M5 June 2013 Q5
5.
A uniform triangular lamina \(ABC\), of mass \(M\), has \(AB = AC\) and \(BC = 2a\). The mid-point of \(BC\) is \(D\) and \(AD = h\), as shown in Figure 1.
Show, using integration, that the moment of inertia of the lamina about an axis through \(A\), perpendicular to the plane of the lamina, is
\[\frac{M}{6}(a^2 + 3h^2)\][You may assume without proof that the moment of inertia of a uniform rod, of length 2l and mass m, about an axis through its midpoint and perpendicular to the rod, is \(\tfrac{1}{3}ml^2\).] (10)
| Scheme | Marks |
|---|---|
| Taking strips parallel to \(BC\): | |
| \(l_x = \dfrac{2ax}{h}\) | M1 A1 |
| \(\delta m = \dfrac{2ax}{h}.\dfrac{M\delta x}{ah} = \dfrac{2Mx\delta x}{h^2}\) | M1 A1 |
| \(\delta I = \dfrac{1}{3}\delta m\left(\dfrac{ax}{h}\right)^2 + \delta m.x^2\) | M1 A1A1 |
| \(= \dfrac{2M}{3h^4}(a^2 + 3h^2)x^3\delta x\) | A1 |
| \(I = \dfrac{2M}{3h^4}(a^2 + 3h^2)\displaystyle\int_0^h x^3\,\mathrm{d}x\) | DM1 |
| \(= \dfrac{2M}{3h^4}(a^2 + 3h^2)\left[\dfrac{x^4}{4}\right]_0^h\) | |
| \(= \dfrac{M}{6}(a^2 + 3h^2)\) ** | A1 |
| (10 marks) |
Notes
First M1 for attempt to find length of strip in terms of \(x\), \(h\) and \(a\), using similar triangles or equivalent (must be dim correct)
First A1 for a correct expression
Second M1 for attempt to find mass of strip in terms of \(x\), \(h\), \(M\) and \(\delta x\) (must be dim correct) (this mark is not available if \(\rho\) is not found)
Second A1 for a correct expression.
Third M1 for use of parallel axes rule on strip
Third A1 for \(\tfrac{1}{12}\delta m l_x^2\) term
Fourth A1 for \(\delta m x^2\) term
Fifth A1 for a correct expression in \(a\), \(x\), \(h\), \(M\) and \(\delta x\)
Fourth M1 dependent on Third M1, for integrating their \(\delta I\) (which must have a \(\delta x\))
Sixth A1 for PRINTED ANSWER
N.B. Third M1 They may use perpendicular axes on whole triangle, with same working.
(N.B. if \(\rho\) is used but never eliminated, can score max M1A1M0A0M1A1A0M1A0)