| 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 | Energy method angular speed |
| Difficulty | Standard +0.3 This is a standard M4 rotational dynamics problem requiring moment of inertia calculation using parallel axis theorem, then energy methods with friction. The steps are routine for this module: calculate I using standard formulas, apply energy conservation with work done against friction. While multi-step, it follows a predictable template with no novel insight required, making it slightly easier than average. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Moment of inertia calculation | M1, A1 | \(I = 124 \times 6^2 + (\frac{1}{3} \times 75 \times 3 \cdot 6^2 + 75 \times 2 \cdot 4^2) = 4464 + 756 = 5220 \text{ kg m}^2\) (shown) |
| Energy equation | M1 | Gain in K.E. = loss in G.P.E. – work done by frictional couple; \(\frac{1}{2}I\omega^2 = mgh - C\theta\) |
| Substitution and solution | M1, A1 | \(2610\omega^2 = 75 \times 9 \cdot 8 \times 2 \cdot 4 - 850 \times \frac{2}{3}\); \(\omega = 1 \cdot 72 \text{ rad s}^{-1}\) |
**Moment of inertia calculation** | M1, A1 | $I = 124 \times 6^2 + (\frac{1}{3} \times 75 \times 3 \cdot 6^2 + 75 \times 2 \cdot 4^2) = 4464 + 756 = 5220 \text{ kg m}^2$ (shown)
**Energy equation** | M1 | Gain in K.E. = loss in G.P.E. – work done by frictional couple; $\frac{1}{2}I\omega^2 = mgh - C\theta$
**Substitution and solution** | M1, A1 | $2610\omega^2 = 75 \times 9 \cdot 8 \times 2 \cdot 4 - 850 \times \frac{2}{3}$; $\omega = 1 \cdot 72 \text{ rad s}^{-1}$
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\includegraphics[max width=\textwidth, alt={}, center]{98647526-b52a-4316-9a09-48d756b8f599-3_117_913_251_630}
An arm on a fairground ride is modelled as a uniform rod $A B$, of mass 75 kg and length 7.2 m , with a particle of mass 124 kg attached at $B$. The arm can rotate about a fixed horizontal axis perpendicular to the rod and passing through the point $P$ on the rod, where $A P = 1.2 \mathrm {~m}$.\\
(i) Show that the moment of inertia of the arm about the axis is $5220 \mathrm {~kg} \mathrm {~m} ^ { 2 }$.\\
(ii) The arm is released from rest with $A B$ horizontal, and a frictional couple of constant moment 850 N m opposes the motion. Find the angular speed of the arm when $B$ is first vertically below $P$.
\hfill \mbox{\textit{OCR M4 2002 Q6 [8]}}