| Exam Board | SPS |
|---|---|
| Module | SPS FM Mechanics (SPS FM Mechanics) |
| Year | 2021 |
| Session | January |
| Marks | 11 |
| Topic | Centre of Mass 2 |
| Type | Centre of mass of lamina by integration |
| Difficulty | Challenging +1.2 This is a standard Further Maths mechanics question on centres of mass requiring integration to find ȳ (with area given), then applying equilibrium conditions for a suspended lamina. Part (a) involves routine integration of a polynomial and division by area. Part (b) requires setting up moments about the suspension point using geometry and trigonometry. While multi-step, these are well-practiced techniques with no novel insight required, placing it moderately above average difficulty. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids |
Numerical (calculator) integration is not acceptable in this question.
\includegraphics{figure_4}
The shaded region $OAB$ in Figure 2 is bounded by the $x$-axis, the line with equation $x = 4$ and the curve with equation $y = \frac{1}{4}(x-2)^3 + 2$. The point $A$ has coordinates $(4, 4)$ and the point $B$ has coordinates $(4, 0)$.
A uniform lamina $L$ has the shape of $OAB$. The unit of length on both axes is one centimetre. The centre of mass of $L$ is at the point with coordinates $(\bar{x}, \bar{y})$.
Given that the area of $L$ is $8$cm²,
\begin{enumerate}[label=(\alph*)]
\item show that $\bar{y} = \frac{8}{7}$.
[4]
\item The lamina is freely suspended from $A$ and hangs in equilibrium with $AB$ at an angle $\theta°$ to the downward vertical.
Find the value of $\theta$.
[7]
\end{enumerate}
\hfill \mbox{\textit{SPS SPS FM Mechanics 2021 Q6 [11]}}