| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2003 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Force at pivot/axis |
| Difficulty | Standard +0.3 This is a standard M4 rotation question requiring straightforward application of parallel axis theorem, center of mass calculation, torque equation (Iα = Γ), and force resolution. All steps follow routine procedures with no novel insight required, making it slightly easier than average for Further Maths M4 content. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| \(I = \frac{1}{2}mr^2 + Mr^2 = \frac{1}{2} \times 0.08 \times 0.35^2 + 0.24 \times 0.35^2 = 0.0343\) kg m² | M1 A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{x} = 0.2625\) | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\alpha = 24\) rad s⁻² | M1 A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Hence the magnitude of the force \(= \sqrt{2.1^2 + 1.12^2} = 2.38\) N | M1 A1 | [6] |
**Part (a): Moment of Inertia**
$I = \frac{1}{2}mr^2 + Mr^2 = \frac{1}{2} \times 0.08 \times 0.35^2 + 0.24 \times 0.35^2 = 0.0343$ kg m² | M1 A1 | [3]
**Part (b): Centre of Mass Position**
$0.32\bar{x} = 0.08 \times 0 + 0.24 \times 0.35$
$\bar{x} = 0.2625$ | A1 | [2]
**Part (c): Angular Acceleration (when OP is horizontal)**
$C = I\alpha$
$0.32 \times 9.8 \times 0.2625 = 0.0343\alpha$
$\alpha = 24$ rad s⁻² | M1 A1 | [2]
**Part (d): Force at Bearing**
N2(←): $H = m(r\omega^2) = 0.32 \times 0.2625 \times 5^2 = 2.1$ N
N2(↓): $3.136 - V = m(r\alpha)$
$V = 3.136 - 0.32 \times 0.2625 \times 24 = 1.12$ N
Hence the magnitude of the force $= \sqrt{2.1^2 + 1.12^2} = 2.38$ N | M1 A1 | [6]6\\
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A wheel consists of a uniform circular disc, with centre $O$, mass 0.08 kg and radius 0.35 m , with a particle $P$ of mass 0.24 kg attached to a point on the circumference. The wheel is rotating without resistance in a vertical plane about a fixed horizontal axis through $O$ (see diagram).\\
(i) Find the moment of inertia of the wheel about the axis.\\
(ii) Find the distance of the centre of mass of the wheel from the axis.
At an instant when $O P$ is horizontal and the angular speed of the wheel is $5 \mathrm { rad } \mathrm { s } ^ { - 1 }$, find\\
(iii) the angular acceleration of the wheel,\\
(iv) the magnitude of the force acting on the wheel at $O$.
\hfill \mbox{\textit{OCR M4 2003 Q6 [13]}}