Vardaan Learning Institute
System of Particles & Rotational Motion
References
Compiled from: NCERT Physics Part-1 (System of Particles and
Rotational Motion) • H.C. Verma — Concepts of Physics Vol-1 (Ch. 9 & 10: Centre of
Mass, Rotational Mechanics) • D.C. Pandey — Understanding Physics: Mechanics Part-2
• Pradeep's Fundamental Physics XI • S.L. Arora • Cengage (B.M. Sharma) Mechanics-II
• Errorless Physics. Target exams: Boards • JEE Main • JEE Advanced • NEET
• NDA.
1. Centre of Mass (COM)
A real body is a system of particles. The centre of mass is the single
point that moves as if the entire mass were concentrated there and all external forces acted there. It lets
us split any complicated motion into (i) translation of the COM + (ii) rotation about the COM.
COM — Discrete & Continuous
$$x_{cm} = \frac{m_1x_1 + m_2x_2 + \dots}{m_1 + m_2 + \dots} = \frac{\sum m_ix_i}{M}
\qquad \vec{R}_{cm} = \frac{\sum m_i\vec{r}_i}{M}$$
$$\text{Continuous body: } x_{cm} = \frac{1}{M}\int x\,dm$$
1.1 Two-Particle System (Master Result)
Key Results
- COM lies on the line joining the particles, closer to the heavier mass, dividing
the separation d in the inverse ratio of masses: $r_1 = \dfrac{m_2d}{m_1+m_2}$,
$r_2 = \dfrac{m_1d}{m_1+m_2}$, so $m_1r_1 = m_2r_2$.
- Equal masses → COM at the midpoint.
- COM need not lie inside the material of the body: ring, L-shaped lamina, horseshoe — COM lies in
empty space (a favourite 1-marker).
- For symmetric uniform bodies, COM is at the geometric centre.
| Body (uniform) |
Position of COM |
| Rod, ring, disc, sphere, cylinder, cube |
Geometric centre |
| Semicircular ring (radius R) |
$\dfrac{2R}{\pi}$ from the centre, on the axis of symmetry |
| Semicircular disc |
$\dfrac{4R}{3\pi}$ from the centre |
| Solid hemisphere |
$\dfrac{3R}{8}$ from the flat face |
| Hollow hemisphere |
$\dfrac{R}{2}$ from the flat face |
| Solid cone (height h) |
$\dfrac{h}{4}$ from the base |
| Triangular lamina |
Centroid (intersection of medians), h/3 from any base |
If a piece is cut out of a body, treat the hole as
negative mass: $x_{cm} = \dfrac{M x_M - m x_m}{M - m}$. Example: disc of radius R with
a hole of radius R/2 touching the edge → COM shifts by R/6 away from the hole.
Image Placeholder — Fig. 1
COM of a two-particle system and the negative-mass (hole) trick (Landscape •
Background #ffffff)
AI IMAGE PROMPT: Create a two-panel physics textbook diagram side by side, LANDSCAPE
orientation (16:9), pure white background (#ffffff). LEFT PANEL titled "Two-particle system": draw a
horizontal thin black line; on it a LARGE blue filled circle on the left labelled "m₁ = 4 kg" and a
SMALL red filled circle on the right labelled "m₂ = 2 kg", separated by a dimension arrow below
labelled "d = 6 m". Mark the COM with a green (#43a047) crosshair dot on the line CLOSER to the big
mass, with dimension arrows: "r₁ = m₂d/(m₁+m₂) = 2 m" from m₁ to COM, and "r₂ = m₁d/(m₁+m₂) = 4 m"
from COM to m₂; caption "m₁r₁ = m₂r₂ — COM divides d in inverse ratio of masses". RIGHT PANEL titled
"Hole = negative mass": draw a large blue circle (disc) of radius R centred at O; inside it, touching
the right edge, a smaller WHITE circle of radius R/2 with a dashed outline centred at O′ (at distance
R/2 to the right of O), labelled "removed disc, treat as mass −m at O′"; mark O with a black dot
labelled "O (centre of full disc)" and mark the new COM with a green crosshair dot a little to the LEFT
of O labelled "shifted COM, x = R/6 toward the heavy side"; boxed formula below: "x_cm = (M·0 −
m·R/2)/(M − m) = −R/6". Style: flat minimalist vector textbook figure, thin black lines, blue bodies,
white cut-out with dashed edge, green COM crosshairs, dimension arrows with clean sans-serif labels, no
gradients, no shadows, pure white background.
Both marked distances are computed from the formulas in the
panel — reuse them as a worked example.
Q. Particles of 1 kg, 2 kg and 3 kg are placed at the corners (0,0), (4,0) and (0,4) m of
a right triangle. Find the COM.
x: $x_{cm} = \dfrac{1(0)+2(4)+3(0)}{6} = \dfrac{8}{6} =
\mathbf{\tfrac{4}{3} \text{ m}}$
y: $y_{cm} = \dfrac{1(0)+2(0)+3(4)}{6} = \mathbf{2 \text{
m}}$ → COM at (4/3, 2) m.
2. Motion of the Centre of Mass
COM Dynamics
$$\vec{v}_{cm} = \frac{\sum m_i\vec{v}_i}{M} \qquad M\vec{a}_{cm} = \vec{F}_{ext} \qquad \vec{p}_{total} =
M\vec{v}_{cm}$$
The Big Idea
- Internal forces cannot move the COM. If F_ext = 0, the COM moves with constant
velocity (or stays at rest) no matter how the parts fly about.
- Exploding shell: the COM of the fragments continues along the original parabola.
- Man walking on a boat (frictionless water): COM of (man + boat) stays fixed; the
boat shifts backward by $\dfrac{m_{man}\,d}{m_{man}+m_{boat}}$ when the man walks distance d
relative to the boat.
- Both fragments of a bursting projectile land such that $m_1x_1 + m_2x_2 = M x_{cm(unexploded)}$ —
solve "where does the second piece land" questions this way.
Q. A shell fired for a range of 120 m explodes at the top into two equal pieces. One piece
falls straight down and lands at the midpoint of the range. Where does the other land?
COM lands at 120 m regardless: $\dfrac{m(60) + m(x)}{2m} = 120$
$x = \mathbf{180 \text{ m}}$ from the launch point.
3. Torque & Angular Momentum
3.1 Torque (Moment of Force)
3.2 Angular Momentum and τ = dL/dt
Definition (particle): $\vec{L} = \vec{r}\times\vec{p}$, $|L| = mvr\sin\theta = mv
\times$ (perpendicular distance). Unit kg·m²/s; dimension $[ML^2T^{-1}]$.
Derivation of the rotational second law:
$$\frac{d\vec{L}}{dt} = \frac{d}{dt}(\vec{r}\times\vec{p}) = \underbrace{\vec{v}\times m\vec{v}}_{=0} +
\vec{r}\times\frac{d\vec{p}}{dt} = \vec{r}\times\vec{F} = \vec{\tau}$$
$\boxed{\vec{\tau}_{ext} = \dfrac{d\vec{L}}{dt}}$ — the exact rotational
analogue of F = dp/dt. For a rigid body about a fixed axis: $L = I\omega$ and $\tau = I\alpha$.
3.3 Conservation of Angular Momentum
Law + Applications
If $\tau_{ext} = 0$, then L = Iω = constant ⟹ $I_1\omega_1 = I_2\omega_2$.
- Ice skater / ballet dancer: pulling arms in → I decreases → ω increases (spins
faster). KE = L²/2I actually increases (muscular work!).
- Diver / gymnast: tucks to somersault faster, stretches to enter water slowly.
- Planet around the Sun: gravitational torque about the Sun is zero → L constant →
Kepler's equal-areas law; planets speed up at perihelion.
- Person on a rotating stool with dumbbells — NCERT's demonstration of I₁ω₁ =
I₂ω₂.
Image Placeholder — Fig. 2
Torque as r × F with lever arm, and the spinning skater conserving L (Landscape
• Background #ffffff)
AI IMAGE PROMPT: Create a two-panel physics textbook diagram side by side, LANDSCAPE
orientation (16:9), pure white background (#ffffff). LEFT PANEL titled "Torque τ = r F sin θ": draw a
door viewed from above as a long thin black rectangle hinged at the left end (mark the hinge with a
small circle labelled "axis O"); a blue (#1e88e5) position vector arrow r from O along the door to a
point near the free end labelled "r"; at that point a red (#d32f2f) force arrow F at angle θ ≈ 60° to
the door with the angle arc labelled "θ"; a dashed grey perpendicular from O to the LINE of F labelled
"lever arm = r sin θ"; a curved black arrow around O showing the rotation sense; boxed formula "τ = r F
sin θ = F × (lever arm)"; small caption "F through the hinge (θ = 0) ⟹ τ = 0 — push at the handle!".
RIGHT PANEL titled "Conservation of angular momentum": two simple front-view figure sketches of an ice
skater side by side — first with ARMS STRETCHED wide holding small weights, labelled "I large, ω small"
with a short curved rotation arrow above the head; second with ARMS PULLED IN, labelled "I small, ω
large" with a much longer/faster curved rotation arrow; between them a large grey arrow "⟶" and the
boxed law "I₁ω₁ = I₂ω₂ (τ_ext = 0)". Style: flat minimalist vector textbook figure, thin black
outlines, blue r and red F arrows, dashed grey lever arm, curved rotation arrows, clean sans-serif
labels, no gradients, no shadows, pure white background.
Lever arm = perpendicular distance to the LINE of action —
the geometric heart of every torque problem.
Q. A dancer spinning at 30 rpm folds her arms, reducing her moment of inertia to 2/5 of
its value. Find her new spin rate and the ratio of new to old kinetic energy.
L conserved: $I\omega = \tfrac{2I}{5}\omega' \Rightarrow \omega'
= \tfrac{5}{2}\times30 = \mathbf{75 \text{ rpm}}$
KE ratio: $K = \dfrac{L^2}{2I} \propto \dfrac1I \Rightarrow
\dfrac{K'}{K} = \mathbf{\dfrac{5}{2}}$ — energy rises (work done by her muscles).
4. Equilibrium of a Rigid Body, Couple & Principle of Moments
Conditions
A rigid body is in mechanical equilibrium when BOTH: (i) ΣF = 0
(translational equilibrium) and (ii) Στ = 0 about any axis (rotational equilibrium).
Either can hold without the other.
- Couple: two equal, opposite, non-collinear forces. Net force = 0 but net torque = F ×
(perpendicular separation) — produces pure rotation (opening a tap, turning a steering wheel, winding a
clock key). The torque of a couple is the same about every point.
- Principle of moments (lever): at equilibrium, sum of clockwise moments = sum of
anticlockwise moments; for an ideal lever $F_1d_1 = F_2d_2$; Mechanical Advantage $= \dfrac{load}{effort}
= \dfrac{effort\ arm}{load\ arm}$.
- Centre of gravity = point where total gravitational torque is zero; coincides with COM
in a uniform field g. A body balanced on a knife edge at its CG stays horizontal — the standard method
to locate CG of an irregular lamina.
Q. A uniform metre scale balances at the 50 cm mark. When a 20 g mass hangs at the 20 cm
mark, the balance point shifts to the 40 cm mark. Find the mass of the scale.
Moments about the 40 cm pivot: scale's weight (M g) acts at 50
cm → arm 10 cm; the 20 g at 20 cm → arm 20 cm.
$20\times20 = M\times10 \Rightarrow M = \mathbf{40 \text{ g}}$
5. Moment of Inertia (Rotational Mass)
5.1 The Two Theorems (State + Use)
Parallel axis theorem (any body): $I = I_{cm} + Md^{2}$ — axis parallel to one through
the COM at distance d. I is minimum about the COM axis.
Perpendicular axis theorem (planar bodies ONLY): $I_z = I_x +
I_y$ — z ⊥ to the lamina, x and y in its plane through the same point. Example: disc: $I_z =
\tfrac12MR^2$ ⟹ about a diameter $I_x = I_y = \tfrac14MR^2$.
5.2 Standard MOI Table (Must Memorise — asked directly)
| Body |
Axis |
I |
K²/R² (for rolling) |
| Thin rod (length L) |
⊥ through centre |
$\frac{ML^2}{12}$ |
— |
| Thin rod |
⊥ through one end |
$\frac{ML^2}{3}$ |
— |
| Ring / hollow cylinder |
Central symmetry axis |
$MR^2$ |
1 |
| Ring |
Diameter |
$\frac{MR^2}{2}$ |
— |
| Disc / solid cylinder |
Central symmetry axis |
$\frac{MR^2}{2}$ |
1/2 |
| Disc |
Diameter |
$\frac{MR^2}{4}$ |
— |
| Solid sphere |
Diameter |
$\frac{2}{5}MR^2$ |
2/5 |
| Hollow sphere (shell) |
Diameter |
$\frac{2}{3}MR^2$ |
2/3 |
| Ring, tangent in plane |
Tangential (in plane) |
$\frac{3}{2}MR^2$ |
— |
| Disc, tangent ⊥ to plane |
Tangential (⊥ plane) |
$\frac{3}{2}MR^2$ |
— |
| Solid sphere, tangent |
Tangential |
$\frac{7}{5}MR^2$ |
— |
Image Placeholder — Fig. 3
The two theorems + MOI chart of the five standard bodies (Landscape •
Background #ffffff)
AI IMAGE PROMPT: Create a two-row physics textbook diagram, LANDSCAPE orientation
(16:9), pure white background (#ffffff), thin grey divider between rows. TOP ROW, two panels: PANEL A
titled "Parallel axis theorem": a blue disc with TWO parallel vertical dashed axis lines — one through
its centre labelled "I_cm" and one at the edge at distance labelled "d", the second labelled "I = I_cm
+ Md²"; a dimension arrow between the axes labelled "d". PANEL B titled "Perpendicular axis theorem
(planar bodies only)": a flat blue disc drawn in slight perspective lying in a horizontal plane with
three mutually perpendicular dashed axes through its centre labelled "x (in plane)", "y (in plane)",
"z (⊥ to plane)", and the boxed relation "I_z = I_x + I_y". BOTTOM ROW titled "Standard bodies —
memorise": five simple blue shapes in a row, each with a vertical dashed axis through its centre and
its formula below in black text: a RING labelled "Ring: I = MR²"; a DISC labelled "Disc: I = ½MR²"; a
SOLID SPHERE labelled "Solid sphere: I = 2/5 MR²"; a HOLLOW SPHERE (drawn with a dashed inner circle)
labelled "Shell: I = 2/3 MR²"; a ROD (horizontal bar with axis through centre) labelled "Rod (centre):
I = ML²/12; (end): ML²/3". Style: flat minimalist vector textbook figure, thin black outlines, blue
shapes, dashed axis lines, clean sans-serif labels, no gradients, no shadows, pure white background.
One image = the whole MOI toolkit; sketch it from memory
before every test.
Q. Using theorems, find the MOI of (a) a disc about a tangent perpendicular to its plane,
(b) a disc about a tangent in its plane.
(a) Parallel axis on $I_z = \tfrac12MR^2$: $I = \tfrac12MR^2 +
MR^2 = \mathbf{\tfrac{3}{2}MR^2}$
(b) Diameter first (perpendicular axis): $I_{dia} =
\tfrac14MR^2$; then parallel axis: $I = \tfrac14MR^2 + MR^2 = \mathbf{\tfrac{5}{4}MR^2}$
6. Rotational Kinematics & Dynamics (τ = Iα)
6.1 Rotational Kinematic Equations (constant α)
6.2 τ = Iα, Rotational KE, Work & Power
Rotational Dynamics
$$\tau = I\alpha \qquad K_{rot} = \tfrac12I\omega^2 = \frac{L^2}{2I} \qquad W = \tau\theta \qquad P =
\tau\omega$$
Classic Setup — Falling Mass Unwinding a Pulley/Cylinder
Mass m hangs from a string wound on a pulley of moment of inertia I and radius R:
$$mg - T = ma \qquad TR = I\alpha = I\frac{a}{R}$$
$$\boxed{a = \frac{g}{1 + \dfrac{I}{mR^2}}} \qquad T = \frac{mg}{1 + \dfrac{mR^2}{I}}$$
For a disc-pulley (I = ½MR²): $a = \dfrac{2mg}{M+2m}$. Note a < g always —
part of gravity's "effort" goes into spinning the pulley.
Q. A wheel accelerates uniformly from rest to 300 rpm in 10 s. Find α and the number of
revolutions made.
Convert: $\omega = 300\times\dfrac{2\pi}{60} = 10\pi$ rad/s →
$\alpha = \dfrac{10\pi}{10} = \mathbf{\pi \text{ rad/s}^2}$
Angle: $\theta = \tfrac12\alpha t^2 = \tfrac12\pi(100) = 50\pi$
rad → revolutions $= \dfrac{50\pi}{2\pi} = \mathbf{25}$
Q. A 2 kg block hangs from a light string wound around a solid cylinder (M = 4 kg, R = 0.2
m) free to rotate about its axis (g = 10 m/s²). Find the block's acceleration and the string tension.
Disc formula: $a = \dfrac{2mg}{M+2m} = \dfrac{2\times2\times10}{4+4}
= \mathbf{5 \text{ m/s}^2}$
Tension: $T = m(g-a) = 2\times5 = \mathbf{10 \text{ N}}$
7. Rolling Motion — The Chapter's Crown
7.1 Rolling Without Slipping
The Rolling Condition
Rolling = translation of the COM + rotation about the COM, locked together:
$$\boxed{v_{cm} = \omega R} \qquad (a_{cm} = \alpha R)$$
- Contact point: instantaneously at REST (v − ωR = 0) — that's why static (not
kinetic) friction acts, and why an ideal rolling wheel doesn't dissipate energy.
- Top point: speed = v + ωR = 2v (fastest point).
- Any point's velocity = vector sum of v (translation) and ωr (rotation about COM); the contact point
acts as the instantaneous axis of rotation.
7.2 Kinetic Energy of Rolling & the Energy Split
7.3 Rolling Down an Incline (Full Derivation + The Great
Race)
Energy method: from height h, $mgh = \tfrac12mv^2(1 + K^2/R^2)$:
$$v = \sqrt{\frac{2gh}{1 + K^2/R^2}} \qquad
a = \frac{g\sin\theta}{1 + K^2/R^2} \qquad
t = \frac{1}{\sin\theta}\sqrt{\frac{2h}{g}\left(1+\frac{K^2}{R^2}\right)}$$
The race (all released together): smaller K²/R² wins. Order of
arrival: sliding block (frictionless) > solid sphere (2/5) > disc (1/2) > hollow
sphere (2/3) > ring (1). Results are independent of mass and radius! Friction needed
for rolling: $f = \dfrac{mg\sin\theta}{1 + R^2/K^2}$, requiring $\mu \ge
\dfrac{\tan\theta}{1+R^2/K^2}$ (JEE Advanced refinement); this friction is static and does no
work.
Image Placeholder — Fig. 4
Rolling wheel velocity picture + the incline race of four bodies (Landscape •
Background #ffffff)
AI IMAGE PROMPT: Create a two-panel physics textbook diagram side by side, LANDSCAPE
orientation (16:9), pure white background (#ffffff). LEFT PANEL titled "Velocities on a rolling wheel
(v = ωR)": a large blue circle (wheel) resting on a horizontal black ground line, a curved arrow inside
showing clockwise spin labelled "ω", and a green (#43a047) arrow at the CENTRE pointing right labelled
"v (COM)". Mark three points with dots and velocity arrows: TOP of the wheel — long green arrow
pointing right labelled "2v (fastest)"; CENTRE — medium arrow "v"; CONTACT POINT at the ground — no
arrow, just a label "v = 0 (instantaneously at rest — instantaneous axis)"; add a small note "static
friction, no slipping, no energy loss". RIGHT PANEL titled "Race down the incline": a long incline
(angle θ marked at the base) with FOUR bodies drawn on it at different heights as if mid-race, from
lowest (winning) to highest (losing): a solid blue sphere labelled "solid sphere (K²/R² = 2/5) — 1st",
a blue disc labelled "disc (1/2) — 2nd", a dashed-outline hollow sphere labelled "hollow sphere (2/3)
— 3rd", and a ring labelled "ring (1) — last"; below the incline the boxed formula "a = g sin θ / (1 +
K²/R²) — smaller K²/R² wins; independent of m and R". Style: flat minimalist vector textbook figure,
thin black lines, blue bodies, green velocity arrows with lengths 0 : v : 2v proportioned correctly,
clean sans-serif labels, no gradients, no shadows, pure white background.
Two most-tested rolling pictures in one figure: the 0–v–2v
ladder and the K²/R² race order.
Q. A solid sphere rolls without slipping. What fraction of its total kinetic energy is
rotational? A ring of the same mass and speed rolls beside it — whose total KE is larger?
Sphere: fraction $= \dfrac{2/5}{1+2/5} = \mathbf{\dfrac{2}{7}}$
Comparison: $K = \tfrac12mv^2(1+K^2/R^2)$: ring factor 2 vs
sphere factor 7/5 → the ring carries more total KE at the same v.
Q. A solid sphere rolls from rest down a 30° incline of height 7 m (g = 10 m/s²). Find its
speed at the bottom and its acceleration on the incline.
Speed: $v = \sqrt{\dfrac{2gh}{1+2/5}} = \sqrt{\dfrac{140}{1.4}}
= \sqrt{100} = \mathbf{10 \text{ m/s}}$
Acceleration: $a = \dfrac{g\sin30°}{1.4} = \dfrac{5}{1.4}
\approx \mathbf{3.57 \text{ m/s}^2}$
8. Linear vs Rotational — The Grand Analogy Table
| Linear quantity |
Rotational analogue |
Connecting relation |
| Displacement s |
Angle θ |
s = rθ |
| Velocity v |
Angular velocity ω |
v = rω |
| Acceleration a |
Angular acceleration α |
a_t = rα |
| Mass m |
Moment of inertia I |
I = ΣmR² |
| Force F = ma |
Torque τ = Iα |
τ = r × F |
| Momentum p = mv |
Angular momentum L = Iω |
L = r × p |
| F = dp/dt |
τ = dL/dt |
— |
| KE = ½mv² |
KE = ½Iω² |
Rolling: sum of both |
| Work = Fs, Power = Fv |
Work = τθ, Power = τω |
— |
| p conserved if F_ext = 0 |
L conserved if τ_ext = 0 |
— |
9. Solved Examples — Every Question Type in the Chapter
One worked model of each distinct question type asked from this chapter, tagged by exam.
Attempt each yourself first, then check.
Q. Prove τ = dL/dt (3 marks). / State and prove parallel & perpendicular axis theorems
(3 marks each). / Derive a = g sin θ/(1 + K²/R²) for a body rolling down an incline (5 marks).
Where to answer from: Section 3.2 (differentiate r × p), Section
5.1 with Fig. 3, Section 7.3 (energy method — quickest full-marks route).
Q. From a uniform disc of radius R, a circular hole of radius R/2 is punched with its edge
passing through the centre. Locate the COM of the remaining plate.
Masses ∝ areas: full disc 4m at O; hole m at R/2 from O.
Negative mass: $x = \dfrac{4m(0) - m(R/2)}{3m} =
\mathbf{-\dfrac{R}{6}}$ — i.e., R/6 on the opposite side of the hole.
Q. A 60 kg man walks 3 m (relative to the boat) toward the shore on a 120 kg boat floating
on still water. How far does the boat move?
COM fixed: boat shift $= \dfrac{m_{man}\,d}{m_{man}+m_{boat}} =
\dfrac{60\times3}{180} = \mathbf{1 \text{ m}}$ backward (man advances 2 m relative to shore).
Q. A force $\vec{F} = (2\hat{i} + 3\hat{j})$ N acts at the point $\vec{r} = (\hat{i} -
\hat{j})$ m. Find the torque about the origin.
Cross product (z-component): $\tau_z = x F_y - y F_x = (1)(3) -
(-1)(2) = \mathbf{5\hat{k} \text{ N·m}}$
Q. A disc of I₁ = 0.1 kg·m² spinning at 60 rad/s is dropped coaxially onto a stationary
disc I₂ = 0.2 kg·m². Find the common angular velocity and the energy lost.
L conserved: $\omega = \dfrac{I_1\omega_1}{I_1+I_2} =
\dfrac{0.1\times60}{0.3} = \mathbf{20 \text{ rad/s}}$
Loss: $\Delta K = \tfrac12(0.1)(3600) - \tfrac12(0.3)(400) = 180
- 60 = \mathbf{120 \text{ J}}$ (frictional coupling heat — the rotational analogue of a perfectly
inelastic collision: $\Delta K = \frac{I_1I_2}{2(I_1+I_2)}\omega_1^2$ ✓).
Q. A flywheel (I = 2 kg·m²) spinning at 240 rpm is brought to rest in 8 s by a constant
braking torque. Find the torque and the revolutions made while stopping.
ω₀: $240\times\frac{2\pi}{60} = 8\pi$ rad/s; $\alpha = -\pi$
rad/s².
Torque: $|\tau| = I|\alpha| = 2\pi \approx \mathbf{6.28 \text{
N·m}}$
Revolutions: $\theta = \omega_0t + \tfrac12\alpha t^2 = 64\pi -
32\pi = 32\pi$ rad → $\mathbf{16}$ revolutions.
Q. A 2 kg particle moves along the line y = 4 m with a constant speed 3 m/s. Find its
angular momentum about the origin. Is it conserved?
L = mv × (⊥ distance): $L = 2\times3\times4 = \mathbf{24 \text{
kg·m}^2\text{/s}}$, constant.
Conserved? Yes — no force, so no torque; a straight-line mover
can still HAVE angular momentum (common misconception).
Q. A ring, a disc and a solid sphere are released together from the top of an incline (all
roll without slipping). Rank their (i) accelerations, (ii) arrival times, (iii) speeds at the bottom.
Use K²/R² (ring 1, disc ½, sphere 2/5):
(i) $a_{sphere} > a_{disc} > a_{ring}$; (ii)
times reverse: sphere first, ring last; (iii) same order of speeds:
$v_{sphere} > v_{disc} > v_{ring}$ — all independent of m and R.
Q. Find the radius of gyration of a solid sphere of radius R about (a) its diameter, (b) a
tangent.
(a) $K = \sqrt{I/M} = \sqrt{\tfrac{2}{5}}R =
\mathbf{R\sqrt{2/5} \approx 0.63R}$
(b) $I = \tfrac{7}{5}MR^2 \Rightarrow K = \mathbf{R\sqrt{7/5}
\approx 1.18R}$
Q. A uniform rod of length L pivoted at one end is released from the horizontal position.
Find its angular velocity and the speed of its free end as it passes through the vertical.
Energy (COM falls L/2): $mg\dfrac{L}{2} =
\tfrac12\left(\dfrac{mL^2}{3}\right)\omega^2$
ω: $\omega = \sqrt{\dfrac{3g}{L}}$; tip speed:
$v = \omega L = \mathbf{\sqrt{3gL}}$ — faster than a freely falling particle's $\sqrt{gL\cdot2}
\times$... compare: $\sqrt{3gL} > \sqrt{2gL}$, a favourite conceptual twist.
10. Common Mistakes & Misconceptions — Final Checklist
Night Before Exam
Read this the night before the exam:
- The COM can lie outside the body (ring, L-shape). Internal forces can never move the COM.
- Removed-portion problems: treat the hole as negative mass — don't recompute from scratch.
- Torque and work share the unit N·m ↔ J, but torque is a vector; never call torque "joules".
- Lever arm is the ⊥ distance to the LINE of action of the force, not to its point of application.
- Perpendicular axis theorem is for planar (2-D) bodies ONLY — never apply it to a sphere or solid
cylinder across axes.
- Parallel axis theorem needs one axis THROUGH THE COM: I = I_cm + Md² (not between two arbitrary
axes).
- I depends on the axis: quote the axis every time you quote an I.
- A particle moving in a straight line can have non-zero, constant angular momentum about an off-line
point.
- When the skater pulls in her arms, L is constant but KE increases (muscle work) — don't conserve
energy there; conversely in coupled-disc problems energy is LOST, momentum-like L is not.
- Rolling without slipping: contact point is at rest; friction is static and does no work; top point
moves at 2v.
- Rolling race depends only on K²/R² — never on mass or radius.
- On a frictionless incline nothing can roll — everything slides (and the slider beats all rollers).
- In τ = Iα, τ and I must be about the SAME axis.
- Angular quantities need radians: convert rpm → rad/s (× 2π/60) before any formula.
| Concept |
Formula |
Condition / Note |
| COM (discrete) |
$x_{cm} = \frac{\sum m_ix_i}{M}$ |
Two bodies: $m_1r_1 = m_2r_2$ |
| Removed portion |
$x = \frac{Mx_M - mx_m}{M-m}$ |
Hole = negative mass |
| COM motion |
$M\vec{a}_{cm} = \vec{F}_{ext}$ |
Internal forces can't move COM |
| Boat shift |
$\frac{m_{man}d}{m_{man}+m_{boat}}$ |
d = walk relative to boat |
| Torque |
$\tau = rF\sin\theta$ |
$\vec{\tau} = \vec{r}\times\vec{F}$; = F × lever arm |
| Angular momentum |
$L = mvr\sin\theta = I\omega$ |
$\tau = dL/dt$; conserved if τ = 0 |
| Coupled discs |
$\omega = \frac{I_1\omega_1}{I_1+I_2}$ |
ΔK lost $= \frac{I_1I_2\omega_1^2}{2(I_1+I_2)}$ |
| Parallel axis |
$I = I_{cm} + Md^2$ |
Any body; I minimum through COM |
| Perpendicular axis |
$I_z = I_x + I_y$ |
Planar bodies only |
| Key MOIs |
Ring MR²; disc ½MR²; sphere ⅖MR² |
Shell ⅔MR²; rod centre ML²/12, end ML²/3 |
| Radius of gyration |
$K = \sqrt{I/M}$ |
I = MK² |
| Rotational dynamics |
$\tau = I\alpha$; $K = \tfrac12I\omega^2 = \frac{L^2}{2I}$ |
W = τθ; P = τω |
| Falling mass on pulley |
$a = \frac{g}{1+I/mR^2}$ |
Disc pulley: $a = \frac{2mg}{M+2m}$ |
| Rolling condition |
$v = \omega R$ |
Contact pt at rest; top pt at 2v |
| Rolling KE |
$\tfrac12mv^2(1+K^2/R^2)$ |
Sphere splits 5/7 : 2/7; disc 2/3 : 1/3; ring ½ : ½ |
| Rolling down incline |
$a = \frac{g\sin\theta}{1+K^2/R^2}$ |
$v = \sqrt{\frac{2gh}{1+K^2/R^2}}$; sphere beats disc beats ring |
| Rod from horizontal |
$\omega = \sqrt{3g/L}$ |
Tip speed √(3gL) at the vertical |
| rpm → rad/s |
× 2π/60 |
Always convert first |
Study Plan
How to use these notes: Day 1: Sections 1–2
(COM + its motion) + Fig 1; practise three negative-mass problems. Day 2: Sections 3–4
(torque, L, equilibrium) + Fig 2; prove τ = dL/dt unaided. Day 3: Sections 5–6 (MOI
table + theorems + τ = Iα) + Fig 3; write the MOI table from memory. Day 4: Sections
7–8 (rolling + analogy table) + Fig 4, then all ten Types of Section 9 without looking, followed by the
mistakes checklist. Finish every session by writing the formula sheet from memory.
VARDAAN COMET