System of Particles & Rotational Motion: Full Notes

6.1 Centre of Mass

A rigid body or a cloud of particles is, mechanically, an enormous headache: every particle in it can push and pull on every other, and tracking all of it directly is hopeless. The centre of mass is the way out. For a system of $n$ particles of masses $m_1,m_2,\ldots,m_n$ at position vectors $\mathbf{r}_1,\mathbf{r}_2,\ldots$ , it is the mass-weighted average position:

\mathbf{R} = \dfrac{\sum_i m_i\mathbf{r}_i}{\sum_i m_i} = \dfrac{1}{M}\sum_i m_i\mathbf{r}_i

For a continuous body, the sum becomes an integral over infinitesimal mass elements $dm$ , but the idea is identical: it is a single point that summarises where the body’s mass “is”, on average. For a uniform (symmetric) body it sits at the geometric centre, a ring, a square plate, a solid sphere all have their centre of mass at the obvious middle point, and notice this point need not be inside the body’s material at all, for a ring or a hollow spherical shell, the centre of mass lies in the empty space at the centre.

6.1.1 Why the centre of mass matters

The whole point of defining it is this: however complicated the internal forces inside a system are, the centre of mass moves exactly as if the entire mass of the system were concentrated there and the entire external force acted there. Differentiating $\mathbf{R}$ twice and using Newton’s second law on every particle, the internal forces cancel in pairs by the third law (every internal push on one particle is matched by an equal, opposite push on another), leaving only the external forces:

\mathbf{F}_{ext} = M\mathbf{a}_{cm}

This is why a spanner tossed spinning through the air has such a messy individual-point trajectory but its centre of mass traces out a clean parabola, exactly like a single projectile, under gravity alone. Every internal detail of the spin and the tumbling is invisible to this one equation.

Worked example

Centre of mass of a three-particle system

Point masses of 1 kg, 2 kg and 3 kg sit at $(0,0)$ , $(2,0)$ and $(1,3)$ metres. Locate the centre of mass.

x_{cm} = \dfrac{1(0)+2(2)+3(1)}{1+2+3} = \dfrac{7}{6} \approx \textcolor{#e08a1e}{1.17\text{ m}}

y_{cm} = \dfrac{1(0)+2(0)+3(3)}{6} = \dfrac{9}{6} = \textcolor{#e08a1e}{1.5\text{ m}}

Notice the centre of mass, at roughly $(1.17,\,1.5)$ , sits inside the triangle formed by the three masses but is pulled toward the 3 kg corner, exactly as the mass weighting demands, it is nowhere near the simple geometric centroid $(1,1)$ of the triangle.

6.2 Torque and Angular Momentum

Force tells a particle how to accelerate linearly; it says nothing about spin. For that, mechanics needs two new quantities, each built the same way, by crossing a position vector into a linear one. Torque (or moment of force) about a chosen origin is:

\boldsymbol{\tau} = \mathbf{r}\times\mathbf{F}

with magnitude $\tau = rF\sin\theta$ , where $\theta$ is the angle between $\mathbf{r}$ and $\mathbf{F}$ , and direction given by the right-hand rule, curl the fingers from $\mathbf{r}$ toward $\mathbf{F}$ , the thumb points along $\boldsymbol{\tau}$ . The farther from the pivot a force is applied, and the closer it is to perpendicular to $\mathbf{r}$ , the greater its turning effect, which is exactly why doorknobs sit at the edge of the door and not next to the hinge.

Angular momentum about the same origin is built from linear momentum the same way:

\mathbf{L} = \mathbf{r}\times\mathbf{p}

Both $\boldsymbol{\tau}$ and $\mathbf{L}$ are only ever defined relative to a chosen point, unlike force and momentum, change the origin and both quantities generally change too (for a straight-line-moving free particle, $\mathbf{L}$ about a point off its path is nonzero and constant, which already hints that angular momentum is conserved whenever no torque acts).

6.2.1 τ = dL/dt: the rotational twin of F = dp/dt

This parallel is not a coincidence, it is a theorem. Differentiate $\mathbf{L}=\mathbf{r}\times\mathbf{p}$ with respect to time using the product rule:

\dfrac{d\mathbf{L}}{dt} = \dfrac{d\mathbf{r}}{dt}\times\mathbf{p} + \mathbf{r}\times\dfrac{d\mathbf{p}}{dt}

The first term is $\mathbf{v}\times m\mathbf{v}$ , the cross product of a vector with itself (scaled), which is always zero. The second term is just $\mathbf{r}\times\mathbf{F}$ , the torque. So:

Rotational analogue of Newton's second law

\textcolor{#e08a1e}{\boldsymbol{\tau} = \dfrac{d\mathbf{L}}{dt}}

the net external torque equals the rate of change of angular momentum, line for line the rotational mirror of $\mathbf{F}=d\mathbf{p}/dt$ . For a rigid body spinning about a fixed axis, this reduces to the more familiar $\tau = I\alpha$ , once the next section pins down what $I$ (the moment of inertia, standing in for mass) is, and $L = I\omega$ for that same rotation.

Worked example

Torque on a wrench bolt

A mechanic applies 150 N at the end of a 0.30 m wrench handle, at 60° to the handle. Find the torque about the bolt, and state which way it turns the bolt if the force is applied as shown turning it anticlockwise.

\tau = rF\sin\theta = (0.30)(150)\sin60^\circ = 45\times0.866 \approx \textcolor{#e08a1e}{39\text{ N m, anticlockwise}}

Pushing at 90° instead (perpendicular to the handle) would give the maximum possible torque for that same force and reach, $45\text{ N m}$ , which is why the most effective way to use a wrench is to push square across it, not along its length.

6.3 Moment of Inertia

In $\mathbf{F}=m\mathbf{a}$ , mass is the body’s resistance to a change in velocity. The rotational counterpart is the moment of inertia, the body’s resistance to a change in angular velocity, defined for a system of particles about a chosen axis as:

I = \sum_i m_ir_i^2

where $r_i$ is each particle’s perpendicular distance from the axis, not from any point. Two things make $I$ trickier than mass: it depends on how the mass is distributed (mass far from the axis counts far more, by $r^2$ , than the same mass close in), and it depends on which axis is chosen, the very same rod has a small $I$ about an axis through its centre and a much larger one about an axis through its end. It is often written as $I = MK^2$ , where $K$ is the radius of gyration, the distance from the axis at which the entire mass, if concentrated there as a point, would give the same $I$ .

Some standard results, all derivable by direct integration, worth keeping on hand:

Ring about its central (symmetry) axis: $I = MR^2$
Disc about its central axis: $I = \tfrac{1}{2}MR^2$
Solid sphere about a diameter: $I = \tfrac{2}{5}MR^2$
Thin rod about an axis through its centre, perpendicular to its length: $I = \tfrac{1}{12}ML^2$

6.3.1 The perpendicular axes theorem

For any flat, planar lamina, with two perpendicular axes $x$ and $y$ lying in its plane and crossing at a point $O$ , and a third axis $z$ through $O$ perpendicular to the plane:

I_z = I_x + I_y

This follows immediately from $r^2 = x^2+y^2$ for every particle in the plane (its distance from the out-of-plane axis $z$ is the hypotenuse of its distances from $x$ and $y$ ), summed over the whole lamina. It only holds for genuinely flat bodies.

6.3.2 The parallel axes theorem

For any rigid body, relating the moment of inertia about an axis through the centre of mass to a second axis parallel to it, a distance $d$ away:

I = I_{cm} + Md^2

The centre-of-mass axis always gives the smallest possible $I$ among all parallel axes, moving the axis any distance away can only add to it, never subtract, since $Md^2$ is never negative.

Worked example

Moment of inertia of a rod about its end

A uniform rod has mass 1.2 kg and length 1.6 m. Its moment of inertia about a perpendicular axis through its centre is $\tfrac{1}{12}ML^2$ . Find its moment of inertia about a parallel axis through one end.

I_{cm} = \dfrac{ML^2}{12} = \dfrac{1.2\times(1.6)^2}{12} = \dfrac{3.072}{12} = 0.256\text{ kg m}^2

The end axis is $d=L/2=0.8$ m from the centre:

I_{end} = I_{cm}+Md^2 = 0.256 + 1.2(0.8)^2 = 0.256+0.768 = \textcolor{#e08a1e}{1.024\text{ kg m}^2}

exactly matching the direct formula $I_{end}=ML^2/3$ , which gives $1.2\times2.56/3=1.024$ kg m² too, four times the centre value, all from moving the axis half a rod-length.

6.4 Conservation of Angular Momentum

Set $\boldsymbol{\tau}_{ext}=0$ in §6.2’s $\boldsymbol{\tau}=d\mathbf{L}/dt$ and the conclusion is immediate and powerful:

Conservation of angular momentum

\textcolor{#e08a1e}{\mathbf{L} = \text{constant}}\quad\text{whenever}\quad \boldsymbol{\tau}_{ext}=0

Exactly as with linear momentum, internal torques (forces a body exerts on its own parts) can shuffle angular momentum around inside the system, between a spinning skater’s torso and her arms, say, but can never change the total, because by the third law every internal torque is matched by an equal and opposite partner. For a rigid body rotating about a fixed axis, $L=I\omega$ , so if no external torque acts:

I_1\omega_1 = I_2\omega_2

This is precisely the two-mass system in the interactive simulation for this chapter: two equal point masses, connected through a central axis, are dragged inward or outward while nothing external torques the system. Dragging them inward shrinks $I=2mr^2$ , and since $L=I\omega$ is pinned at a fixed value, $\omega$ must shoot up to compensate, a skater or a diver pulling in their arms and legs is doing exactly this with their own body. It is tempting to think this spin-up is “free” energy, it is not: kinetic energy $KE=\tfrac{1}{2}I\omega^2=L^2/2I$ actually increases as $I$ shrinks (since $L$ is fixed), and that extra energy is supplied by the real muscular work done pulling the arms in against the outward pull they feel in the spinning frame, there is no violation of energy conservation anywhere in this.

Worked example

A figure skater pulling her arms in

A skater spinning with arms outstretched has moment of inertia 4.8 kg·m² at 2.0 rad/s. She pulls her arms in, reducing her moment of inertia to 1.2 kg·m². Find her new angular speed and the ratio of her final to initial kinetic energy.

\omega_2 = \dfrac{I_1\omega_1}{I_2} = \dfrac{4.8\times2.0}{1.2} = \textcolor{#e08a1e}{8.0\text{ rad/s}}

KE_1 = \tfrac{1}{2}(4.8)(2.0)^2 = 9.6\text{ J},\qquad KE_2 = \tfrac{1}{2}(1.2)(8.0)^2 = 38.4\text{ J}

\dfrac{KE_2}{KE_1} = \dfrac{38.4}{9.6} = \textcolor{#e08a1e}{4.0}

Her kinetic energy quadruples, exactly matching $I_1/I_2=4.8/1.2=4.0$ , as it must since $KE=L^2/2I$ with $L$ unchanged. That extra energy came from her own muscles, not from nowhere.

6.5 Rolling Motion

A wheel rolling without slipping is doing two things at once, translating (its centre moves at $v_{cm}$ ) and rotating (about that same centre, at $\omega$ ), and the “without slipping” condition locks the two together: the point of the wheel touching the ground must be instantaneously at rest, otherwise it would be scraping or skidding. That gives:

v_{cm} = \omega R

Its total kinetic energy is the straightforward sum of a translational piece and a rotational piece, with no cross term between them:

KE = \tfrac{1}{2}Mv_{cm}^2 + \tfrac{1}{2}I\omega^2

Writing $I=MK^2$ and substituting $\omega=v_{cm}/R$ turns this into a single compact form:

KE = \tfrac{1}{2}Mv_{cm}^2\left(1+\dfrac{K^2}{R^2}\right)

which is the key to every “race down an incline” question: released from the same height $h$ on the same slope, energy conservation ( $Mgh=KE$ ) gives:

Speed at the bottom of a rolling incline

\textcolor{#e08a1e}{v_{cm}} = \sqrt{\dfrac{2gh}{1+K^2/R^2}}

A body with a smaller $K^2/R^2$ , meaning its mass sits closer to the rotation axis, needs less of its total energy budget to get itself spinning, so more is left over for translation, and it arrives faster and sooner. This is exactly why a solid sphere ( $K^2/R^2=2/5$ ) reliably beats a hollow cylinder ( $K^2/R^2=1$ ) down an identical incline: the cylinder’s mass sits out at its rim, demanding a much larger share of the available energy just to spin up.

Worked example

A rolling race: solid sphere vs. hollow cylinder

A solid sphere and a hollow cylinder, each released from rest at the top of the same 1.0 m high incline, roll down without slipping. Compare their speeds at the bottom. ( $g=9.8\text{ m/s}^2$ )

v_{sphere} = \sqrt{\dfrac{2(9.8)(1.0)}{1+2/5}} = \sqrt{\dfrac{19.6}{1.4}} = \sqrt{14.0} \approx \textcolor{#e08a1e}{3.74\text{ m/s}}

v_{cylinder} = \sqrt{\dfrac{2(9.8)(1.0)}{1+1}} = \sqrt{\dfrac{19.6}{2}} = \sqrt{9.8} \approx \textcolor{#e08a1e}{3.13\text{ m/s}}

Neither speed depends on the mass or the radius, only on the shape (through $K^2/R^2$ ), which is why this race has the same winner regardless of the exact sphere or cylinder used, the sphere always wins.

— From the NCERT exercises

A few of the chapter’s own exercise questions, with original worked solutions.

NCERT Exercise 6.9

A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g, are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?

Solution

A uniform metre stick balances alone at its own centre of mass, the 50.0 cm mark. Once the coins are added, the new balance point at 45.0 cm is the pivot, so take torques about it. The stick’s own weight still acts at the 50.0 cm mark, a distance $50.0-45.0=5.0$ cm from the new pivot, while the coins act at 12.0 cm, a distance $45.0-12.0=33.0$ cm away, on the opposite side. Balancing these two torques ( $g$ cancels, and the two 5 g coins together are 10 g):

M(5.0) = (10)(33.0)

M = \dfrac{330}{5.0} = \textcolor{#e08a1e}{66\text{ g}}

A nice check on the method: the far smaller coin mass (10 g) balances the much larger stick mass (66 g) only because it sits so much farther from the new pivot, torque cares about the product of force and distance, not either alone.

NCERT Exercise 6.26

Two discs of moments of inertia I1 and I2 about their respective axes (normal to the disc and passing through the centre), rotating with angular speeds ω1 and ω2, are brought into contact face to face with their axes of rotation coincident. (a) What is the angular speed of the two-disc system? (b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy?

Solution

Once the discs touch, the only torques exchanged are between the discs themselves (friction at the contact face), so there is no external torque on the two-disc system and its total angular momentum about the shared axis is conserved:

I_1\omega_1 + I_2\omega_2 = (I_1+I_2)\,\omega

\omega = \textcolor{#e08a1e}{\dfrac{I_1\omega_1+I_2\omega_2}{I_1+I_2}}

For part (b), compare the kinetic energies directly:

KE_i = \tfrac{1}{2}I_1\omega_1^2+\tfrac{1}{2}I_2\omega_2^2, \qquad KE_f = \tfrac{1}{2}(I_1+I_2)\omega^2

Substituting $\omega$ from above and simplifying $KE_i-KE_f$ algebraically collapses to a single perfect square:

\textcolor{#e08a1e}{KE_i-KE_f} = \dfrac{I_1I_2(\omega_1-\omega_2)^2}{2(I_1+I_2)} \;\ge\; 0

which is never negative, so $KE_f\le KE_i$ , with equality only if $\omega_1=\omega_2$ (the discs were already spinning together, so nothing changes on contact). The lost energy is exactly the energy angular momentum conservation allows to vanish into heat and sound at the rubbing contact face, the same role ordinary friction plays in an inelastic collision between two blocks.