| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2002 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Equation of motion angular acceleration |
| Difficulty | Challenging +1.2 This is a Further Maths M4 question requiring the parallel axis theorem, rotational dynamics, and force analysis. Part (i) is standard application of parallel axis theorem (I = ½ma² + m(a/3)²). Parts (ii-iii) require torque equation and constraint forces, which are routine M4 techniques but involve multiple connected steps and careful vector consideration, making it moderately challenging even for Further Maths students. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| Parallel axes rule | M1, A1 | \(I_A = I_c + m(\frac{a}{3})^2 = \frac{1}{3}ma^2 + \frac{1}{9}ma^2 = \frac{11}{9}ma^2\) |
| On release - Moment equation | M1 | \(M(A): C = I\ddot{\theta}\); \(mg(\frac{a}{4}) = (\frac{11}{9}ma^2)\ddot{\theta}\) |
| Angular acceleration | A1 | \(\ddot{\theta} = \frac{9g}{44a}\) |
| Force on disc at A | M1 | When released from rest, no 'central' force needed to maintain circular motion about A, so force on disc is purely vertical (upwards) |
| Vertical force equation | M1 | \(N2(1): mg - F = m(r\ddot{\theta})\) |
| Centripetal force calculation | A1 | \(F = mg - m(\frac{a}{3}a)(\frac{9g}{44a}) = \frac{8}{11}mg\) |
**Parallel axes rule** | M1, A1 | $I_A = I_c + m(\frac{a}{3})^2 = \frac{1}{3}ma^2 + \frac{1}{9}ma^2 = \frac{11}{9}ma^2$
**On release - Moment equation** | M1 | $M(A): C = I\ddot{\theta}$; $mg(\frac{a}{4}) = (\frac{11}{9}ma^2)\ddot{\theta}$
**Angular acceleration** | A1 | $\ddot{\theta} = \frac{9g}{44a}$
**Force on disc at A** | M1 | When released from rest, no 'central' force needed to maintain circular motion about A, so force on disc is purely vertical (upwards)
**Vertical force equation** | M1 | $N2(1): mg - F = m(r\ddot{\theta})$
**Centripetal force calculation** | A1 | $F = mg - m(\frac{a}{3}a)(\frac{9g}{44a}) = \frac{8}{11}mg$
---4 A uniform circular disc has mass $m$, radius $a$ and centre $C$. The disc is free to rotate in a vertical plane about a fixed horizontal axis passing through a point $A$ on the disc, where $C A = \frac { 1 } { 3 } a$.\\
(i) Find the moment of inertia of the disc about this axis.
The disc is released from rest with $C A$ horizontal.\\
(ii) Find the initial angular acceleration of the disc.\\
(iii) State the direction of the force acting on the disc at $A$ immediately after release, and find its magnitude.
\hfill \mbox{\textit{OCR M4 2002 Q4 [8]}}