| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Prove MI by integration |
| Difficulty | Challenging +1.2 This is a standard M5 moment of inertia proof using integration with a given formula. While it requires setting up coordinates, integrating over the triangle using parallel rods, and applying the perpendicular axis theorem or parallel axis theorem, the approach is methodical and follows textbook procedures. The given rod formula significantly simplifies the work. Harder than routine C3 calculus but typical for Further Maths mechanics. |
| Spec | 6.04b Find centre of mass: using symmetry |
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\begin{figure}[h]
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\caption{Figure 1}
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A uniform triangular lamina $A B C$, of mass $M$, has $A B = A C$ and $B C = 2 a$. The mid-point of $B C$ is $D$ and $A D = h$, as shown in Figure 1.
Show, using integration, that the moment of inertia of the lamina about an axis through $A$, perpendicular to the plane of the lamina, is
$$\frac { M } { 6 } \left( a ^ { 2 } + 3 h ^ { 2 } \right)$$
[You may assume without proof that the moment of inertia of a uniform rod, of length $2 l$ and mass $m$, about an axis through its midpoint and perpendicular to the rod, is $\frac { 1 } { 3 } m l ^ { 2 }$.]
\hfill \mbox{\textit{Edexcel M5 2013 Q5 [10]}}