A2 June 2023 Q3
3. [In this question you may quote, without proof, the formula for the distance of the centre of mass of a uniform circular arc from its centre.]
Five pieces of a uniform wire are joined together to form the rigid framework \(OABCO\) shown in Figure 1, where
- \(OA\), \(OB\) and \(BC\) are straight, with \(OA = OB = BC = r\)
- arc \(AB\) is one quarter of a circle with centre \(O\) and radius \(r\)
- arc \(OC\) is one quarter of a circle of radius \(r\)
- all five pieces of wire lie in the same plane
Given that the distance of the centre of mass of the framework from \(OA\) is \(d\),
The framework is freely pivoted at \(A\).
The framework is held in equilibrium, with \(AO\) vertical, by a horizontal force of magnitude \(F\) which is applied to the framework at \(C\).
Given that the weight of the framework is \(W\)
| Scheme | Marks | AO |
|---|---|---|
| \(\alpha = \dfrac{\pi}{4} \Rightarrow \dfrac{r\sin\alpha}{\alpha} = r \times \dfrac{1}{\sqrt{2}} \times \dfrac{4}{\pi}\ \left(= \dfrac{2\sqrt{2}r}{\pi}\right)\) | B1 | 1.1b |
| Distance from \(OA = \dfrac{2\sqrt{2}r}{\pi} \times \cos\dfrac{\pi}{4} = \dfrac{2\sqrt{2}r}{\pi \times \sqrt{2}} = \dfrac{2r}{\pi}\) * | B1* | 2.2a |
| (2) |
Notes
B1: Correct use of given formula. Seen or implied.
B1*: Obtain given result from correct working, e.g. by use of trigonometry or use of Pythagoras.
| Scheme | Marks | AO |
|---|---|---|
| Moments about \(OA\): | M1 | 3.1a |
| \(r \times \dfrac{r}{2} + r \times r + \dfrac{2r}{\pi} \times \dfrac{\pi r}{2} + \dfrac{2r}{\pi} \times \dfrac{\pi r}{2} = (3r + \pi r)d\) | A1 A1 | 1.1b 1.1b |
| \(\left(\dfrac{7r^2}{2} = (3r + \pi r)d\right) \Rightarrow d = \dfrac{7r}{2(3 + \pi)}\) * | A1* | 2.2a |
| (4) |
Notes
M1: Condone a dimension error in the arc length, but otherwise must be a dimensionally correct equation. Need all terms. Allow use of a parallel axis.
A1: Unsimplified equation with at most one error.
A1: Correct unsimplified equation.
An error in the arc length affects 3 terms – count it as a single error.
A1*: Obtain given answer from correct working.
| Scheme | Marks | AO |
|---|---|---|
| Moments about \(A\) or any other complete method to obtain \(F\) in terms of \(W\) | M1 | 3.1a |
| \(W \times \dfrac{7r}{2(3 + \pi)} = F \times 2r\) | A1 | 1.1b |
| \(F = \dfrac{7}{4(3 + \pi)}W\) | A1 | 1.1b |
| (3) | ||
| (9 marks) |
Notes
M1: Dimensionally correct equation with distances perpendicular to forces.
A1: Correct unsimplified equation.
A1: Any equivalent form. \(0.28W\) or better \((0.28494...W)\)