| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Prove MI by integration |
| Difficulty | Challenging +1.8 This M5 question requires proving a standard moment of inertia result using integration (non-trivial setup with variable mass elements), applying it to deduce a radius of gyration, and calculating MOI for a composite shape using parallel axis theorem. The integration in part (a) requires careful coordinate setup and the composite shape in (c) demands decomposition strategy and multiple applications of theorems—significantly above average difficulty but standard for M5. |
| Spec | 6.04b Find centre of mass: using symmetry6.04d Integration: for centre of mass of laminas/solids |
\begin{enumerate}[label=(\alph*)]
\item A uniform lamina of mass $m$ is in the shape of a triangle $ABC$. The perpendicular distance of $C$ from the line $AB$ is $h$. Prove, using integration, that the moment of inertia of the lamina about $AB$ is $\frac{1}{6}mh^2$.
[7]
\end{enumerate}
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Deduce the radius of gyration of a uniform square lamina of side $2a$, about a diagonal.
[3]
\end{enumerate}
The points $X$ and $Y$ are the mid-points of the sides $RQ$ and $RS$ respectively of a square $PQRS$ of side $2a$. A uniform lamina of mass $M$ is in the shape of $PQXYS$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Show that the moment of inertia of this lamina about $XY$ is $\frac{79}{84}Ma^2$.
[6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 Q7 [16]}}