Edexcel M5 2014 June — Question 4 8 marks

Exam BoardEdexcel
ModuleM5 (Mechanics 5)
Year2014
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments of inertia
TypeProve MI by integration
DifficultyChallenging +1.8 This is a challenging M5 question requiring 3D integration using the perpendicular axis theorem for discs. While the disc MOI is given, students must set up the integral correctly (slicing the sphere, expressing mass elements in terms of density and volume, handling the variable radius as a function of position), then integrate. This goes beyond routine mechanics and requires spatial reasoning and careful calculus, making it significantly harder than average A-level questions.
Spec6.04b Find centre of mass: using symmetry
4. A uniform solid sphere has mass \(M\) and radius \(a\). Prove, using integration, that the moment of inertia of the sphere about a diameter is \(\frac { 2 M a ^ { 2 } } { 5 }\) [0pt] [You may assume without proof that the moment of inertia of a uniform circular disc, of mass \(m\) and radius \(r\), about an axis through its centre and perpendicular to its plane is \(\frac { 1 } { 2 } m r ^ { 2 }\).]
Question 4:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Consider sphere as stack of discs; disc at height \(x\) from centre has radius \(r = \sqrt{a^2 - x^2}\)M1 Setting up integration with disc element
Mass of disc: \(dm = \rho \pi r^2\, dx = \rho\pi(a^2-x^2)\,dx\)B1 Correct element of mass
Total mass: \(M = \rho \cdot \frac{4}{3}\pi a^3\), so \(\rho = \frac{3M}{4\pi a^3}\)B1 Correct density
MI of disc about diameter (axis through centre of sphere): \(\frac{1}{2}dm\cdot r^2 + dm\cdot x^2\) by parallel axisM1 Using parallel axis theorem for each disc
\(I = \int_{-a}^{a} \left(\frac{1}{2}(a^2-x^2) + x^2\right)\rho\pi(a^2-x^2)\,dx\)M1 Setting up full integral
\(= \rho\pi\int_{-a}^{a}\left(\frac{1}{2}(a^2-x^2)^2 + x^2(a^2-x^2)\right)dx\)A1 Correct integrand
\(= \rho\pi\int_{-a}^{a}\left(\frac{1}{2}a^4 - a^2x^2 + \frac{1}{2}x^4 + a^2x^2 - x^4\right)dx\) Expanding
\(= \rho\pi\int_{-a}^{a}\left(\frac{1}{2}a^4 - \frac{1}{2}x^4\right)dx\)M1 Simplifying integrand
\(= \rho\pi\left[\frac{1}{2}a^4 x - \frac{x^5}{10}\right]_{-a}^{a} = \rho\pi\left(a^5 - \frac{a^5}{5}\right) = \frac{4\rho\pi a^5}{5}\)A1 Correct integration
\(= \frac{4}{5}\cdot\frac{3M}{4\pi a^3}\cdot\pi a^5 = \frac{2Ma^2}{5}\)A1 Completing proof 
# Question 4:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Consider sphere as stack of discs; disc at height $x$ from centre has radius $r = \sqrt{a^2 - x^2}$ | M1 | Setting up integration with disc element |
| Mass of disc: $dm = \rho \pi r^2\, dx = \rho\pi(a^2-x^2)\,dx$ | B1 | Correct element of mass |
| Total mass: $M = \rho \cdot \frac{4}{3}\pi a^3$, so $\rho = \frac{3M}{4\pi a^3}$ | B1 | Correct density |
| MI of disc about diameter (axis through centre of sphere): $\frac{1}{2}dm\cdot r^2 + dm\cdot x^2$ by parallel axis | M1 | Using parallel axis theorem for each disc |
| $I = \int_{-a}^{a} \left(\frac{1}{2}(a^2-x^2) + x^2\right)\rho\pi(a^2-x^2)\,dx$ | M1 | Setting up full integral |
| $= \rho\pi\int_{-a}^{a}\left(\frac{1}{2}(a^2-x^2)^2 + x^2(a^2-x^2)\right)dx$ | A1 | Correct integrand |
| $= \rho\pi\int_{-a}^{a}\left(\frac{1}{2}a^4 - a^2x^2 + \frac{1}{2}x^4 + a^2x^2 - x^4\right)dx$ | | Expanding |
| $= \rho\pi\int_{-a}^{a}\left(\frac{1}{2}a^4 - \frac{1}{2}x^4\right)dx$ | M1 | Simplifying integrand |
| $= \rho\pi\left[\frac{1}{2}a^4 x - \frac{x^5}{10}\right]_{-a}^{a} = \rho\pi\left(a^5 - \frac{a^5}{5}\right) = \frac{4\rho\pi a^5}{5}$ | A1 | Correct integration |
| $= \frac{4}{5}\cdot\frac{3M}{4\pi a^3}\cdot\pi a^5 = \frac{2Ma^2}{5}$ | A1 | Completing proof |
4. A uniform solid sphere has mass $M$ and radius $a$. Prove, using integration, that the moment of inertia of the sphere about a diameter is $\frac { 2 M a ^ { 2 } } { 5 }$\\[0pt]
[You may assume without proof that the moment of inertia of a uniform circular disc, of mass $m$ and radius $r$, about an axis through its centre and perpendicular to its plane is $\frac { 1 } { 2 } m r ^ { 2 }$.]\\

\hfill \mbox{\textit{Edexcel M5 2014 Q4 [8]}}
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