Challenging +1.2 This is a standard moment of inertia calculation for a lamina using integration. While it requires setting up the integral correctly (using horizontal strips dy or vertical strips with the parallel axis theorem) and involves some algebraic manipulation, it follows a well-established procedure taught in Further Maths M5. The region is straightforward to visualize and the integration, though requiring care, is routine for this level.
4. A uniform lamina of mass \(M \mathrm {~kg}\) is modelled as the region which is bounded by the curve with equation \(y = x ^ { 2 }\), the positive \(x\)-axis and the line with equation \(x = 2\). The unit of length on both axes is the metre. Find the moment of inertia of the lamina about the \(x\)-axis.
(6)
M1 for summing MI's from \(x=0\) to \(x=2\); A1 for answer
Total: [6]
## Question 4:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_0^2 x^2\,\mathrm{d}x = \frac{8}{3}$ | M1 | M1 for finding area of lamina |
| $\rho = \frac{3M}{8}$ | A1 | A1 for correct mass per unit area |
| $\delta m = y\rho\,\delta x = \frac{3Mx^2\,\delta x}{8}$ | M1 | M1 for expression for mass of strip in terms of $M$ and $x$ only |
| $\delta I = \frac{1}{3}\delta m\, y^2 = \frac{M}{8}x^6\,\delta x$ | M1 | M1 for correct expression for MI of strip in terms of $M$ and $x$ only |
| $I = \frac{M}{8}\int_0^2 x^6\,\mathrm{d}x = \frac{16M}{7}\ \text{kg m}^2$ | M1 A1 | M1 for summing MI's from $x=0$ to $x=2$; A1 for answer |
**Total: [6]**
4. A uniform lamina of mass $M \mathrm {~kg}$ is modelled as the region which is bounded by the curve with equation $y = x ^ { 2 }$, the positive $x$-axis and the line with equation $x = 2$. The unit of length on both axes is the metre. Find the moment of inertia of the lamina about the $x$-axis.\\
(6)\\
\hfill \mbox{\textit{Edexcel M5 2018 Q4 [6]}}