System of Particles & Rotational Motion

For a continuous body, replace the sum by an integral: \(X=\dfrac{1}{M}\int x\,dm\). Write dm in terms of a coordinate (dm = λ dx for a rod, σ dA for a sheet) and integrate.

Example: non-uniform rod. A rod of length L has linear density λ = kx (heavier toward x = L). M = ∫₀ᴸ kx dx = kL²/2 and ∫x dm = ∫₀ᴸ kx² dx = kL³/3, so X = 2L/3. The CM moves toward the heavier end.

Standard results (measured along the symmetry axis) are collected in the formula card. For any uniform body with a symmetry axis or centre, the CM lies on it.

Treat the removed piece as a "negative mass". If the full body has mass M₁ with CM at x₁, and the removed piece has mass M₂ with CM at x₂, the remainder has

\[X=\frac{M_1x_1-M_2x_2}{M_1-M_2}.\]

For a uniform plate, mass is proportional to area. A disc of radius R with a hole of radius R/2 touching its rim: the hole has ¼ of the mass, centred R/2 from O. The remainder’s CM is at −(¼ × R/2)/(¾) = −R/6, on the side away from the hole (see figure).

• \(M\vec V_{CM}=\sum m_i\vec v_i=\vec P\) and \(M\vec A_{CM}=\vec F_{\rm ext}\). Internal forces cancel in pairs, so they cannot move the CM.
• No external force: the CM keeps its velocity. A person walking on a boat on still water moves the boat the other way; if the CM was at rest, m × (person’s displacement) = M × (boat’s displacement), measured relative to the ground.
• Projectile that explodes: explosion forces are internal, so the CM continues on the original parabola until a fragment hits the ground.
• The CM can accelerate only if there is a net external force (gravity, friction from the ground, and so on).

• 1 revolution = 2π rad = 360°. A speed of N revolutions per minute (rpm) is ω = 2πN/60 rad/s.
• Angular velocity is a vector along the axis. Curl the fingers of your right hand in the sense of rotation; the thumb gives the direction of ω. Linear velocity of a point is \(\vec v=\vec\omega\times\vec r\).
• Clock hands: second hand 2π/60 rad/s, minute hand 2π/3600 rad/s, hour hand 2π/43 200 rad/s. Ratios: minute : hour = 12 : 1, second : minute = 60 : 1.

A point at distance r from the axis has two acceleration parts:

• Tangential: a_t = rα, along the velocity. Zero if ω is constant.
• Centripetal (radial): a_c = ω²r = v²/r, toward the axis. Present whenever the body rotates.
• They are perpendicular, so the total is \(a=\sqrt{a_t^2+a_c^2}=r\sqrt{\alpha^2+\omega^4}\).

Angle turned in the nth second (constant α): θₙ = ω₀ + α(2n − 1)/2, the rotational copy of the distance in the nth second.

1. Formula: τ = rF sinθ, where θ is the angle between r and F.
2. Lever arm: τ = F × d⊥, the force times the perpendicular distance from the point to the line of action (figure). Sliding a force along its line of action does not change its torque.
3. Cross product: \(\vec\tau=\vec r\times\vec F\). In the xy-plane, τ_z = xF_y − yF_x. Positive means anticlockwise (along +z).

Torque is zero when F = 0, when r = 0 (force applied at the axis) or when F passes through the axis (θ = 0 or 180°).

A couple is two equal, opposite, parallel forces with different lines of action. Net force is zero, so the CM does not accelerate, but the torque is F × (separation) about every point. A couple therefore produces pure rotation. Turning a tap or a steering wheel with two hands applies a couple.

Principle of moments: for a lever in equilibrium, clockwise moments equal anticlockwise moments about the fulcrum: F₁d₁ = F₂d₂. The weight of a uniform rod acts at its midpoint (centre of gravity). A metre rule balanced on a knife edge is the standard laboratory application.

1. Forces (figure): weight mg at the middle; N₂ horizontal from the smooth wall at the top; N₁ upward and friction f toward the wall at the foot.
2. Vertical balance: N₁ = mg. Horizontal balance: f = N₂.
3. Torques about the foot (N₁ and f have no torque there): N₂ × L sinθ = mg × (L/2) cosθ, so N₂ = mg/(2 tanθ).
4. No slip needs f ≤ μN₁: mg/(2 tanθ) ≤ μmg, so tanθ ≥ 1/(2μ).

A steeper ladder (larger θ) is safer. When a person climbs higher, the torque of the load about the foot grows, so the friction needed grows too; ladders usually slip when someone is near the top.

Horizontal push at height h on a block of width b: sliding needs F = μmg; tipping about the front lower edge needs F h = mg(b/2). It tips first if μ > b/(2h).

Block on a tilting incline (width b, height h, CG at the centre): it topples when the vertical line through the CG passes the lower edge, tanθ = b/h. It slides when tanθ = μ. Whichever angle is smaller happens first.

Vehicle on a level curve: it skids if v² > μrg, and overturns if v² > grd/(2h), where d is the distance between the wheels and h the height of the CG. A low CG and wide track resist overturning.

Thin rod about its centre (perpendicular axis). Mass per length λ = M/L. An element dx at distance x has dm = λ dx: \[I=\int_{-L/2}^{L/2}x^2\lambda\,dx=\lambda\frac{L^3}{12}=\frac{ML^2}{12}.\] About one end, integrate from 0 to L: I = λL³/3 = ML²/3.

Ring about its central axis. Every element is at distance R, so I = R²∫dm = MR².

Disc about its central axis. Split it into rings of radius r and width dr. Area 2πr dr, so dm = (M/πR²)(2πr dr) = (2M/R²) r dr: \[I=\int_0^R r^2\,\frac{2M}{R^2}r\,dr=\frac{2M}{R^2}\cdot\frac{R^4}{4}=\frac{MR^2}{2}.\] A solid cylinder about its own axis is a stack of such discs, so it also has MR²/2.

Write I = Mk². The radius of gyration k is the distance at which the whole mass could be placed (as a ring) to give the same I. Ring: k = R. Disc: k = R/√2. Solid sphere: k = √(2/5) R. Rod about centre: k = L/(2√3).

I depends on the mass, how the mass is spread, and the axis. It does not depend on the angular velocity or the torque applied. Spreading the same mass farther out (a ring instead of a disc, a hollow sphere instead of a solid one) increases I.

Parallel-axis theorem. Put the CM at the origin and the new axis at distance d. A mass element at (x, y) is at distance² (x − d)² + y² from the new axis. Summing: Σm(x² + y²) − 2dΣmx + d²Σm. The first term is I_CM, the middle term is zero because Σmx = 0 about the CM, and the last is Md². So I = I_CM + Md².

Perpendicular-axis theorem. For a lamina in the xy-plane, every element has z = 0, so its distance² from the z-axis is x² + y². Thus I_z = Σm(x² + y²) = Σmy² + Σmx² = I_x + I_y. It fails for a 3-D body because then I_x = Σm(y² + z²) contains z terms.

• Ring about a diameter: I_z = MR² and I_x = I_y by symmetry, so I_diameter = MR²/2.
• Disc about a diameter: MR²/4. About a tangent in its plane: MR²/4 + MR² = 5MR²/4. About a tangent perpendicular to its plane: MR²/2 + MR² = 3MR²/2.
• Ring about a tangent in its plane: MR²/2 + MR² = 3MR²/2. Perpendicular to its plane: 2MR².
• Solid sphere about a tangent: 2MR²/5 + MR² = 7MR²/5. Hollow sphere: 5MR²/3.
• Square plate (side a): I_z = Ma²/6 through the centre. Every in-plane axis through the centre, including a diagonal, has Ma²/12.

Every linear result has a rotational copy. The work–energy theorem becomes W = ½Iω² − ½Iω₀².

A block m hangs from a string wound on a pulley (disc or cylinder) of moment of inertia I and radius R. The string does not slip, so a = Rα.

1. Block: mg − T = ma.
2. Pulley: TR = Iα = Ia/R, so T = Ia/R².
3. Add: mg = (m + I/R²)a, so \(a=\dfrac{mg}{m+I/R^2}\).

For a uniform disc, I/R² = M/2, so a = mg/(m + M/2). The pulley acts like an extra mass I/R² that must be accelerated. With two hanging masses (Atwood machine), a = (m₁ − m₂)g/(m₁ + m₂ + I/R²), and the tensions on the two sides are no longer equal.

• Torque about the pivot = Mg(L/2); I = ML²/3. Initial α = 3g/(2L).
• The free end starts with acceleration αL = 3g/2, more than g. A coin placed on the free end lifts off.
• Energy at the vertical position: Mg(L/2) = ½(ML²/3)ω², so ω = √(3g/L) and the free end moves at √(3gL).

A particle moving in a straight line has angular momentum about any point not on that line: L = mv d⊥, where d⊥ is the perpendicular distance from the point to the line. If no force acts, v and d⊥ stay fixed, so L is constant.

This links to Kepler’s second law. A planet moves under a central force, which has zero torque about the Sun, so L is conserved. The area swept per unit time is dA/dt = L/(2m), a constant: equal areas in equal times.

• Skater, diver, dancer: pulling arms in reduces I and increases ω. Rotational KE = L²/2I rises; the person’s muscles do the extra work.
• Person on a rotating platform walking toward the centre: I falls, ω rises.
• Earth shrinking with the same mass: I ∝ R², so ω ∝ 1/R² and the length of the day ∝ R².
• Two discs coupled on a common axis: ω = (I₁ω₁ + I₂ω₂)/(I₁ + I₂). Kinetic energy is lost, like a perfectly inelastic collision: ΔK = I₁I₂(ω₁ − ω₂)²/[2(I₁ + I₂)].

Angular impulse is ∫τ dt = ΔL, the rotational copy of impulse = change in momentum.

When a particle hits a rod that is pivoted, the pivot exerts a large force during the impact. So linear momentum is not conserved. But the pivot force passes through the pivot and has no torque about it, so angular momentum about the pivot is conserved:

\[mvd=(I_{\rm rod}+md^2)\,\omega\quad(\text{particle sticks at distance }d)\]

Every point of a rolling wheel has the centre’s velocity v plus a rotational velocity of size ωR tangent to the rim. With v = ωR (no slipping):

• Bottom (contact) point: v − ωR = 0. It is momentarily at rest.
• Top point: v + ωR = 2v.
• Points level with the centre: v forward and v vertical, so √2 v at 45°.
• In general, a rim point at angle φ from the contact point (measured at the centre) moves at 2v sin(φ/2).

Equivalent picture: the wheel rotates about the contact point P with angular speed ω. Each point’s speed is ω × (distance from P), perpendicular to the line from P (figure).

With I = Mk² and ω = v/R:

\[K=\tfrac12Mv^2+\tfrac12Mk^2\frac{v^2}{R^2}=\tfrac12Mv^2\left(1+\frac{k^2}{R^2}\right).\]

A ball thrown along a rough floor sliding without spin (v = u, ω = 0) feels kinetic friction backward. Friction slows the centre and its torque spins the ball up, until v = ωR. Then friction stops acting and the ball rolls.

The contact force passes through the contact point, so angular momentum about a point on the floor is conserved: MuR = MvR + Iv/R. This gives \(v=\dfrac{u}{1+k^2/R^2}\): a ring ends at u/2, a disc at 2u/3, a solid sphere at 5u/7.

Forces and torque (figure):

1. Along the slope: Mg sinθ − f = Ma.
2. Torque about the centre: fR = Iα = Mk²(a/R), so f = Ma(k²/R²).
3. Add: Mg sinθ = Ma(1 + k²/R²), so a = g sinθ/(1 + k²/R²).

Energy: after a vertical drop h, Mgh = ½Mv²(1 + k²/R²), so v² = 2gh/(1 + k²/R²). Since v² = 2a(h/sinθ), this gives the same a.

Time for a slope length L: \(t=\sqrt{\dfrac{2L(1+k^2/R^2)}{g\sin\theta}}\). A block sliding on a frictionless incline (a = g sinθ) beats all of them. With equal masses, all rolling bodies reach the bottom with the same total KE (Mgh); they differ only in how it is split.

Pure rolling needs f ≤ μN = μMg cosθ, which gives

\[\mu\ \ge\ \tan\theta\,\frac{k^2/R^2}{1+k^2/R^2}:\quad \text{sphere } \tfrac27\tan\theta,\ \text{disc } \tfrac13\tan\theta,\ \text{ring } \tfrac12\tan\theta.\]

If μ is smaller, the body rolls with slipping: kinetic friction μMg cosθ acts, a = g(sinθ − μcosθ), and energy is lost.

Rolling up an incline with no driving torque, friction again acts up the slope: it must slow the spin as gravity slows the centre. The height reached is h = v²(1 + k²/R²)/(2g), higher than a sliding block’s v²/2g, because the rotational energy is also converted to height.