JEE Physics System Of Particles and Rotational Motion Notes Set 07

Centre of Mass

1. For particles, \( \vec{r}_{cm} = \dfrac{\sum m_i\vec{r}_i}{\sum m_i} \). For a continuous body, \( x_{cm} = \dfrac{\int x\,dm}{\int dm} \). With equal masses it is the plain average of positions. The centre of mass divides the line joining two particles in the inverse ratio of their masses.
2. A body with a hole is treated as the full body plus a negative mass at the hole. For a disc of radius R with a hole of radius r whose centre is at distance d from the centre, the shift is \( \dfrac{r^2 d}{R^2 - r^2} \) on the side opposite the hole.
3. Standard values from the centre: semicircular wire \( \dfrac{2R}{\pi} \), semicircular disc \( \dfrac{4R}{3\pi} \), hemispherical shell \( \dfrac{R}{2} \), solid hemisphere \( \dfrac{3R}{8} \), triangle at one third of the height from the base.
4. The centre of mass can lie outside the body, as for a ring. It depends only on masses and positions, not on internal forces.

Motion of the Centre of Mass

1. \( M\vec{a}_{cm} = \vec{F}_{ext} \). Internal forces never change \( \vec{v}_{cm} \). For a system starting at rest with no external force, the centre of mass stays at rest, so a person walking on a floating plank shifts the plank by \( \dfrac{mL}{M+m} \).
2. For pulley systems, find the common acceleration of the masses first and then take the mass-weighted vector average to get \( \vec{a}_{cm} \).
3. Momentum is conserved when the net external force is zero (also along a direction in which it is zero). Kinetic energy can change under internal forces, as in an explosion.

Explosions

1. A body at rest breaking into pieces keeps total momentum zero. For two pieces, \( m_1v_1 = m_2v_2 \), kinetic energies are in the inverse ratio of masses and the energy released is \( \dfrac{P^2}{2}\left(\dfrac{1}{m_1}+\dfrac{1}{m_2}\right) \).
2. When a projectile explodes at its highest point, the horizontal momentum is conserved and the vertical momentum stays zero, so fragments with no vertical velocity reach the ground together.
3. The centre of mass of the fragments continues on the original parabolic path until a fragment lands.

Collisions

Momentum is conserved in every collision when no external force acts. Kinetic energy is conserved only in elastic collisions. In a bullet and block (ballistic) problem, use momentum for the impact and energy for the swing that follows.

Worked Examples

1. A 2 kg ball at 3 m/s hits a 1 kg ball at rest with e = 0.5. Momentum: \( 6 = 2v_1 + v_2 \); restitution: \( v_2 - v_1 = 1.5 \). So \( v_1 = 1.5 \) m/s and \( v_2 = 3 \) m/s.
2. A man of 60 kg stands at one end of a 120 kg boat of length 6 m. He walks to the other end. The boat moves \( \dfrac{60\times 6}{180} = 2 \) m the other way.
3. A 6 kg body at rest explodes into 4 kg and 2 kg pieces, and the 4 kg piece moves at 3 m/s. The 2 kg piece moves at 6 m/s the other way, and the energy released is \( 18 + 36 = 54 \) J.