} \).
2. The maximum extension x satisfies \( \frac12kx^2 = K_{initial} - \frac12(m_1+m_2)v_{cm}^2 = \frac12\mu v_{rel}^2 \).
3. The angular frequency of relative oscillation is \( \omega = \sqrt{\frac{k}{\mu}} \).
4. Energy conservation and momentum conservation are both used. Momentum is conserved only along the direction with no external force.

Worked Examples

1. Masses 1 kg at (0, 0), 2 kg at (3, 0) and 3 kg at (0, 4). Then \( x_{cm} = \frac{0 + 6 + 0}{6} = 1 \) and \( y_{cm} = \frac{0+0+12}{6} = 2 \). The centre of mass is at (1, 2).
2. A uniform disc of radius R has a circular hole of radius R/2 centred at R/2 from the centre. The centre of mass of the remainder is R/6 from the centre of the disc on the side opposite to the hole.
3. A man of mass 60 kg stands at one end of a 3 m boat of mass 120 kg at rest on still water. He walks to the other end. The boat moves \( \frac{60}{180}\times3 = 1 \) m in the opposite direction.
4. A rod of length L has linear density \( \lambda = \lambda_0\left(1 + \frac xL\right) \). Then \( x_{cm} = \frac{\int_0^L x(1 + x/L)dx}{\int_0^L(1+x/L)dx} = \frac{L^2/2 + L^2/3}{3L/2} = \frac{5L}{9} \).
5. A projectile of mass 3 kg explodes at the top of its path into two parts. A 1 kg part falls vertically from rest. The 2 kg part must carry the whole original horizontal momentum. If the projectile was moving at 12 m/s at the top, the second part moves at \( \frac{3\times12}{2} = 18 \) m/s horizontally.
6. A 2 kg block at 6 m/s moves towards a 4 kg block at rest on a smooth surface, joined by a light spring. The centre of mass speed is 2 m/s. At the maximum compression both move at 2 m/s. The energy stored is \( \frac12(2)(36) - \frac12(6)(4) = 36 - 12 = 24 \) J.