Edexcel M5 2008 June — Question 6 10 marks

Exam BoardEdexcel
ModuleM5 (Mechanics 5)
Year2008
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments of inertia
TypeProve MI by integration
DifficultyChallenging +1.8 This is a Further Maths M5 question requiring integration to find moment of inertia using the perpendicular axis theorem and parallel axis theorem. While the disc result is given, students must set up the integral correctly, apply the parallel axis theorem to each disc element at distance x from the base, and integrate over the cylinder's height. This requires solid understanding of multiple concepts and careful algebraic manipulation, making it significantly harder than average A-level questions but standard for FM mechanics.
Spec6.04d Integration: for centre of mass of laminas/solids

6. A uniform solid right circular cylinder has mass \(M\), height \(h\) and radius \(a\). Find, using integration, its moment of inertia about a diameter of one of its circular ends.
[0pt] [You may assume without proof that the moment of inertia of a uniform circular disc, of mass \(m\) and radius \(a\), about a diameter is \(\frac { 1 } { 4 } m a ^ { 2 }\).]

Question 6:
AnswerMarks Guidance
WorkingMarks Notes
\(\delta m = \pi a^2 \delta x \cdot \dfrac{M}{\pi a^2 h} = \dfrac{M\,\delta x}{h}\)M1 A1
\(\delta I = \frac{1}{4}\delta m \cdot a^2 + \delta m \cdot x^2\)M1 A1
\(= \dfrac{M}{4h}(a^2 + 4x^2)\delta x\)M1 A1
\(I = \displaystyle\int_0^h \dfrac{M}{4h}(a^2 + 4x^2)\,dx\)M1 A1
\(= \dfrac{M}{4h}\!\left[a^2 x + \dfrac{4}{3}x^3\right]_0^h\)M1
\(= \dfrac{M}{4}(a^2 + \tfrac{4}{3}h^2)\)
\(= \dfrac{M}{12}(3a^2 + 4h^2)\)A1 Total: 10
## Question 6:

| Working | Marks | Notes |
|---|---|---|
| $\delta m = \pi a^2 \delta x \cdot \dfrac{M}{\pi a^2 h} = \dfrac{M\,\delta x}{h}$ | M1 A1 | |
| $\delta I = \frac{1}{4}\delta m \cdot a^2 + \delta m \cdot x^2$ | M1 A1 | |
| $= \dfrac{M}{4h}(a^2 + 4x^2)\delta x$ | M1 A1 | |
| $I = \displaystyle\int_0^h \dfrac{M}{4h}(a^2 + 4x^2)\,dx$ | M1 A1 | |
| $= \dfrac{M}{4h}\!\left[a^2 x + \dfrac{4}{3}x^3\right]_0^h$ | M1 | |
| $= \dfrac{M}{4}(a^2 + \tfrac{4}{3}h^2)$ | | |
| $= \dfrac{M}{12}(3a^2 + 4h^2)$ | A1 | **Total: 10** |

---
6. A uniform solid right circular cylinder has mass $M$, height $h$ and radius $a$. Find, using integration, its moment of inertia about a diameter of one of its circular ends.\\[0pt]
[You may assume without proof that the moment of inertia of a uniform circular disc, of mass $m$ and radius $a$, about a diameter is $\frac { 1 } { 4 } m a ^ { 2 }$.]\\

\hfill \mbox{\textit{Edexcel M5 2008 Q6 [10]}}