| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2006 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Force at pivot/axis |
| Difficulty | Challenging +1.2 This is a standard M5 rotation question requiring systematic application of parallel axis theorem, energy conservation, and Newton's second law for rotation. While it involves multiple parts and careful bookkeeping of four rods, each step follows established procedures without requiring novel insight. The moment of inertia calculation is methodical, and parts (b)-(d) are routine applications of standard mechanics principles, making it moderately above average difficulty for A-level. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force4.10c Integrating factor: first order equations6.04c Composite bodies: centre of mass |
| Answer | Marks |
|---|---|
| (a) \(I_{AB} = 2 \times \frac{2}{3}ma^2 + m(2a)^2 = \frac{20}{3}ma^2\) | M1, A1, A1 |
| By 1r axis: \(I_A = 2 \times \frac{20}{3}ma^2 = \frac{40}{3}ma^2\) (✓) | M1, A1 (5) |
| Or \(I_A = 2 \times \frac{4}{3}ma^2 + 2\left(\frac{1}{3}ma^2 + 5ma^2\right) = \frac{40}{3}ma^2\) | M1, M1, A1, A1 |
| (b) \(M(A): \frac{40ma^2}{3}\ddot{\theta} = 4mg.a\sqrt{2}\) | M1, A1 |
| \(\ddot{\theta} = \frac{3g\sqrt{2}}{10a}\) | A1 (3) |
| (c) \(\frac{1}{2}.\frac{40ma^2}{3}\dot{\theta}^2 = 4mg.a\sqrt{2}\) | M1, A1 |
| \(\theta = \sqrt{\frac{3g\sqrt{2}}{5a}}\) | A1 (3) |
| (d) \(R(E): x = 4ma\sqrt{2}\dot{\theta}^2 = 4ma\sqrt{2}.\frac{3g\sqrt{2}}{5a} = \frac{24mg}{5}\) | M1, A1 |
| \(R(Y): 4mg - Y = 4ma\sqrt{2}\ddot{\theta} = 4ma\sqrt{2}.\frac{3g\sqrt{2}}{10a}\) | M1 |
| \(Y = 4mg - 4ma\sqrt{2}.\frac{3g\sqrt{2}}{10a} = \frac{8mg}{5}\) | A1 |
| \(R = \sqrt{x^2 + y^2} = \frac{8mg}{5}\sqrt{1^2 + 3^2} = \frac{8\sqrt{10}mg}{5}\) | M1, A1 |
(a) $I_{AB} = 2 \times \frac{2}{3}ma^2 + m(2a)^2 = \frac{20}{3}ma^2$ | M1, A1, A1 |
By 1r axis: $I_A = 2 \times \frac{20}{3}ma^2 = \frac{40}{3}ma^2$ (✓) | M1, A1 (5) |
Or $I_A = 2 \times \frac{4}{3}ma^2 + 2\left(\frac{1}{3}ma^2 + 5ma^2\right) = \frac{40}{3}ma^2$ | M1, M1, A1, A1 |
(b) $M(A): \frac{40ma^2}{3}\ddot{\theta} = 4mg.a\sqrt{2}$ | M1, A1 |
$\ddot{\theta} = \frac{3g\sqrt{2}}{10a}$ | A1 (3) |
(c) $\frac{1}{2}.\frac{40ma^2}{3}\dot{\theta}^2 = 4mg.a\sqrt{2}$ | M1, A1 |
$\theta = \sqrt{\frac{3g\sqrt{2}}{5a}}$ | A1 (3) |
(d) $R(E): x = 4ma\sqrt{2}\dot{\theta}^2 = 4ma\sqrt{2}.\frac{3g\sqrt{2}}{5a} = \frac{24mg}{5}$ | M1, A1 |
$R(Y): 4mg - Y = 4ma\sqrt{2}\ddot{\theta} = 4ma\sqrt{2}.\frac{3g\sqrt{2}}{10a}$ | M1 |
$Y = 4mg - 4ma\sqrt{2}.\frac{3g\sqrt{2}}{10a} = \frac{8mg}{5}$ | A1 |
$R = \sqrt{x^2 + y^2} = \frac{8mg}{5}\sqrt{1^2 + 3^2} = \frac{8\sqrt{10}mg}{5}$ | M1, A1 |8. Four uniform rods, each of mass $m$ and length $2 a$, are joined together at their ends to form a plane rigid square framework $A B C D$ of side $2 a$. The framework is free to rotate in a vertical plane about a fixed smooth horizontal axis through $A$. The axis is perpendicular to the plane of the framework.
\begin{enumerate}[label=(\alph*)]
\item Show that the moment of inertia of the framework about the axis is $\frac { 40 m a ^ { 2 } } { 3 }$.
The framework is slightly disturbed from rest when $C$ is vertically above $A$. Find
\item the angular acceleration of the framework when $A C$ is horizontal,
\item the angular speed of the framework when $A C$ is horizontal,
\item the magnitude of the force acting on the framework at $A$ when $A C$ is horizontal.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2006 Q8 [17]}}