Challenging +1.2 This is a standard rotational dynamics problem requiring energy conservation (PE to KE) with given moment of inertia. Students must identify the geometry (center of mass location for equilateral triangle), apply conservation of energy, and relate angular to linear velocity. While it involves multiple steps and careful geometry, it follows a well-established method taught in M5 with no novel insight required—moderately above average difficulty.
2. A uniform equilateral triangular lamina \(A B C\) has mass \(m\) and sides of length \(\sqrt { } 3 a\). The lamina is free to rotate in a vertical plane about a fixed smooth horizontal axis \(L\), which passes through \(A\) and is perpendicular to the lamina. The midpoint of \(B C\) is \(D\). The lamina is held with \(A D\) making an angle of \(60 ^ { \circ }\) with the upward vertical through \(A\) and released from rest. The moment of inertia of the lamina about the axis \(L\) is \(\frac { 5 m a ^ { 2 } } { 4 }\)
Find the speed of \(D\) when \(A D\) is vertical.
(8)
Speed of \(D = \omega \times \sqrt{3}a = \sqrt{3}a\sqrt{\frac{4\sqrt{3}g}{15a}}\)
M1
\(v = \omega \times AD\)
\(v = \sqrt{\frac{4\sqrt{3}ga}{5}}\)
A1
Accept equivalent exact forms
# Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $AD = \sqrt{3}a$ (height of equilateral triangle, from $A$ to midpoint $D$ of $BC$) | B1 | Correct length of $AD$ |
| Centre of mass of triangle is at $\frac{1}{3} \times \sqrt{3}a = \frac{\sqrt{3}a}{3}$ from $A$ along $AD$ | B1 | Position of centre of mass from $A$ |
| Initial position: $AD$ at $60°$ to upward vertical, so height of $G$ above $A$: $\frac{\sqrt{3}a}{3}\cos60° = \frac{\sqrt{3}a}{6}$ | M1 A1 | Height of centre of mass in initial position |
| When $AD$ vertical, height of $G$ above $A$: $\frac{\sqrt{3}a}{3}$ | M1 | Height of centre of mass in final position |
| Drop in height of $G$: $\frac{\sqrt{3}a}{3} - \frac{\sqrt{3}a}{6} = \frac{\sqrt{3}a}{6}$ | A1 | Correct change in height |
| Energy equation: $\frac{1}{2} \cdot \frac{5ma^2}{4} \cdot \omega^2 = mg \cdot \frac{\sqrt{3}a}{6}$ | M1 | Applying conservation of energy |
| $\omega^2 = \frac{4\sqrt{3}g}{15a}$ | A1 | |
| Speed of $D = \omega \times \sqrt{3}a = \sqrt{3}a\sqrt{\frac{4\sqrt{3}g}{15a}}$ | M1 | $v = \omega \times AD$ |
| $v = \sqrt{\frac{4\sqrt{3}ga}{5}}$ | A1 | Accept equivalent exact forms |
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2. A uniform equilateral triangular lamina $A B C$ has mass $m$ and sides of length $\sqrt { } 3 a$. The lamina is free to rotate in a vertical plane about a fixed smooth horizontal axis $L$, which passes through $A$ and is perpendicular to the lamina. The midpoint of $B C$ is $D$. The lamina is held with $A D$ making an angle of $60 ^ { \circ }$ with the upward vertical through $A$ and released from rest. The moment of inertia of the lamina about the axis $L$ is $\frac { 5 m a ^ { 2 } } { 4 }$
Find the speed of $D$ when $A D$ is vertical.\\
(8)\\
\hfill \mbox{\textit{Edexcel M5 2014 Q2 [8]}}