| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2013 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Prove MI by integration |
| Difficulty | Hard +2.3 This is a challenging M5 question requiring 3D integration of moments of inertia using the perpendicular axis theorem and parallel axis theorem, combined with careful setup of variable mass elements. It demands strong spatial reasoning, manipulation of multiple integration steps, and synthesis of several mechanical principles—significantly harder than standard A-level questions but within the scope of Further Maths M5. |
| Spec | 6.04b Find centre of mass: using symmetry |
Show, using integration, that the moment of inertia of a uniform solid right circular cone of mass $M$, height $h$ and base radius $a$, about an axis through the vertex, parallel to the base, is
$$\frac{3M}{20}(a^2 + 4h^2)$$
[You may assume without proof that the moment of inertia of a uniform circular disc, of radius $r$ and mass $m$, about a diameter is $\frac{1}{4}mr^2$.]
(13 marks)
---4. Show, using integration, that the moment of inertia of a uniform solid right circular cone of mass $M$, height $h$ and base radius $a$, about an axis through the vertex, parallel to the base, is
$$\frac { 3 M } { 20 } \left( a ^ { 2 } + 4 h ^ { 2 } \right)$$
[You may assume without proof that the moment of inertia of a uniform circular disc, of radius $r$ and mass $m$, about a diameter is $\frac { 1 } { 4 } m r ^ { 2 }$.]
\hfill \mbox{\textit{Edexcel M5 2013 Q4 [13]}}