Motion of rigid bodies in two dimensions

At a Glance

• Frequency: 4 sub-parts across 3 of 13 years (2019, 2023, 2024)
• Priority tier: T3
• Marks (count): 10 (3), 15 (1)
• Average solve time: ~10 min
• Difficulty mix: medium 3, hard 1
• Section: B | Dominant type: application

Why This Chapter Matters

Rigid-body motion questions in Section B span three types: (a) steady precession of a rotating rod (take moments about the fixed end); (b) conservation of horizontal centre of mass when a body slides on a smooth surface; and (c) harder problems involving rolling or pendulum-like motion. The steady-precession template is identical every time — take moments about the hinge, integrate the centrifugal moment using $I_O=\frac{1}{3}ML^2$ or $\frac{4}{3}Ma^2$ as appropriate, and balance against the gravity moment. The momentum-conservation template has no integration: write the CG before and after.

Minimum Theory

Moment of inertia. For a uniform rod of mass $M$, length $L$, about one end: $I_O=\frac{1}{3}ML^2$. About the centre: $I_{\text{cm}}=\frac{1}{12}ML^2$.

Steady precession of a rod. A uniform rod of mass $M$, length $2a$, hinged at $O$ and revolving at steady angular velocity $\omega$ making angle $\alpha$ with the vertical:

• Gravity moment about $O$ (tending to decrease $\alpha$): $Mga\sin\alpha$.
• Centrifugal moment about $O$ (tending to increase $\alpha$): $\omega^2\sin\alpha\cos\alpha\int_0^{2a}r^2\lambda\,dr = \omega^2\sin\alpha\cos\alpha\cdot I_O$.

For rod of length $2a$: $I_O=\frac{1}{3}M(2a)^2=\frac{4}{3}Ma^2$.

Balancing at steady state: $\dfrac{4}{3}Ma^2\omega^2\sin\alpha\cos\alpha=Mga\sin\alpha$, giving

$\cos\alpha=\frac{3g}{4a\omega^2}.$

Conservation of horizontal CG. If no external horizontal force acts (smooth floor), the horizontal centre of mass of the system is constant. Use this to find the displacement of one body when another moves.

Question Archetypes

steady-precession (1 question(s); 2019)

Solution Template

1. Identify: rod mass $M$, length $2a$, hinged at $O$, angle $\alpha$, angular velocity $\omega$.
2. Gravity moment about $O$: $M\cdot g\cdot(\text{horizontal distance to CG}) = Mga\sin\alpha$.
3. Centrifugal moment about $O$: $\omega^2\sin\alpha\cos\alpha\cdot I_O$ where $I_O=\frac{4}{3}Ma^2$.
4. Balance: $\frac{4}{3}Ma^2\omega^2\sin\alpha\cos\alpha=Mga\sin\alpha$; cancel $M$, $a$, $\sin\alpha$ (for non-vertical steady state).
5. Solve: $\cos\alpha=\frac{3g}{4a\omega^2}$.
6. State existence condition: requires $4a\omega^2\ge3g$.

Worked Example

2019 Paper 2, 2019-P2-Q5c (10 marks)

A uniform rod $OA$ of length $2a$ is hinged at $O$ and revolves at angular velocity $\omega$ at constant angle $\alpha$ to the vertical. Find $\alpha$.

Centrifugal moment: $\displaystyle\int_0^{2a}\lambda\omega^2(r\sin\alpha)(r\cos\alpha)\,dr=\omega^2\sin\alpha\cos\alpha\cdot\frac{M}{2a}\cdot\frac{(2a)^3}{3}=\frac{4}{3}Ma^2\omega^2\sin\alpha\cos\alpha$.

Gravity moment: $Mga\sin\alpha$.

Balance ($\sin\alpha\ne0$): $\frac{4}{3}a\omega^2\cos\alpha=g$.

$\boxed{\cos\alpha=\frac{3g}{4a\omega^2}\quad(\text{valid for }4a\omega^2\ge3g).}$

Physical meaning: equivalent to a simple conical pendulum of effective length $4a/3 = I_O/(Ma)$.

Common Traps

• Two factors of $r$ in the centrifugal integral. The centrifugal force on element $dm$ is $dm\cdot\omega^2(r\sin\alpha)$ (proportional to horizontal radius $r\sin\alpha$), and the moment arm about the horizontal axis through $O$ is $r\cos\alpha$ (vertical distance). Together: $dm\cdot\omega^2(r\sin\alpha)(r\cos\alpha)$, giving $\int r^2\,dm=I_O$.
• $I_O$ for length $2a$ rod. $I_O=\frac{1}{3}M(2a)^2=\frac{4}{3}Ma^2$, not $\frac{1}{3}Ma^2$.
• Check validity. If $4a\omega^2<3g$, the equation $\cos\alpha=3g/(4a\omega^2)>1$ has no solution; the only steady configuration is $\alpha=0$ (rod hangs down). Always report the existence condition.

momentum-conservation (1 question(s); 2024)

Solution Template

1. No external horizontal force → CG of the whole system is conserved.
2. Write: initial CG $=$ final CG.
3. Initial positions of all bodies; final positions after displacement.
4. Solve for the unknown displacement.

Worked Example

2024 Paper 2, 2024-P2-Q5d (10 marks)

A board of mass $m$, length $2a$ rests on a smooth horizontal plane. A man of mass $M$ walks from one end to the other. Find the distance moved by the board.

Let board’s CG initially at $x=0$; man initially at $x=a$ (one end). Let the board shift by $\Delta$.

Initial CG: $\frac{m\cdot0+M\cdot a}{m+M}=\frac{Ma}{m+M}$.

Final CG (board shifted by $\Delta$, man at the other end $x=\Delta-a$): $\frac{m\Delta+M(\Delta-a)}{m+M}=\Delta-\frac{Ma}{m+M}$.

Setting equal: $\frac{Ma}{m+M}=\Delta-\frac{Ma}{m+M}$, giving $\Delta=\frac{2Ma}{m+M}$.

$\boxed{\text{Board moves }\frac{2Ma}{m+M}\text{ opposite to the man's direction.}}$

Common Traps

• The floor is smooth (no external horizontal force). “Rough board” refers to friction between man and board (internal); it’s the floor that must be smooth for CG conservation to apply.
• The man walks a distance $2a$ relative to the board. His displacement relative to the ground is $2a-\Delta$ (he moves $2a$ on the board while the board moves $\Delta$ opposite). The CG equation handles this automatically.
• Direction: the board moves opposite to the man. If the man walks in the $+x$ direction, the board shifts in the $-x$ direction (momentum conservation: no net horizontal momentum initially).

Marks-Aware Writing

For steady precession (10 marks): draw or describe the forces (gravity, hinge reaction, centrifugal); take moments about $O$ (state why: to eliminate hinge reaction); write the centrifugal integral with both $r\sin\alpha$ and $r\cos\alpha$ factors; evaluate $I_O$; balance and solve. Show the existence condition.

For momentum conservation (10 marks): state “no external horizontal force → CG conserved”; write the CG equation; solve for $\Delta$; state direction. Four clear steps.

Practice Set

• 2023-P2-Q7b (15 m) — — Rolling rigid body on a surface; more complex but uses the same inertia integrals.