| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2004 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Energy method angular speed |
| Difficulty | Challenging +1.2 This is a standard M4 rotation problem requiring parallel axis theorem, torque equation, and energy methods. While it involves multiple steps and careful application of mechanics principles, the techniques are well-practiced at this level with no novel insights required. The presence of a constant frictional couple adds moderate complexity beyond basic textbook exercises. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| \(I_C = \frac{1}{2}mr^2 = \frac{1}{2} \times 4 \times 0.6^2 = 0.72\) | B1 | Moment of inertia of cylinder |
| \(I_A = I_C + m \times 0.4^2 = 1.36\) kg m² | M1 A1 | Parallel axes theorem applied correctly |
| Answer | Marks | Guidance |
|---|---|---|
| \(\alpha = 8.00\) rad s⁻¹ | M1 A1 | [3] Torque equation at release |
| Answer | Marks | Guidance |
|---|---|---|
| \(\omega = 3.46\) rad s⁻¹ | M1 M1 A1 | [4] Energy equation with all three terms; solving for \(\omega\) |
$I_C = \frac{1}{2}mr^2 = \frac{1}{2} \times 4 \times 0.6^2 = 0.72$ | B1 | Moment of inertia of cylinder
$I_A = I_C + m \times 0.4^2 = 1.36$ kg m² | M1 A1 | Parallel axes theorem applied correctly
$C = I\alpha$
$39.2 \times 0.4 - 4.8 = 1.36 \times \alpha$
$\alpha = 8.00$ rad s⁻¹ | M1 A1 | [3] Torque equation at release
Gain in K.E. = loss in G.P.E. - work done by friction
$\frac{1}{2}I\omega^2 = 4 \times 9.8 \times 0.4 - 4.8 \times \frac{3}{4}$
$\omega^2 = 11.9708...$
$\omega = 3.46$ rad s⁻¹ | M1 M1 A1 | [4] Energy equation with all three terms; solving for $\omega$
---5\\
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A uniform circular disc has mass 4 kg , radius 0.6 m and centre $C$. The disc can rotate in a vertical plane about a fixed horizontal axis which is perpendicular to the disc and which passes through the point $A$ on the disc, where $A C = 0.4 \mathrm {~m}$. A frictional couple of constant moment 4.8 Nm opposes the motion. The disc is released from rest with $A C$ horizontal (see diagram).\\
(i) Find the moment of inertia of the disc about the axis through $A$.\\
(ii) Find the angular acceleration of the disc immediately after it is released.\\
(iii) Find the angular speed of the disc when $C$ is first vertically below $A$.
\hfill \mbox{\textit{OCR M4 2004 Q5 [9]}}