JEE Physics Centre of Mass Notes Set 03

Centre of Mass of Discrete Particles

The centre of mass (COM) is the single point whose motion represents the translational motion of a whole system: it moves as if the total mass were concentrated there and the net external force acted on it. For particles of masses \( m_i \) at positions \( \vec r_i \), \( \vec r_{cm}=\dfrac{\sum m_i\vec r_i}{\sum m_i} \), applied separately to each coordinate. If the origin is placed at the COM, the sum of the mass moments \( \sum m_i\vec r_i \) is zero. For two particles separated by r, the COM lies on the line joining them, nearer the heavier one, at distances \( r_1=\dfrac{m_2}{m_1+m_2}r \) from \( m_1 \) and \( r_2=\dfrac{m_1}{m_1+m_2}r \) from \( m_2 \). Example: masses 2 kg and 3 kg at x = 0 and x = 5 m have \( x_{cm}=\dfrac{2(0)+3(5)}{5}=3 \) m, which is 3 m from the lighter mass and 2 m from the heavier.

Continuous Bodies and Symmetry

For a continuous body replace the sums by integrals: \( x_{cm}=\dfrac{\int x\,dm}{\int dm} \), and similarly for y and z, writing dm as density times length, area or volume of a small element. If the body is symmetric about a line or plane, the COM lies on it, so a uniform rod, ring, disc, sphere or rectangular plate has its COM at the geometric centre. For a rod of length L whose linear density grows as \( \lambda=kx \), \( x_{cm}=\dfrac{\int_0^Lkx^2dx}{\int_0^Lkx\,dx}=\dfrac{L^3/3}{L^2/2}=\dfrac{2L}{3} \) from the light end, so the COM moves towards the heavier end.

Standard Results to Remember

Distances are measured from the centre of the flat face or circle unless stated. Semicircular ring of radius R: \( \dfrac{2R}{\pi} \) from the centre along the axis of symmetry. Semicircular disc: \( \dfrac{4R}{3\pi} \). Solid hemisphere: \( \dfrac{3R}{8} \). Hollow hemisphere (thin shell): \( \dfrac R2 \). Solid cone of height h: \( \dfrac h4 \) from the base, i.e. \( \dfrac{3h}{4} \) from the apex. Uniform triangular lamina: \( \dfrac h3 \) from the base along the median, at the centroid. A uniform wire bent into an arc, and the full shapes, follow from the same integral by slicing into rings or discs of known mass.

Composite Bodies and Removed Portions

A composite body can be treated as point masses placed at the COMs of its parts, \( \vec r_{cm}=\dfrac{\sum M_i\vec r_i}{\sum M_i} \). A hole is handled by adding a body of negative mass at the position of the hole. Example: a uniform disc of radius R has a circular hole of radius R/2 whose centre is at R/2 from the disc's centre. With mass proportional to area, the hole has mass \( \tfrac14 \) of the full disc, the remaining part \( \tfrac34 \), so \( x_{cm}=\dfrac{M(0)-\tfrac M4\cdot\tfrac R2}{M-\tfrac M4}=-\dfrac R6 \), i.e. R/6 from the centre on the side opposite to the hole.

Motion of the Centre of Mass

Differentiating the COM definition gives \( M\vec v_{cm}=\sum m_i\vec v_i=\vec P \) and \( M\vec a_{cm}=\vec F_{ext} \): internal forces, which occur in equal and opposite pairs, never change the COM motion. When the net external force is zero the total momentum is conserved and the COM stays at rest or moves with constant velocity. Hence a projectile that explodes in flight leaves its COM moving along the original parabola until a fragment lands, and a man walking on a boat floating on still water makes the boat move so that the COM of the system does not shift. Example: a 60 kg man walks the full 6 m length of a 120 kg boat. The COM is fixed, so the boat moves backwards by \( \dfrac{60\times6}{60+120}=2 \) m relative to the water.

Conservation of Momentum and Impulse

For an isolated system \( \sum m_i\vec v_i \) is constant, and this can be applied along any direction in which the net external force is zero. Impulse is the integral of force over time, \( \vec J=\int\vec F\,dt=\Delta\vec p \), and equals the area under the force–time graph; the average force is the impulse divided by the contact time. For steady streams of mass hitting a surface, the force is \( \dfrac{dm}{dt}\) times the change in velocity per unit mass, which explains machine-gun recoil and the force of a water jet. A spring between two blocks exerts equal and opposite forces, so the momentum of the pair is conserved while energy is exchanged between the springs and the blocks.

Collisions

In every collision the momentum of the system is conserved, because the impact forces are internal. The coefficient of restitution is \( e=\dfrac{\text{speed of separation}}{\text{speed of approach}} \) along the line of impact, with \( e=1 \) for an elastic collision (kinetic energy conserved) and \( e=0 \) for a perfectly inelastic one (bodies move together). For a head-on collision of \( m_1 \) with \( u_1 \) and \( m_2 \) at rest: \( v_1=\dfrac{(m_1-em_2)u_1}{m_1+m_2} \), \( v_2=\dfrac{(1+e)m_1u_1}{m_1+m_2} \). Equal masses exchange velocities in an elastic collision. The kinetic energy lost is \( \tfrac12\mu v_{rel}^2(1-e^2) \) with reduced mass \( \mu=\dfrac{m_1m_2}{m_1+m_2} \). Example: a 2 kg ball at 6 m/s strikes a 4 kg ball at rest elastically. Then \( v_1=\dfrac{2-4}{6}\times6=-2 \) m/s and \( v_2=\dfrac{2\times2}{6}\times6=4 \) m/s, and the kinetic energy \( 36 \) J is unchanged \( \left(\tfrac12\cdot2\cdot4+\tfrac12\cdot4\cdot16=36\right) \). For a 1 kg and 3 kg pair approaching with relative speed 4 m/s and sticking together, the loss is \( \tfrac12\cdot\tfrac34\cdot16=6 \) J.

Oblique Collisions and Bouncing

Resolve velocities along the common normal (line of centres or normal to the wall) and along the tangent. Momentum is conserved along the normal; the tangential velocity of each smooth sphere is unchanged, and the normal components are related by the coefficient of restitution. For a ball hitting a smooth fixed floor the horizontal velocity is unchanged and the vertical velocity becomes \( e \) times its value, so the angle of rebound satisfies \( \tan\theta'=\dfrac{\tan\theta}{e} \) with the normal. A ball dropped from height H and bouncing with coefficient e reaches \( e^2H \) after the first rebound and \( e^{2n}H \) after n rebounds. Two equal smooth spheres colliding elastically with one initially at rest move off at right angles to each other.

Variable Mass and Rocket Motion

When mass is gained or lost continuously, apply momentum conservation to the system over a short interval dt, keeping track of the velocity of the added or ejected mass relative to the body. For a body of mass m moving at v while ejecting mass at relative speed u, \( m\dfrac{dv}{dt}=F_{ext}+u\left(-\dfrac{dm}{dt}\right) \), where the second term is the thrust. With no external force the rocket equation is \( v-v_0=u\ln\dfrac{m_0}{m} \). Example: a rocket in free space ejects gas at 2 km/s relative to itself; when its mass falls to \( 1/e \) of the initial value, its speed gain is \( 2\ln e=2 \) km/s. For a cart collecting or spilling sand, write the momentum before and after for the system including the sand, and remember that sand dropped vertically with no horizontal velocity relative to the cart exerts no horizontal force on it.