OCR M4 2004 January — Question 4 7 marks

Exam BoardOCR
ModuleM4 (Mechanics 4)
Year2004
SessionJanuary
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments of inertia
TypeProve MI by integration
DifficultyChallenging +1.2 This is a standard M4 moment of inertia calculation requiring integration with a straightforward parabolic boundary. While it involves multiple steps (finding mass per unit area, setting up the integral with y² element, integrating x⁴/a², and simplifying), the technique is routine for this module and follows a well-practiced formula. The algebra is manageable and the result is given to verify, making it moderately above average but not requiring novel insight.
Spec6.04d Integration: for centre of mass of laminas/solids

4 The region between the curve \(y = \frac { x ^ { 2 } } { a }\) and the \(x\)-axis for \(0 \leqslant x \leqslant a\) is occupied by a uniform lamina with mass \(m\). Show that the moment of inertia of this lamina about the \(x\)-axis is \(\frac { 1 } { 7 } m a ^ { 2 }\).

4 The region between the curve $y = \frac { x ^ { 2 } } { a }$ and the $x$-axis for $0 \leqslant x \leqslant a$ is occupied by a uniform lamina with mass $m$. Show that the moment of inertia of this lamina about the $x$-axis is $\frac { 1 } { 7 } m a ^ { 2 }$.

\hfill \mbox{\textit{OCR M4 2004 Q4 [7]}}