Centres of Mass and Elastic Collisions

This topic explores fundamental concepts that shape our understanding of the world.

This topic covers two major areas of further mechanics: finding the centre of mass of laminas, Solids, and composite bodies, and analysing elastic and inelastic collisions between particles Including oblique impacts.

Definition. The centre of mass of a lamina bounded by $y = f(x)$, $x = a$, $x = b$And the $x$-axis is the point $(\bar{x}, \bar{y})$ where:

$\boxed{\bar{x} = \frac{\displaystyle\int_a^b x \cdot f(x)\,dx}{\displaystyle\int_a^b f(x)\,dx}}$

$\boxed{\bar{y} = \frac{\displaystyle\int_a^b \frac{1}{2}[f(x)]^2\,dx}{\displaystyle\int_a^b f(x)\,dx}}$

The denominator is the total area of the lamina: $A = \displaystyle\int_a^b f(x)\,dx$.

Consider a triangle with vertices at $(0, 0)$, $(b, 0)$And $(c, h)$.

The line from $(0, 0)$ to $(c, h)$ is $y = \dfrac{h}{c}\,x$ and the line from $(b, 0)$ to $(c, h)$ Is $y = \dfrac{h}{c - b}(x - b)$.

For simplicity, take a right triangle with vertices (0, 0)$$(b, 0)$$(0, h)Where $f(x) = h - \dfrac{h}{b}x = h\!\left(1 - \dfrac{x}{b}\right)$.

$\bar{x} = \frac{\displaystyle\int_0^b x \cdot h\!\left(1 - \frac{x}{b}\right)dx}{\displaystyle\int_0^b h\!\left(1 - \frac{x}{b}\right)dx}$

Numerator: $\displaystyle h\int_0^b \left(x - \frac{x^2}{b}\right)dx = h\left[\frac{x^2}{2} - \frac{x^3}{3b}\right]_0^b = h\left(\frac{b^2}{2} - \frac{b^2}{3}\right) = \frac{hb^2}{6}$.

Denominator: $\displaystyle h\int_0^b \left(1 - \frac{x}{b}\right)dx = h\left[x - \frac{x^2}{2b}\right]_0^b = \frac{hb}{2}$.

$\bar{x} = \frac{hb^2/6}{hb/2} = \frac{b}{3}$

$\bar{y} = \frac{\displaystyle\int_0^b \frac{1}{2}h^2\!\left(1 - \frac{x}{b}\right)^2 dx}{hb/2} = \frac{\dfrac{h^2}{2}\displaystyle\int_0^b \left(1 - \frac{2x}{b} + \frac{x^2}{b^2}\right)dx}{hb/2}$

$= \frac{\dfrac{h^2}{2}\!\left[b - b + \dfrac{b}{3}\right]}{hb/2} = \frac{h^2 b / 6}{hb/2} = \frac{h}{3}$

For a general triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$:

$\boxed{\bar{x} = \frac{x_1 + x_2 + x_3}{3}, \qquad \bar{y} = \frac{y_1 + y_2 + y_3}{3}}$

$\square$

$\boxed{\bar{x} = \frac{x_1 + x_2 + x_3}{3}, \qquad \bar{y} = \frac{y_1 + y_2 + y_3}{3}}$

For a triangle of base $b$ and height $h$ with base on the $x$-axis: $\bar{y} = \dfrac{h}{3}$.

For a uniform semicircular lamina of radius $r$:

$\boxed{\bar{y} = \frac{4r}{3\pi}}$

The centre of mass lies on the axis of symmetry, a distance $\dfrac{4r}{3\pi}$ from the Diameter.

For a sector of a circle of radius $r$ with half-angle $\alpha$ (so the sector subtends $2\alpha$ at The centre):

$\boxed{\bar{x} = \frac{2r\sin\alpha}{3\alpha}}$

This lies on the axis of symmetry. For a semicircle ($\alpha = \pi/2$): $\bar{x} = \dfrac{2r}{3(\pi/2)} = \dfrac{4r}{3\pi}$Consistent with Section 2.2.

For a uniform circular arc of radius $r$ subtending angle $2\alpha$ at the centre:

$\boxed{\bar{x} = \frac{r\sin\alpha}{\alpha}}$

Definition. For a body composed of $n$ parts with masses $m_1, m_2, \ldots, m_n$ and centres of Mass at $(x_1, y_1), (x_2, y_2), \ldots, (x_n, y_n)$:

$\boxed{\bar{x} = \frac{\displaystyle\sum_{i=1}^{n} m_i x_i}{\displaystyle\sum_{i=1}^{n} m_i}}$

$\boxed{\bar{y} = \frac{\displaystyle\sum_{i=1}^{n} m_i y_i}{\displaystyle\sum_{i=1}^{n} m_i}}$

For a composite body, negative masses can be used for holes or removed sections.

A framework is made of uniform wires (rods). Each rod has its centre of mass at its midpoint. The Total mass is proportional to the total length.

$\boxed{\bar{x} = \frac{\displaystyle\sum m_i x_i}{\displaystyle\sum m_i} = \frac{\displaystyle\sum \ell_i x_i}{\displaystyle\sum \ell_i}}$

Where $\ell_i$ is the length of the $i$-th rod and $x_i$ is the $x$-coordinate of its midpoint.

When a lamina is freely suspended from a point, it hangs with its centre of mass directly below the Point of suspension. This means:

• The line of action of the weight passes through the point of suspension.
• The lamina is in equilibrium when the point of suspension is vertically above the centre of mass.

For a body suspended from a point $P$ to hang in equilibrium, the centre of mass $G$ must be Directly below $P$. If suspended from a second point $Q$, $G$ must be directly below $Q$. The Intersection of the two vertical lines through $P$ and $Q$ gives $G$.

Definition. The impulse $J$ delivered by a force $F$ acting for a time $\Delta t$ is:

$\boxed{J = F \cdot \Delta t = \Delta p = m(v - u)}$

Where $u$ is the initial velocity and $v$ is the final velocity.

Definition. The coefficient of restitution $e$ for a collision between two bodies is:

$\boxed{e = -\frac{v_1 - v_2}{u_1 - u_2}}$

Where $u_1, u_2$ are the velocities before collision and $v_1, v_2$ are the velocities after Collision, with all velocities measured in the same direction.

• $e = 1$: perfectly elastic collision (kinetic energy conserved)
• $e = 0$: perfectly inelastic collision (bodies coalesce)
• $0 < e < 1$: inelastic collision

For two particles of masses $m_1$ and $m_2$ with velocities $u_1$ and $u_2$:

Conservation of momentum:

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

Newton’s experimental law:

$v_2 - v_1 = e(u_1 - u_2)$

Solving simultaneously:

$v_1 = \frac{m_1 u_1 + m_2 u_2 - m_2 e(u_1 - u_2)}{m_1 + m_2}$

$v_2 = \frac{m_1 u_1 + m_2 u_2 + m_1 e(u_1 - u_2)}{m_1 + m_2}$

Consider two particles of masses $m_1$ and $m_2$ with velocities $u_1$ and $u_2$ colliding with Coefficient of restitution $e$.

The loss in kinetic energy is:

$\Delta KE = \left(\frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2\right) - \left(\frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2\right)$

Using the solutions for $v_1$ and $v_2$ and defining the reduced mass $\mu = \dfrac{m_1 m_2}{m_1 + m_2}$:

$\boxed{\Delta KE = \frac{1}{2}\mu(u_1 - u_2)^2(1 - e^2)}$

Where $\mu = \dfrac{m_1 m_2}{m_1 + m_2}$ is the reduced mass.

When $e = 1$: $\Delta KE = 0$ (perfectly elastic, no energy loss). ✓ When $e = 0$: $\Delta KE = \dfrac{1}{2}\mu(u_1 - u_2)^2$ (maximum energy loss for coalescence). ✓

$\square$

When a smooth sphere hits a smooth wall, the component of velocity parallel to the wall is Unchanged, and the component perpendicular to the wall is reversed and reduced by the coefficient of Restitution.

If the wall is along the $y$-axis and the sphere approaches with velocity $(u_x, u_y)$:

$v_x = -e \cdot u_x, \qquad v_y = u_y$

The angle of incidence $\alpha$ and angle of reflection $\beta$ satisfy:

$\tan\beta = \frac{u_y}{e \cdot u_x} = \frac{\tan\alpha}{e}$

Since $e \leq 1$We have $\tan\beta \geq \tan\alpha$So the angle of reflection is greater than or Equal to the angle of incidence.

When two smooth spheres collide obliquely, we resolve velocities into the normal direction (along The line of centres) and the tangential direction (perpendicular to the line of centres).

• Tangential components are unchanged (smooth spheres).
• Normal components obey conservation of momentum and Newton’s restitution law.

Steps:

1. Resolve the velocities of both spheres into normal and tangential components.
2. Apply conservation of momentum in the normal direction.
3. Apply Newton’s restitution law in the normal direction.
4. Combine to find the final velocities.

$\boxed{\bar{x} = \frac{\displaystyle\sum m_i x_i}{\displaystyle\sum m_i}, \qquad \bar{y} = \frac{\displaystyle\sum m_i y_i}{\displaystyle\sum m_i}}$

$\boxed{\bar{y}_{\mathrm{semicircle}} = \frac{4r}{3\pi}, \qquad \bar{x}_{\mathrm{sector}} = \frac{2r\sin\alpha}{3\alpha}}$

$\boxed{e = -\frac{v_1 - v_2}{u_1 - u_2}}$

$\boxed{\Delta KE = \frac{1}{2}\mu(u_1 - u_2)^2(1 - e^2), \quad \mu = \frac{m_1 m_2}{m_1 + m_2}}$

$\boxed{\mathrm{Oblique wall impact: } v_{\mathrm{normal}} = -e \cdot u_{\mathrm{normal}}, \quad v_{\mathrm{tangential}} = u_{\mathrm{tangential}}}$

Problem. Two smooth spheres $A$ and $B$ have equal mass $m$. Before collision, $A$ moves with Velocity $(5\mathbf{i} + 3\mathbf{j})$ m/s and $B$ is stationary. The line of centres at impact Makes angle $\alpha$ with $\mathbf{i}$Where $\tan\alpha = 3/4$. The coefficient of restitution is $e = 1/2$. Find the velocities after collision.

Solution. The normal direction is along the line of centres: $\hat{\mathbf{n}} = \dfrac{4}{5}\mathbf{i} + \dfrac{3}{5}\mathbf{j}$.

The tangential direction: $\hat{\mathbf{t}} = -\dfrac{3}{5}\mathbf{i} + \dfrac{4}{5}\mathbf{j}$.

Resolving $A$‘s velocity: $u_{An} = (5)(4/5) + (3)(3/5) = 4 + 9/5 = 29/5$.

$u_{At} = (5)(-3/5) + (3)(4/5) = -3 + 12/5 = -3/5$.

$u_{Bn} = 0$, $u_{Bt} = 0$.

Tangential components unchanged: $v_{At} = -3/5$, $v_{Bt} = 0$.

Normal direction (conservation of momentum): $m(29/5) + m(0) = mv_{An} + mv_{Bn} \implies v_{An} + v_{Bn} = 29/5$ … (i)

Restitution: $v_{Bn} - v_{An} = (1/2)(29/5) = 29/10$ … (ii)

From (i) and (ii): $2v_{Bn} = 29/5 + 29/10 = 87/10 \implies v_{Bn} = 87/20$.

$v_{An} = 29/5 - 87/20 = 116/20 - 87/20 = 29/20$.

$\mathbf{v}_A = v_{An}\hat{\mathbf{n}} + v_{At}\hat{\mathbf{t}} = \dfrac{29}{20}\!\left(\dfrac{4}{5}\mathbf{i} + \dfrac{3}{5}\mathbf{j}\right) + \dfrac{-3}{5}\!\left(-\dfrac{3}{5}\mathbf{i} + \dfrac{4}{5}\mathbf{j}\right)$

$= \left(\dfrac{116}{100} + \dfrac{9}{25}\right)\mathbf{i} + \left(\dfrac{87}{100} - \dfrac{12}{25}\right)\mathbf{j}$

$= \left(\dfrac{29}{25} + \dfrac{9}{25}\right)\mathbf{i} + \left(\dfrac{87}{100} - \dfrac{48}{100}\right)\mathbf{j} = \dfrac{38}{25}\mathbf{i} + \dfrac{39}{100}\mathbf{j}$

$\mathbf{v}_B = \dfrac{87}{20}\!\left(\dfrac{4}{5}\mathbf{i} + \dfrac{3}{5}\mathbf{j}\right) = \dfrac{87}{25}\mathbf{i} + \dfrac{261}{100}\mathbf{j}$

Problem. A uniform square lamina $ABCD$ has side $6a$. An equilateral triangle of side $2a$ is Removed with one vertex at the centre of the square and the opposite side on $AB$. Find the centre Of mass of the remaining lamina.

Solution. Square: area $= 36a^2$Centre of mass at $(3a, 3a)$.

Equilateral triangle with side $2a$: area $= \dfrac{\sqrt{3}}{4}(2a)^2 = \sqrt{3}\,a^2$.

Height $= \sqrt{3}\,a$. The triangle’s centroid is at distance $\sqrt{3}\,a/3$ from the base.

Assuming the square has vertices at (0, 0)$$(6a, 0)$$(6a, 6a)$$(0, 6a)And the triangle has Its base on the top edge $y = 6a$ with centroid at $(3a, 6a - \sqrt{3}\,a/3)$:

Using negative mass:

$\bar{y} = \frac{36a^2 \times 3a - \sqrt{3}\,a^2 \times (6a - a\sqrt{3}/3)}{36a^2 - \sqrt{3}\,a^2} = \frac{108a^3 - 6\sqrt{3}\,a^3 + a^3}{a^2(36 - \sqrt{3})}$

$= \frac{a(109 - 6\sqrt{3})}{36 - \sqrt{3}}$

Problem. A ball is projected from point $O$ with speed $u$ at angle $\alpha$ to the horizontal Towards a smooth vertical wall at horizontal distance $d$. The coefficient of restitution between The ball and the wall is $e$. Show that the horizontal distance from the wall to the point where the Ball next hits the ground is $ed$.

Solution. Time to reach the wall: $t_1 = d/(u\cos\alpha)$.

After impact with the wall:

• Horizontal velocity reverses and reduces: $v_x' = e \cdot u\cos\alpha$ (moving away from wall).
• Vertical velocity unchanged: $v_y' = u\sin\alpha - gd/(u\cos\alpha)$.

The ball follows a parabolic trajectory after bouncing. By the reversibility of projectile motion And the scaling of horizontal velocity by factor $e$The horizontal range from the wall is $ed$. $\blacksquare$

Problem. Find the centre of mass of a uniform solid right circular cone of height $h$ and base Radius $r$.

Solution. Place the cone with its vertex at the origin and axis along the $z$-axis, extending to $z = h$.

At height $z$The cross-section is a disc of radius $\dfrac{rz}{h}$With volume $dV = \pi\!\left(\dfrac{rz}{h}\right)^2 dz$.

$\bar{z} = \frac{\displaystyle\int_0^h z \cdot \pi r^2 z^2/h^2\,dz}{\displaystyle\int_0^h \pi r^2 z^2/h^2\,dz} = \frac{h^4/4}{h^3/3} = \frac{3h}{4}$

The centre of mass is at distance $\dfrac{3h}{4}$ from the vertex (or $\dfrac{h}{4}$ from the base).

Problem. Three identical particles A$$B$$C of mass $m$ are at rest in a straight line on a Smooth surface with equal spacing $d$. Particle $A$ is given velocity $u$ towards $B$. If all Collisions are perfectly elastic ($e = 1$), describe the subsequent motion.

Solution. Collision 1 ($A$ hits $B$): By symmetry of equal masses with e = 1$$A stops And $B$ moves with velocity $u$ towards $C$.

Collision 2 ($B$ hits $C$): Similarly, $B$ stops and $C$ moves with velocity $u$ away.

After both collisions: $A$ at rest, $B$ at rest, $C$ moves with velocity $u$. No further collisions Occur.

Problem. A uniform rectangular lamina $ABCD$ with $AB = 8$ cm and $BC = 6$ cm is freely Suspended from vertex $A$ and hangs in equilibrium. Find the angle that diagonal $AC$ makes with the Vertical.

Solution. Centre of mass $G$ is at the centre of the rectangle. With $A$ at the origin and $B$ Along the positive $x$-axis, $G = (4, 3)$.

When suspended from $A$The line $AG$ is vertical. The vector $AG = (4, 3)$ makes angle $\arctan(3/4)$ with the horizontal.

The diagonal $AC = (8, 6)$ also makes angle $\arctan(6/8) = \arctan(3/4)$ with the horizontal.

Since $AG$ is parallel to $AC$The angle between the diagonal $AC$ and the vertical is the same as The angle between $AG$ and the vertical: $90^\circ - \arctan(3/4) = \arctan(4/3) \approx 53.1^\circ$.

Finding centres of mass by integration requires the same techniques as volumes of revolution: Substitution, integration by parts, and definite integrals. See Further Calculus.

The kinetic energy loss formula $\Delta KE = \frac{1}{2}\mu(u_1 - u_2)^2(1-e^2)$ connects to the Work-energy principle and conservation of momentum. See Projectile Motion.

Resolving velocities in oblique collisions requires vector decomposition and dot products. See Vectors in 3D.

A uniform lamina is in the shape of a semicircle of radius $a$ with a circle of radius $a/2$ Removed. The centre of the removed circle lies on the diameter of the semicircle, at distance $a/2$ From the centre of the semicircle. Find the centre of mass of the remaining lamina.

A particle of mass $2$ kg moving at $6$ m/s collides directly with a stationary particle of mass $m$ Kg. After collision, the 2 kg particle rebounds with speed $1$ m/s and the coefficient of Restitution is $e = 2/3$. Find $m$.

Prove that in any elastic collision between two particles (with $e = 1$), the relative speed of Separation equals the relative speed of approach.

A uniform solid hemisphere of radius $r$ and a uniform solid cone of base radius $r$ and height $h$ Are joined base-to-base. Both are made of the same material. For what value of $h$ does the Composite body have its centre of mass exactly at the join?

Problem. A uniform lamina consists of a semicircle of radius $a$ attached to a rectangle of Width $2a$ and height $h$. The flat side of the semicircle coincides with one edge of the rectangle. Find the distance of the centre of mass from the base of the rectangle.

Solution. Semicircle: area $= \dfrac{\pi a^2}{2}$Centre of mass at $\dfrac{4a}{3\pi}$ above the diameter.

Rectangle: area $= 2ah$Centre of mass at $\dfrac{h}{2}$ above the base.

Taking the base as datum:

$\bar{y} = \frac{\frac{\pi a^2}{2}\!\left(h + \frac{4a}{3\pi}\right) + 2ah \cdot \frac{h}{2}}{\frac{\pi a^2}{2} + 2ah}$

$= \frac{\frac{\pi a^2 h}{2} + \frac{2a^3}{3} + ah^2}{\frac{\pi a^2}{2} + 2ah}$

Problem. A particle of mass $m$ moving at $4\,\mathrm{m\,s^{-1}}$ collides elastically with a Stationary particle of mass $2m$. After the collision, the first particle moves at $60°$ to its Original direction. Find the speeds after collision.

Solution. Conservation of momentum (along original direction): $m \times 4 = mv_1\cos 60° + 2mv_2\cos\theta$.

$4 = \dfrac{v_1}{2} + 2v_2\cos\theta$ … (1)

Perpendicular to original direction: $0 = mv_1\sin 60° - 2mv_2\sin\theta$.

$v_1\sin 60° = 2v_2\sin\theta$ … (2)

Conservation of KE: $\dfrac{1}{2}m \times 16 = \dfrac{1}{2}mv_1^2 + \dfrac{1}{2}(2m)v_2^2$.

$16 = v_1^2 + 2v_2^2$ … (3)

From (2): $v_2\sin\theta = \dfrac{v_1\sqrt{3}}{4}$. From (1): $v_2\cos\theta = 2 - \dfrac{v_1}{4}$.

Squaring and adding: $v_2^2 = \dfrac{3v_1^2}{16} + 4 - \dfrac{v_1}{2} + \dfrac{v_1^2}{16} = \dfrac{v_1^2}{4} - \dfrac{v_1}{2} + 4$.

Substituting into (3): $16 = v_1^2 + 2\!\left(\dfrac{v_1^2}{4} - \dfrac{v_1}{2} + 4\right) = \dfrac{3v_1^2}{2} - v_1 + 8$.

$\dfrac{3v_1^2}{2} - v_1 - 8 = 0 \implies 3v_1^2 - 2v_1 - 16 = 0$.

$v_1 = \dfrac{2 \pm \sqrt{4+192}}{6} = \dfrac{2 \pm 14}{6}$. Taking Positive: $v_1 = \dfrac{16}{6} = \dfrac{8}{3}\,\mathrm{m\,s^{-1}}$.

$v_2^2 = \dfrac{64}{9 \times 4} - \dfrac{4}{3} + 4 = \dfrac{16}{9} - \dfrac{4}{3} + 4 = \dfrac{16-12+36}{9} = \dfrac{40}{9}$.

$v_2 = \dfrac{2\sqrt{10}}{3}\,\mathrm{m\,s^{-1}}$.

Problem. A uniform solid cylinder of radius $r$ and mass $m$ is placed on a rough inclined plane At angle $\alpha$. The coefficient of friction is $\mu$. Determine whether the cylinder slides or Rolls.

Solution. For sliding: $mg\sin\alpha > \mu mg\cos\alpha \implies \tan\alpha > \mu$.