Chapter 6 ยท System of Particles and Rotational Motion

6.1 Centre of Mass and Its Motion

Every system of particles has one special point, the centre of mass, that moves as if all the mass were there and all the external forces acted on it: $M\mathbf{a}_\text{cm}=\mathbf{F}_\text{ext}$.

A firework shell climbs into the night sky and bursts into a hundred glowing fragments that fly off in every direction. Each fragment follows its own path, and the scene looks hopelessly complicated. Yet a physicist on the ground can say exactly where the "average" of all those fragments is at every moment: it carries on along the very parabola the shell was following before it burst, as though nothing had happened. The explosion, violent as it is, is entirely internal to the shell.

Chapters 2โ€“5 treated every body as a particle. Real bodies have size: they spin, tumble and change shape, and a crowd of particles โ€” a gas, a galaxy, a family on a raft โ€” is more complicated still. This section finds the one point of any system, the centre of mass, whose motion is as simple as a particle's. Its position is a mass-weighted average, it can be found by symmetry for many everyday bodies, and Newton's second law for it involves only the external forces. That lets us predict where exploding shells land, how far a boat slides when someone walks along it, and why no amount of internal effort can move a system's centre of mass.

Rigid bodies: translation and rotation

Most solid objects โ€” a cricket bat, a wheel, a door โ€” keep their shape as they move: however they are pushed, the distances between their parts hardly change. We idealise this.

๐Ÿ“˜ Definition 6.1 โ€” Rigid body; translation and rotation

A rigid body is a body in which the distance between every pair of particles stays fixed, whatever forces act on it. Two simple kinds of motion are possible:

  • Pure translation: at every instant all particles of the body have the same velocity. Any line drawn in the body stays parallel to itself.
  • Rotation about a fixed axis: one line of the body (the axis of rotation) stays at rest, and every other particle moves in a circle in a plane perpendicular to the axis, with its centre on the axis.

The general motion of a rigid body is a combination of a translation and a rotation.

A box sliding down a ramp without tipping is in pure translation: its top corner and its bottom corner move along parallel lines at the same speed. A ceiling fan, a potter's wheel, a door on its hinges and the hands of a clock rotate about fixed axes. A cylinder rolling down a slope does both at once: its axis moves forward (translation) while the cylinder turns about that axis (rotation). A spinning top whose point stays on the floor has a fixed point but its axis swings round โ€” a more complicated motion we will not analyse.

Real bodies are never perfectly rigid โ€” a tennis ball squashes, a steel beam bends a little โ€” but for most problems in this chapter the changes of shape are too small to matter. When they do matter (Chapter 8) we study them on their own.

Centre of mass of a system of particles

Start with two particles on the $x$-axis: mass $m_1$ at $x_1$ and $m_2$ at $x_2$. Their centre of mass is at

$$X=\frac{m_1x_1+m_2x_2}{m_1+m_2}.$$

This is an average of $x_1$ and $x_2$ in which each position is weighted by its mass. If the masses are equal, $X=\tfrac12(x_1+x_2)$, the midpoint. If $m_2$ is larger, $X$ moves towards $x_2$. Put the origin at the centre of mass ($X=0$) and the formula says $m_1x_1=-m_2x_2$: the distances $d_1,d_2$ of the particles from the centre of mass satisfy $m_1d_1=m_2d_2$. The centre of mass divides the line joining the particles in the inverse ratio of their masses โ€” exactly where a light rod carrying the two masses would balance on a knife-edge.

๐Ÿ“˜ Definition 6.2 โ€” Centre of mass

For a system of $n$ particles of masses $m_1,m_2,\dots,m_n$ at position vectors $\mathbf{r}_1,\mathbf{r}_2,\dots,\mathbf{r}_n$, with total mass $M=\sum m_i$, the centre of mass is the point with position vector

$$\mathbf{R}_\text{cm}=\frac{\sum m_i\mathbf{r}_i}{M}.$$

In components,

$$X=\frac{\sum m_ix_i}{M},\quad Y=\frac{\sum m_iy_i}{M},\quad Z=\frac{\sum m_iz_i}{M}.$$

The centre of mass is a property of the system, not of the coordinates: moving the origin shifts every $\mathbf{r}_i$ and $\mathbf{R}_\text{cm}$ by the same vector, so the physical point is the same.

โœ๏ธ Example 1 โ€” Two particles, and the Earth and the Moon
(a) A $2.0\ \text{kg}$ ball is at $x=0$ and a $3.0\ \text{kg}$ ball at $x=1.0\ \text{m}$. Find their centre of mass and check the inverse-ratio rule. (b) The Earth ($5.97\times10^{24}\ \text{kg}$) and the Moon ($7.35\times10^{22}\ \text{kg}$) are $3.84\times10^8\ \text{m}$ apart, centre to centre. Find the distance of their centre of mass from the Earth's centre. Is it inside or outside the Earth ($R_E=6.37\times10^6\ \text{m}$)?
(a) $X=\dfrac{2.0\times0+3.0\times1.0}{2.0+3.0}=0.60\ \text{m}$ โ€” nearer the heavier ball.
Rule $d_1=0.60\ \text{m}$, $d_2=0.40\ \text{m}$: $m_1d_1=1.2=m_2d_2$ โœ“, so $d_1:d_2=3:2=m_2:m_1$.
(b) Origin at the Earth's centre: $X=\dfrac{m_\text{M}\,d}{M_E+m_\text{M}}=\dfrac{7.35\times10^{22}\times3.84\times10^8}{6.04\times10^{24}}=4.67\times10^6\ \text{m}$.
Inside? $4.67\times10^6\ \text{m}\lt R_E$: the centre of mass lies inside the Earth, about $1.7\times10^3\ \text{km}$ below the surface. Both bodies circle this point once a month, so the Earth's centre wobbles round it.
Check $M_E\times X=2.79\times10^{31}$ and $m_\text{M}\times(d-X)=7.35\times10^{22}\times3.79\times10^8=2.79\times10^{31}$ โœ“.
โœ๏ธ Example 2 โ€” Three particles in a plane
Particles of mass $1.0\ \text{kg}$, $2.0\ \text{kg}$ and $3.0\ \text{kg}$ are at the corners $(0,\,0)$, $(4.0,\,0)$ and $(0,\,3.0)$ of a right-angled triangle (coordinates in metres). Find their centre of mass.
Mass $M=1.0+2.0+3.0=6.0\ \text{kg}$.
$X$ $X=\dfrac{1.0(0)+2.0(4.0)+3.0(0)}{6.0}=\dfrac{8.0}{6.0}=1.33\ \text{m}$.
$Y$ $Y=\dfrac{1.0(0)+2.0(0)+3.0(3.0)}{6.0}=1.50\ \text{m}$.
Check The point $(1.33,\,1.50)\ \text{m}$ lies inside the triangle ($\tfrac{1.33}{4.0}+\tfrac{1.50}{3.0}=0.83\lt1$) and is pulled towards the heavy $3.0\ \text{kg}$ corner. It is the default arrangement of Interactive 1 โ€” pause the animation to see the readouts $X=1.33\ \text{m}$, $Y=1.50\ \text{m}$ โœ“.
Interactive 1 โ€” Centre-of-Mass Finder

Drag the masses around the grid (1 square = $0.5\ \text{m}$) and change any mass with the slider. The purple target is $\mathbf{R}_\text{cm}=\sum m_i\mathbf{r}_i/M$, recomputed as you move. Press Play to send the lightest mass round a circle and watch how much โ€” or how little โ€” the centre of mass follows it.

1.0 kg
total mass $M$โ€”
$X=\sum m_ix_i/M$โ€”
$Y=\sum m_iy_i/M$โ€”

Centre of mass of extended and composite bodies

A brick or a cricket bat contains about $10^{25}$ atoms, far too many to add up one by one. Instead imagine the body cut into many small pieces, each small enough to treat as a particle at its own position, and add $\Delta m\,\mathbf{r}$ over the pieces. In the limit of very small pieces the sum becomes an integral, $\mathbf{R}_\text{cm}=\dfrac{1}{M}\int\mathbf{r}\,dm$. For the regular bodies met in this course we never need to evaluate it: symmetry does the work.

Take a uniform rod lying from $x=-\tfrac{L}{2}$ to $x=+\tfrac{L}{2}$. For every small piece at $+x$ there is an identical piece at $-x$; their contributions $\Delta m\,x$ and $\Delta m(-x)$ cancel in pairs, so $X=0$ โ€” the midpoint. The same pairing argument works for any uniform body that has a centre of symmetry, and along any axis or plane of symmetry.

๐Ÿ“ Result 6.1 โ€” Centre of mass by symmetry

For a body of uniform density (uniform thickness for a plate, uniform cross-section for a rod or wire), the centre of mass lies on every axis or plane of symmetry. In particular it is at

  • the midpoint of a uniform rod;
  • the geometric centre of a uniform ring, disc, sphere (solid or hollow) or cube;
  • the intersection of the diagonals of a uniform rectangular plate, and the midpoint of the axis of a uniform cylinder;
  • the intersection of the medians (the centroid) of a uniform triangular plate.

The centre of mass need not lie in the material: for a ring it is at the empty centre.

The triangle needs one more line of reasoning. Slice a uniform triangular plate into thin strips parallel to one side. Each strip is a thin uniform rod whose midpoint lies on the median drawn to that side, so the whole centre of mass lies on that median. The same is true for the other two medians, so it lies where they meet. Placing equal masses at the three vertices gives the same point: the centroid is at the average of the vertex coordinates.

Many objects are built from simple pieces โ€” an L-shaped bracket, a T-square, a hammer with a head and a handle. Each piece can be replaced by a particle carrying the piece's whole mass at the piece's own centre of mass, and then Definition 6.2 is applied to these few "particles".

๐Ÿงญ Problem-solving strategy โ€” composite bodies
  1. Split the body into simple parts (rectangles, discs, rods) and choose axes โ€” along any line of symmetry if possible.
  2. Find the mass of each part. For a uniform plate of thickness $t$ and density $\rho$, mass $=\rho t\times\text{area}$, so you may use areas in place of masses; for a uniform wire, use lengths.
  3. Locate each part's own centre of mass by symmetry (Result 6.1).
  4. Apply $X=\dfrac{\sum m_ix_i}{\sum m_i}$, $Y=\dfrac{\sum m_iy_i}{\sum m_i}$ to the parts.
  5. Holes: treat the complete body as one part and the hole as a part of negative mass (negative area) at the hole's centre.
โœ๏ธ Example 3 โ€” An L-shaped bracket
A uniform flat metal bracket is L-shaped. With the corner at the origin, one arm covers $0\le x\le40\ \text{cm}$, $0\le y\le10\ \text{cm}$, and the other covers $0\le x\le10\ \text{cm}$, $10\le y\le40\ \text{cm}$. Find its centre of mass. Does it lie on the metal?
Parts A: $40\times10$ rectangle, area $400\ \text{cm}^2$, centre $(20,\,5)$. B: $10\times30$ rectangle, area $300\ \text{cm}^2$, centre $(5,\,25)$. Uniform plate: use areas for masses.
$X$ $X=\dfrac{400(20)+300(5)}{700}=\dfrac{9500}{700}=13.6\ \text{cm}$.
$Y$ $Y=\dfrac{400(5)+300(25)}{700}=\dfrac{9500}{700}=13.6\ \text{cm}$.
Check The L is symmetric about the line $y=x$ (each arm is $40\ \text{cm}$ long and $10\ \text{cm}$ wide), so $X=Y$ was expected โœ“. The point $(13.6,\,13.6)\ \text{cm}$ has both coordinates greater than $10\ \text{cm}$, so it lies in the empty corner, off the metal.
โœ๏ธ Example 4 โ€” A disc with a hole
A circular hole of radius $6.0\ \text{cm}$ is cut from a uniform disc of radius $12\ \text{cm}$. The edge of the hole touches the edge of the disc. Where is the centre of mass of the remaining plate?
Set-up Origin at the centre of the full disc, $x$-axis through the centre of the hole, which is at $x=+6.0\ \text{cm}$. By symmetry about the $x$-axis, $Y=0$.
Masses Mass $\propto$ area $\propto r^2$. Full disc: $M$ at $x=0$. Hole: radius halved, so mass $\tfrac14M$, entered as $-\tfrac14M$ at $x=6.0\ \text{cm}$.
$X$ $X=\dfrac{M(0)-\tfrac14M(6.0)}{M-\tfrac14M}=\dfrac{-1.5}{0.75}=-2.0\ \text{cm}$.
Answer $2.0\ \text{cm}$ from the centre of the disc, on the side away from the hole โ€” in general $R/6$ for a hole of radius $R/2$.
Check Put the hole back: the remaining plate ($\tfrac34M$ at $-2.0$) plus the hole's material ($\tfrac14M$ at $+6.0$) gives $\tfrac34(-2.0)+\tfrac14(6.0)=0$, the centre of the full disc โœ“.

Motion of the centre of mass and the momentum of a system

Why is the centre of mass so special? Let the particles move. In a short time $\Delta t$ each $\mathbf{r}_i$ changes by $\Delta\mathbf{r}_i$, so $M\,\Delta\mathbf{R}_\text{cm}=\sum m_i\,\Delta\mathbf{r}_i$ (the total mass is constant). Dividing by $\Delta t$ and letting $\Delta t\to0$,

$$M\mathbf{V}_\text{cm}=\sum m_i\mathbf{v}_i=\mathbf{P},$$

the total linear momentum. Repeating the step with velocities gives $M\mathbf{A}_\text{cm}=\sum m_i\mathbf{a}_i=\sum\mathbf{F}_i$, where $\mathbf{F}_i$ is the total force on particle $i$. Part of each $\mathbf{F}_i$ is external (gravity, a push from outside the system) and part is internal (forces from the other particles of the system). By Newton's third law the internal forces come in equal and opposite pairs โ€” particle 1 pulls particle 2 exactly as hard as 2 pulls 1 โ€” so in the sum they cancel completely. Only the external forces survive.

๐Ÿ“ Result 6.2 โ€” Motion of the centre of mass

For a system of particles of constant total mass $M$,

$$M\mathbf{a}_\text{cm}=\mathbf{F}_\text{ext},$$

where $\mathbf{F}_\text{ext}$ is the vector sum of the external forces on all the particles. The centre of mass moves exactly like a single particle of mass $M$ acted on by the net external force. Internal forces โ€” however large, as in an explosion โ€” cannot change its motion.

๐Ÿ“ Result 6.3 โ€” Linear momentum of a system

The total linear momentum of a system of particles is

$$\mathbf{P}=\sum m_i\mathbf{v}_i=M\mathbf{v}_\text{cm},\qquad \frac{d\mathbf{P}}{dt}=\mathbf{F}_\text{ext}.$$

If the net external force is zero, $\mathbf{P}$ is conserved, so the centre of mass moves with constant velocity; if it starts at rest, it stays at rest. (This is the conservation of momentum of Section 4.2, seen through the centre of mass.)

The motion of a rigid body can therefore be split cleanly: the centre of mass moves under the external forces as a particle would โ€” this is the translational part โ€” and the body turns about the centre of mass, which is the subject of Sections 6.2 and 6.3. The tumbling dumbbell of the chapter introduction is exactly this: a projectile centre of mass with a rotation on top.

โœ๏ธ Example 5 โ€” The exploding shell
A shell is fired from level ground at $30\ \text{m/s}$, $60^\circ$ above the horizontal. At the highest point of its path it explodes into two fragments of equal mass. One fragment falls vertically, starting from rest. Neglecting air resistance, where does the other land?
Idea The explosion forces are internal. The only external force is gravity, before and after, so the centre of mass of the fragments carries on along the original parabola.
Range Unexploded range: $R=\dfrac{u^2\sin2\theta}{g}=\dfrac{900\times\sin120^\circ}{9.8}=79.5\ \text{m}$; the top is above $x=\tfrac12R=39.8\ \text{m}$.
Landing Both fragments start at the top with zero vertical velocity, so they reach the ground together, when the centre of mass does โ€” at $x=R$. Fragment 1 lands at $x_1=\tfrac12R$. Then $\dfrac{m x_1+m x_2}{2m}=R$ gives $x_2=2R-\tfrac12R=\tfrac32R=119\ \text{m}$ from the gun.
Check By momentum: at the top $v_x=30\cos60^\circ=15\ \text{m/s}$, so $2m(15)=m(0)+m v_2$, $v_2=30\ \text{m/s}$. The fall from the top takes $\tfrac12T=\dfrac{30\sin60^\circ}{9.8}=2.65\ \text{s}$, so fragment 2 travels $30\times2.65=79.5\ \text{m}$ beyond $39.8\ \text{m}$: $119\ \text{m}$ โœ“.
Interactive 2 โ€” The Exploding Shell

The shell of Example 5 bursts at the top of its path into fragments A and B. Choose A's share of the mass and A's horizontal velocity just after the burst (as a multiple of the shell's velocity $15\ \text{m/s}$ there; $0$ means A drops from rest). B's velocity follows from conservation of momentum. The purple dot is the centre of mass of the fragments: it never leaves the dashed parabola.

0.500 ร— 15 m/s
B's velocity after burstโ€”
A lands atโ€”
B lands atโ€”
centre of mass lands atโ€”
โœ๏ธ Example 6 โ€” Walking along a boat
A $60\ \text{kg}$ man stands at the back of a $140\ \text{kg}$ boat floating at rest on still water. He walks $4.0\ \text{m}$ towards the front, measured along the boat. Neglect the horizontal force of the water. (a) How far does the boat move, and which way? (b) How far does the man move relative to the shore? (c) If he walks at $1.0\ \text{m/s}$ relative to the boat, how fast does the boat move?
Idea Man + boat is the system. The friction between his feet and the deck is internal; there is no horizontal external force, and the centre of mass starts at rest, so it stays put (Result 6.3).
(a) Let the boat move $D$ (positive = forwards) relative to the shore. The man moves $D+4.0$. Centre of mass fixed: $60(D+4.0)+140D=0$, so $D=-\dfrac{240}{200}=-1.2\ \text{m}$: the boat moves $1.2\ \text{m}$ backwards.
(b) The man moves $-1.2+4.0=2.8\ \text{m}$ forwards relative to the shore.
(c) Total momentum zero: $60(v_b+1.0)+140v_b=0$, $v_b=-0.30\ \text{m/s}$ (backwards); the man moves at $0.70\ \text{m/s}$ forwards.
Check $60\times2.8=168=140\times1.2$ and $60\times0.70=42=140\times0.30$ โœ“ โ€” displacements and velocities both share in the inverse ratio of the masses.
โš ๏ธ Common mistakes

1. "The centre of mass is always inside the body." Not for a ring, a horseshoe, a boomerang or an L-shaped bracket (Example 3). It is a mathematical point defined by the mass distribution.

2. Using distances measured from different origins. Every $x_i$ in $\sum m_ix_i$ must be measured from the same origin. Choose it at one of the masses or on a symmetry line to make zeros.

3. Areas for non-uniform bodies. Replacing masses by areas (or lengths) is allowed only when the plate (or wire) is uniform. A hole is entered with a minus sign and subtracted from the total mass.

4. Thinking internal forces can move the centre of mass. A person cannot lift herself by pulling on her own hair, and an explosion cannot change where the centre of mass of the fragments goes. To accelerate, a car needs an external force โ€” friction from the road on its tyres.

5. Applying the "fragments" argument after one fragment has landed. Once a fragment hits the ground, the ground pushes on it: a new external force acts, and the centre of mass leaves the parabola.

6. Relative displacements. In boat problems the man walks $4.0\ \text{m}$ relative to the boat; his displacement relative to the shore is $D+4.0$, not $4.0$.

Exercises

Take $g=9.8\ \text{m/s}^2$ and neglect air resistance unless told otherwise. All bodies called "uniform" have uniform density and thickness.

I
Centre of mass of particles.
  • 1 A $4.0\ \text{kg}$ mass is at $x=0$ and a $6.0\ \text{kg}$ mass at $x=2.0\ \text{m}$. Find their centre of mass.
  • 2 Two particles of masses $2m$ and $m$ are $0.90\ \text{m}$ apart. How far is their centre of mass from the heavier one?
  • 3 Masses of $1.0$, $2.0$ and $3.0\ \text{kg}$ lie on the $x$-axis at $x=1.0$, $2.0$ and $3.0\ \text{m}$. Find $X$.
  • 4 In a carbon monoxide molecule the carbon atom ($12\ \text{u}$) and the oxygen atom ($16\ \text{u}$) are $1.13\times10^{-10}\ \text{m}$ apart. How far is the centre of mass from the carbon atom?
  • 5 Masses of $1.0$, $2.0$, $3.0$ and $4.0\ \text{kg}$ sit at the corners $(0,0)$, $(2.0,0)$, $(2.0,2.0)$ and $(0,2.0)$ of a square (metres). Find the centre of mass.
  • 6 Find the centre of mass of $2.0\ \text{kg}$ at $(1.0,\,2.0)$, $3.0\ \text{kg}$ at $(-2.0,\,1.0)$ and $5.0\ \text{kg}$ at $(0,\,-3.0)$ (metres).
  • 7 Particles of $1.0\ \text{kg}$ are at $(0,0)$ and $(4.0,\,0)$ m. Where must a $2.0\ \text{kg}$ particle be placed so that the centre of mass of all three is at $(2.0,\,2.0)\ \text{m}$?
  • 8 Prove that the centre of mass of two particles lies on the line joining them and divides it in the inverse ratio of their masses.
1 $X=\dfrac{6.0\times2.0}{10.0}=1.2\ \text{m}$.
2 $d_{2m}=\dfrac{m}{3m}\times0.90=0.30\ \text{m}$ (and $0.60\ \text{m}$ from the lighter one: $2m\times0.30=m\times0.60$).
3 $X=\dfrac{1+4+9}{6.0}=\dfrac{14}{6.0}=2.3\ \text{m}$ ($2.33$).
4 $d_\text{C}=\dfrac{16}{28}\times1.13\times10^{-10}=6.46\times10^{-11}\ \text{m}$ โ€” nearer the heavier oxygen atom.
5 $X=\dfrac{2(2)+3(2)}{10}=1.0\ \text{m}$, $Y=\dfrac{3(2)+4(2)}{10}=1.4\ \text{m}$: $(1.0,\,1.4)\ \text{m}$.
6 $X=\dfrac{2-6+0}{10}=-0.40\ \text{m}$, $Y=\dfrac{4+3-15}{10}=-0.80\ \text{m}$: $(-0.40,\,-0.80)\ \text{m}$.
7 $\dfrac{0+4.0+2x}{4.0}=2.0\Rightarrow x=2.0\ \text{m}$; $\dfrac{2y}{4.0}=2.0\Rightarrow y=4.0\ \text{m}$: at $(2.0,\,4.0)\ \text{m}$.
8 Choose the $x$-axis through both particles: then $y_1=y_2=z_1=z_2=0$, so $Y=Z=0$ โ€” the centre of mass is on the line. Put the origin at the centre of mass: $m_1x_1+m_2x_2=0$, so $x_1,x_2$ have opposite signs (it lies between them) and $\dfrac{|x_1|}{|x_2|}=\dfrac{m_2}{m_1}$.
II
Extended and composite bodies.
  • 1 Where is the centre of mass of (a) a uniform metre stick, (b) a uniform circular ring, (c) a uniform triangular plate? In which case is there no material at the centre of mass?
  • 2 A $50\ \text{g}$ mass is fixed at the $10\ \text{cm}$ mark of a uniform $100\ \text{g}$ metre stick. Find the mark at which the stick balances on a knife-edge.
  • 3 A uniform rectangular plate $0.60\ \text{m}\times0.40\ \text{m}$ lies with one corner at the origin and its edges along the axes. A square $0.20\ \text{m}\times0.20\ \text{m}$ is cut out of the corner at the origin. Find the centre of mass of what remains.
  • 4 A uniform T-shaped plate is made of a vertical stem $10\ \text{cm}$ wide and $30\ \text{cm}$ tall with a crossbar $30\ \text{cm}$ long and $10\ \text{cm}$ tall centred on top of it. How far above the bottom of the stem is the centre of mass?
  • 5 A circular hole of radius $R/3$ is cut from a uniform disc of radius $R$ so that the hole touches the disc's edge. Show that the centre of mass of the remainder is $R/12$ from the disc's centre.
  • 6 A uniform wire is bent into an L with legs $60\ \text{cm}$ (along $+x$) and $40\ \text{cm}$ (along $+y$) meeting at the origin. Find its centre of mass.
  • 7 A uniform rod $1.0\ \text{m}$ long is bent at its midpoint into a right angle. How far is its centre of mass from the bend?
1 (a) The $50\ \text{cm}$ mark; (b) the centre of the ring โ€” no material there; (c) the intersection of the medians.
2 Stick: $100\ \text{g}$ at $50\ \text{cm}$. $X=\dfrac{100(50)+50(10)}{150}=36.7\ \text{cm}$ mark.
3 Areas: full $0.24\ \text{m}^2$ at $(0.30,\,0.20)$; hole $-0.04\ \text{m}^2$ at $(0.10,\,0.10)$. $X=\dfrac{0.072-0.004}{0.20}=0.34\ \text{m}$, $Y=\dfrac{0.048-0.004}{0.20}=0.22\ \text{m}$: $(0.34,\,0.22)\ \text{m}$ โ€” moved away from the hole.
4 Stem: area $300\ \text{cm}^2$, centre $15\ \text{cm}$ up; bar: $300\ \text{cm}^2$, centre $35\ \text{cm}$ up. Equal areas: $Y=\tfrac12(15+35)=25\ \text{cm}$ above the bottom, on the axis of symmetry.
5 Hole mass $\tfrac19M$ at $x=R-\tfrac{R}{3}=\tfrac{2R}{3}$. $X=\dfrac{0-\tfrac19M\cdot\tfrac{2R}{3}}{\tfrac89M}=-\dfrac{2R}{27}\cdot\dfrac98=-\dfrac{R}{12}$, i.e. $R/12$ on the side away from the hole.
6 Masses $\propto$ lengths: $60$ at $(30,\,0)$, $40$ at $(0,\,20)$. $X=\dfrac{60\times30}{100}=18\ \text{cm}$, $Y=\dfrac{40\times20}{100}=8.0\ \text{cm}$: $(18,\,8.0)\ \text{cm}$.
7 Each half ($0.50\ \text{m}$) has its centre $0.25\ \text{m}$ from the bend along its own arm: centre of mass $(0.125,\,0.125)\ \text{m}$, at $0.125\sqrt2=0.18\ \text{m}$ from the bend.
III
Motion of the centre of mass and momentum of a system.
  • 1 A $2.0\ \text{kg}$ ball moves at $3.0\ \text{m/s}$ along $+x$ and a $3.0\ \text{kg}$ ball at $2.0\ \text{m/s}$ along $-x$. Find the velocity of their centre of mass and the total momentum.
  • 2 A $1.0\ \text{kg}$ particle has velocity $(2.0\hat{\mathbf{i}}+3.0\hat{\mathbf{j}})\ \text{m/s}$ and a $2.0\ \text{kg}$ particle $(-1.0\hat{\mathbf{i}}+1.0\hat{\mathbf{j}})\ \text{m/s}$. Find $\mathbf{v}_\text{cm}$.
  • 3 Blocks A ($2.0\ \text{kg}$) and B ($3.0\ \text{kg}$) on a frictionless floor are joined by a spring. A force of $10\ \text{N}$ along $+x$ acts on A and $4.0\ \text{N}$ along $-x$ on B. Find the acceleration of the centre of mass. Does the spring affect it?
  • 4 A bomb at rest explodes into two pieces whose masses are in the ratio $1:2$. The lighter piece flies off at $30\ \text{m/s}$. Find the velocity of the heavier piece. Where is the centre of mass afterwards?
  • 5 A shell moving horizontally at $20\ \text{m/s}$ at a height of $80\ \text{m}$ explodes into two equal halves. One half falls vertically from rest. Where do the two halves and the centre of mass land, measured from the point directly below the explosion?
  • 6 A $50\ \text{kg}$ girl stands at one end of a $150\ \text{kg}$, $4.0\ \text{m}$ raft at rest on still water and walks to the other end. How far does the raft move? How far does she move relative to the water?
  • 7 Two skaters, $50\ \text{kg}$ and $75\ \text{kg}$, at rest on frictionless ice, hold the ends of a $10\ \text{m}$ rope and pull themselves together. How far does each move before they meet?
  • 8 If only internal forces act on a system, explain why its centre of mass cannot accelerate. Then explain what external force makes a car โ€” or a walking person โ€” speed up.
1 $\mathbf{P}=2.0(3.0)+3.0(-2.0)=0$, so $v_\text{cm}=0$: the centre of mass is at rest.
2 $\mathbf{P}=(2.0-2.0)\hat{\mathbf{i}}+(3.0+2.0)\hat{\mathbf{j}}=5.0\hat{\mathbf{j}}\ \text{kg m/s}$; $\mathbf{v}_\text{cm}=\mathbf{P}/3.0=1.7\hat{\mathbf{j}}\ \text{m/s}$ ($1.67$).
3 $a_\text{cm}=\dfrac{10-4.0}{5.0}=1.2\ \text{m/s}^2$ along $+x$. No โ€” the spring forces are internal; they change how A and B move relative to each other, not $a_\text{cm}$.
4 $0=m(30)+2m\,v$, $v=-15\ \text{m/s}$: $15\ \text{m/s}$ opposite to the lighter piece. The centre of mass stays where the bomb was (until a piece hits something).
5 Fall time $t=\sqrt{\dfrac{2\times80}{9.8}}=4.04\ \text{s}$. Momentum: $2m(20)=m(0)+m v$, $v=40\ \text{m/s}$. Landings: $0$ and $40\times4.04=1.6\times10^2\ \text{m}$; centre of mass at $81\ \text{m}$ $=20\times4.04$, where the unexploded shell would have landed โœ“.
6 $50(D+4.0)+150D=0$: $D=-1.0\ \text{m}$ (raft moves $1.0\ \text{m}$ backwards); the girl moves $3.0\ \text{m}$ relative to the water.
7 The centre of mass stays fixed, $\dfrac{75}{125}\times10=6.0\ \text{m}$ from the $50\ \text{kg}$ skater. They meet there: the $50\ \text{kg}$ skater moves $6.0\ \text{m}$, the $75\ \text{kg}$ skater $4.0\ \text{m}$.
8 Internal forces cancel in pairs (third law), so $\mathbf{F}_\text{ext}=0$ and $\mathbf{a}_\text{cm}=0$. A car's engine turns the wheels; the tyres push back on the road and the road's static friction pushes the car forward โ€” an external force. A walker's foot pushes backward on the floor and floor friction pushes her forward.
IV
Applications and harder problems.
  • 1 The Sun ($1.99\times10^{30}\ \text{kg}$, radius $6.96\times10^8\ \text{m}$) and the Earth ($5.97\times10^{24}\ \text{kg}$) are $1.50\times10^{11}\ \text{m}$ apart. Find their centre of mass. Is it inside the Sun?
  • 2 In an Atwood machine, masses of $3.0\ \text{kg}$ and $2.0\ \text{kg}$ hang from a light string over a light frictionless pulley. Find the acceleration of the centre of mass of the two masses, and check it with $M\mathbf{a}_\text{cm}=\mathbf{F}_\text{ext}$.
  • 3 A $3.0\ \text{kg}$ shell flying horizontally at $60\ \text{m/s}$ explodes into pieces of $1.0\ \text{kg}$ and $2.0\ \text{kg}$. The $1.0\ \text{kg}$ piece moves straight backwards at $30\ \text{m/s}$. Find the velocity of the other piece.
  • 4 (Modelling) A firework rocket's motor burns out $50\ \text{m}$ above the ground while it is moving straight up at $30\ \text{m/s}$. At the top of its climb the shell bursts into glowing fragments flying out in all directions. (a) At what height does it burst? (b) Neglecting air resistance, where is the centre of mass of the fragments $2.0\ \text{s}$ after the burst (none has landed)? (c) Real fragments are slowed strongly by the air. Does this break Result 6.2?
  • 5 Balls of $0.20\ \text{kg}$ and $0.30\ \text{kg}$, joined by a light spring, are thrown through the air. At one instant the $0.20\ \text{kg}$ ball is at $(2.0,\,3.0)\ \text{m}$ and the centre of mass at $(2.6,\,2.4)\ \text{m}$. Where is the other ball?
  • 6 A uniform triangular plate has vertices at $(0,0)$, $(6.0,\,0)$ and $(0,\,3.0)$ (cm). Find its centre of mass, and show that it is the point where the medians meet.
1 From the Sun's centre: $X=\dfrac{5.97\times10^{24}\times1.50\times10^{11}}{1.99\times10^{30}}=4.50\times10^5\ \text{m}$ ($450\ \text{km}$) โ€” deep inside the Sun, only $0.065\%$ of its radius from the centre.
2 $a=\dfrac{(3.0-2.0)g}{5.0}=1.96\ \text{m/s}^2$ (3 kg down, 2 kg up). $a_\text{cm}=\dfrac{3.0(-1.96)+2.0(+1.96)}{5.0}=-0.39\ \text{m/s}^2$: $0.39\ \text{m/s}^2$ downwards. Check: $T=\dfrac{2m_1m_2g}{m_1+m_2}=23.5\ \text{N}$; external forces are the weights ($49\ \text{N}$ down) and the pulley's pull $2T=47.0\ \text{N}$ up; net $1.96\ \text{N}$ down $=5.0\times0.392$ โœ“.
3 $3.0(60)=1.0(-30)+2.0\,v$: $v=105\ \text{m/s}\approx1.1\times10^2\ \text{m/s}$ forwards.
4 (a) Extra rise $\dfrac{30^2}{2\times9.8}=45.9\ \text{m}$; bursts at $96\ \text{m}$. (b) The centre of mass starts from rest at the top and falls freely: $95.9-\tfrac12(9.8)(2.0)^2=76\ \text{m}$ above the launch point. (c) No โ€” air resistance is an external force (from the air), so it rightly changes $\mathbf{a}_\text{cm}$; Result 6.2 still holds with it included.
5 $0.50(2.6)=0.20(2.0)+0.30x$: $x=3.0\ \text{m}$; $0.50(2.4)=0.20(3.0)+0.30y$: $y=2.0\ \text{m}$. The other ball is at $(3.0,\,2.0)\ \text{m}$.
6 Centroid = average of the vertices: $(2.0,\,1.0)\ \text{cm}$. The median from $(0,3.0)$ to the midpoint $(3.0,0)$ of the opposite side is $y=3.0-x$, which passes through $(2.0,\,1.0)$; the median from $(0,0)$ to $(3.0,\,1.5)$ is $y=\tfrac12x$, also through $(2.0,\,1.0)$ โœ“.
๐ŸŽฏ Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board ยท 3 marks Define the centre of mass of a system of two particles. Show that it divides the line joining them in the inverse ratio of their masses, and locate it for particles of $2.0\ \text{kg}$ and $6.0\ \text{kg}$ that are $0.40\ \text{m}$ apart.

Definition The point with position vector $\mathbf{R}=\dfrac{m_1\mathbf{r}_1+m_2\mathbf{r}_2}{m_1+m_2}$. [1]
Ratio With the origin at the centre of mass, $m_1\mathbf{r}_1+m_2\mathbf{r}_2=0$, so $\mathbf{r}_1$ and $\mathbf{r}_2$ are oppositely directed along one line and $\dfrac{r_1}{r_2}=\dfrac{m_2}{m_1}$. [1]
Numbers From the $2.0\ \text{kg}$ particle: $\dfrac{6.0}{8.0}\times0.40=0.30\ \text{m}$ (and $0.10\ \text{m}$ from the $6.0\ \text{kg}$ one). [1]
Exam tip Check: $2.0\times0.30=6.0\times0.10$ โœ“. The centre of mass is always nearer the heavier particle.

E2. NEET Particles of mass $1\ \text{kg}$, $2\ \text{kg}$ and $3\ \text{kg}$ are placed at $(0,0)$, $(1,0)$ and $(0,1)$ (metres). Their centre of mass is at (a) $\left(\tfrac12,\tfrac13\right)$ (b) $\left(\tfrac13,\tfrac12\right)$ (c) $\left(\tfrac13,\tfrac13\right)$ (d) $\left(\tfrac12,\tfrac12\right)$

Idea $X=\dfrac{2(1)}{6}=\dfrac13$, $Y=\dfrac{3(1)}{6}=\dfrac12$.
Answer (b).
Exam tip (a) swaps the coordinates; (c) is the plain average of the positions, ignoring the masses (correct only for equal masses).

E3. JEE Main (Numerical value) A uniform square plate has side $12\ \text{cm}$. A square of side $6\ \text{cm}$ is cut away from one corner. Find the distance (in cm, to two decimal places) of the centre of mass of the remaining plate from the centre of the original square.

Set-up Origin at the centre of the full square. Full plate: area $144$ at $(0,0)$; removed square: area $36$ at $(3,\,3)$, entered as $-36$.
Centre $X=Y=\dfrac{0-36\times3}{144-36}=-1\ \text{cm}$.
Answer Distance $\sqrt{1^2+1^2}=1.41\ \text{cm}$, along the diagonal away from the cut corner.
Exam tip Forgetting to subtract $36$ from the denominator gives $0.75\ \text{cm}$ per axis โ€” a favourite wrong option.

E4. JEE Advanced (One or more correct) A projectile of mass $2m$ is fired from level ground with range $R$. At the highest point it splits into two equal parts; one part retraces its path exactly back to the launch point. Air resistance is negligible. Which statements are correct? (A) The other part lands at a distance $2R$ from the launch point. (B) The centre of mass of the two parts lands at distance $R$. (C) The kinetic energy of the system just after the split is greater than just before. (D) The horizontal momentum of the system changes during the split.

Momentum At the top the velocity is horizontal, $v$. Retracing means the first part has $-v$. $2m\,v=m(-v)+m\,v_2$, so $v_2=3v$.
(A), (B) Both parts fall from the top in the same time $t$, during which the unsplit shell would have moved $vt=\tfrac12R$. The second part moves $3vt=\tfrac32R$ beyond $\tfrac12R$: it lands at $2R$ โœ“. The centre of mass lands at $\tfrac12(0+2R)=R$ โœ“ โ€” as it must, since only gravity acts externally.
(C) Before: $\tfrac12(2m)v^2=mv^2$. After: $\tfrac12mv^2+\tfrac12m(3v)^2=5mv^2$ โœ“ โ€” the extra comes from the chemical energy of the charge.
(D) โœ— The split forces are internal; there is no horizontal external force.
Answer (A), (B), (C).
Exam tip Momentum is conserved in an explosion; kinetic energy is not (it increases). Never use energy conservation across an explosion.

E5. JEE Main NEET A $60\ \text{kg}$ man stands at one end of a $90\ \text{kg}$ boat of length $3.0\ \text{m}$ at rest on still water. He walks to the other end. Neglecting water resistance, the boat moves through (a) $1.2\ \text{m}$ (b) $1.8\ \text{m}$ (c) $2.0\ \text{m}$ (d) $3.0\ \text{m}$

Idea No external horizontal force: the centre of mass stays fixed. $60(D+3.0)+90D=0$, so $|D|=\dfrac{60\times3.0}{150}=1.2\ \text{m}$ (opposite to his walk).
Answer (a).
Exam tip (b) is how far the man moves relative to the water; (c) uses $\dfrac{60\times3.0}{90}$, forgetting that the man's own displacement is not $3.0\ \text{m}$. Shortcut: $D=\dfrac{m\,L}{m+M}$.

E6. Board ยท Case study At a festival a firework shell is fired vertically upward at $35\ \text{m/s}$ from the ground. At the top of its flight it bursts into three fragments of equal mass. Two of them fly off horizontally at $20\ \text{m/s}$, one due east and one due west. (i) Find the height at which the shell bursts. [2] (ii) Find the velocity of the third fragment just after the burst. [1] (iii) Where is the centre of mass of the fragments $2.0\ \text{s}$ after the burst (none has landed)? [1]

(i) $h=\dfrac{u^2}{2g}=\dfrac{35^2}{2\times9.8}=62.5\ \text{m}\approx63\ \text{m}$. [2]
(ii) At the top the shell is at rest, so the total momentum after the burst is zero: taking east as $+x$, $m(+20)+m(-20)+m\mathbf{v}_3=0$, giving $\mathbf{v}_3=0$ โ€” the third fragment starts from rest. [1]
(iii) The centre of mass starts from rest at $62.5\ \text{m}$ and falls freely: $62.5-\tfrac12(9.8)(2.0)^2=42.9\ \text{m}\approx43\ \text{m}$, directly above the launch point. [1]
Exam tip Check (iii): all three fragments start with zero vertical velocity, so all are at $42.9\ \text{m}$ after $2.0\ \text{s}$; the east and west ones are $40\ \text{m}$ either side, and their average position is above the launch point โœ“.
๐Ÿ”ฌ Researchโ€“Teaching NexusResearch-tutored ยท optional, about 45 minutes

Research question. Where is a person's centre of mass, and how can a high jumper clear a bar while her centre of mass passes close to โ€” or even below โ€” it?

Why it matters. Sports scientists, physiotherapists, designers of prosthetic limbs and engineers who build walking robots all need to know where the body's centre of mass is, because balance and jumping are controlled by it. Since the "flop" technique swept the world's high-jump competitions in the late 1960s, athletes arch their backs over the bar so that their body wraps around it. Result 6.2 says the take-off fixes the parabola of the centre of mass; posture decides where the body is relative to that point.

What to do. Model the body as six segments, each with its mass at its own midpoint: head and neck $8\%$ of body mass, trunk $50\%$, each arm $5\%$, each leg $16\%$ (rounded values; published biomechanics tables split the body more finely).

  1. Take a side-on photograph of a classmate standing upright against a grid (or draw a stick figure to scale). Record the coordinates of the midpoint of each segment.
  2. Compute $Y=\dfrac{\sum m_iy_i}{M}$ using the percentages as masses. Express the height of the centre of mass as a percentage of the person's height.
  3. Repeat for an arched "flop" posture over a bar (photograph a careful pose on a mat, or draw one). Is the centre of mass above or below the line of the bar?
  4. Ask the subject to touch their toes with heels against a wall. Why is it impossible? Use your model to explain where the centre of mass goes.
  5. Change each percentage by $\pm10\%$ (renormalising to $100\%$). How sensitive is your answer to the segment data?

Tools: Phone camera, squared paper or any image viewer that shows pixel coordinates, a spreadsheet.

Hand in: Your segment table (fraction, $x_i$, $y_i$), the centre of mass for both postures marked on the images, and a paragraph answering the research question.

Going further. A "reaction board" measures the centre of mass directly: a plank of length $L$ rests on two bathroom scales, one at each end. With a person lying on it, the extra reading $W_2$ at the head end gives the distance of the centre of mass from the feet end as $x=\dfrac{W_2L}{W}$, where $W$ is the person's weight. Derive this (it is Definition 6.2 with forces in place of masses โ€” see Section 6.2) and try it.

Instructor note: for an upright adult the model gives a centre of mass at roughly $55$โ€“$57\%$ of standing height, in line with published values. In a well-arched flop the centre of mass can pass a few centimetres below the bar, so for the same take-off speed a higher bar can be cleared than with a straddle or scissors technique. The toe-touch fails because the hips must move backward to keep the centre of mass above the feet. Rounded segment fractions shift the answer by only about $1\%$ of height.

๐Ÿงช Self-check quiz

0/5 correct

Q1. The centre of mass of a uniform circular ring is

Q2. Particles of mass $1.0\ \text{kg}$ and $3.0\ \text{kg}$ are $4.0\ \text{m}$ apart. Their centre of mass is at a distance from the $1.0\ \text{kg}$ particle of

Q3. A shell explodes in mid-air into many fragments. Neglecting air resistance, until the first fragment lands, the centre of mass of the fragments

Q4. The net external force on a system of particles is zero. Which statement must be true?

Q5. A boy walks from one end of a boat to the other on still, frictionless water. During the walk, the centre of mass of the boy + boat

โ† Chapter 6 โ€” System of Particles and Rotational Motion6.2 Torque, Angular Momentum and Equilibrium of a Rigid Body โ†’