| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2018 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Force at pivot/axis |
| Difficulty | Challenging +1.2 This is a standard mechanics problem requiring parallel axis theorem application and force analysis at release. Part (a) is routine calculation using I = ml²/3 for three rods. Part (b) requires resolving forces when angular acceleration is known from torque equation. While multi-step, it follows standard M5 techniques without requiring novel insight—slightly above average due to the geometry and careful component resolution needed. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\text{MI}=2\times\frac{4ma^2}{3}+\left(\frac{ma^2}{3}+m(2a\cos 30°)^2\right)\) | M1 A1, A1 | M1 for attempt at MI dimensionally correct; first A1 for first two terms; second A1 for third term unsimplified |
| \(=6ma^2\) | A1 | A1 for correct GIVEN ANSWER |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(X-3mg\cos 60°=0\) since \(\dot{\theta}=0\) | M1 | M1 for equation of motion inwards with usual rules and RHS \(=0\) |
| \(X=\frac{3mg}{2}\) | A1 | A1 for correct value of \(X\) |
| \(3mg\sin 60°\pm Y=3m\cdot\frac{2}{3}\cdot 2a\cos 30°\,\ddot{\theta}\) | M1 A1 A1 | M1 for equation of motion tangentially; A1A1 for correct equation (A1A0 if one error) |
| \(M(A):\; 3mga=6ma^2\ddot{\theta}\) OR \(6a\ddot{\theta}=2\sqrt{3}g\sin 60°\) | M1 A1 | M1 for rotational equation about \(A\) giving equation in \(\ddot{\theta}\) only |
| Eliminating \(\ddot{\theta}\): \(Y=\frac{mg\sqrt{3}}{2}\) | DM1 A1 | DM1 dependent on previous 2 M's for eliminating to give \(Y\) in terms of \(mg\) |
| \(R=\sqrt{X^2+Y^2}=\frac{mg}{2}\sqrt{3+3^2}=mg\sqrt{3}\) | M1 A1 | M1 independent, for finding magnitude using \(X\) and \(Y\) in terms of \(mg\) only; A1 for correct answer |
# Question 6:
## Part 6(a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\text{MI}=2\times\frac{4ma^2}{3}+\left(\frac{ma^2}{3}+m(2a\cos 30°)^2\right)$ | M1 A1, A1 | M1 for attempt at MI dimensionally correct; first A1 for first two terms; second A1 for third term unsimplified |
| $=6ma^2$ | A1 | A1 for correct GIVEN ANSWER |
## Part 6(b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $X-3mg\cos 60°=0$ since $\dot{\theta}=0$ | M1 | M1 for equation of motion inwards with usual rules and RHS $=0$ |
| $X=\frac{3mg}{2}$ | A1 | A1 for correct value of $X$ |
| $3mg\sin 60°\pm Y=3m\cdot\frac{2}{3}\cdot 2a\cos 30°\,\ddot{\theta}$ | M1 A1 A1 | M1 for equation of motion tangentially; A1A1 for correct equation (A1A0 if one error) |
| $M(A):\; 3mga=6ma^2\ddot{\theta}$ OR $6a\ddot{\theta}=2\sqrt{3}g\sin 60°$ | M1 A1 | M1 for rotational equation about $A$ giving equation in $\ddot{\theta}$ only |
| Eliminating $\ddot{\theta}$: $Y=\frac{mg\sqrt{3}}{2}$ | DM1 A1 | DM1 dependent on previous 2 M's for eliminating to give $Y$ in terms of $mg$ |
| $R=\sqrt{X^2+Y^2}=\frac{mg}{2}\sqrt{3+3^2}=mg\sqrt{3}$ | M1 A1 | M1 independent, for finding magnitude using $X$ and $Y$ in terms of $mg$ only; A1 for correct answer |
---6. Three equal uniform rods, each of mass $m$ and length $2 a$, form the sides of a rigid equilateral triangular frame $A B C$. The frame is free to rotate in a vertical plane about a fixed smooth horizontal axis $L$ which passes through $A$ and is perpendicular to the plane of the frame.
\begin{enumerate}[label=(\alph*)]
\item Show that the moment of inertia of the frame about $L$ is $6 m a ^ { 2 }$.
The frame is held with $A B$ horizontal and $C$ below $A B$, and released from rest.
Given that the centre of mass of the frame is two thirds of the way along a median from a vertex,
\item find the magnitude of the force exerted by the axis on the frame at $A$ at the instant when the frame is released.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2018 Q6 [15]}}