# Question 6:

## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I_{\text{rod}} = \frac{1}{3}(2m)a^2 + (2m)a^2\ \left(= \frac{8}{3}ma^2\right)$ | B1 | |
| $I_{\text{particle}} = m(2a)^2$ | B1 | |
| $I = 4ma^2 + \frac{8}{3}ma^2\ \left(= \frac{20}{3}ma^2\right)$ | B1 **[3]** | |

## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}\left(\frac{20}{3}ma^2\right)\omega^2 = 4mga(1 - \cos\theta)$ | M1 | Equation involving KE (must involve $I$) and PE (two terms) |
| | A1 A1 | A1 for KE term, A1 for PE term |
| $\omega^2 = \dfrac{6g}{5a}(1 - \cos\theta)$ | A1 **[4]** | **AG** Correctly obtained. If M0 then SC B1 for either KE or PE term correct |

## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2\omega\alpha = \dfrac{6g}{5a}(\sin\theta)\omega$ | M1 | Differentiating $\omega$ with respect to $t$ or for applying $C = I\alpha$ |
| $\alpha = \dfrac{3g}{5a}\sin\theta$ | A1 **[2]** | |

## Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Centre of mass of pendulum is $\dfrac{4a}{3}$ from $A$ | B1 | |
| $Y + 3mg\cos\theta = 3mr\omega^2$ | *M1 | For radial acceleration $r\omega^2$ – must sub for $\omega^2$ - allow incorrect $m$ and $r$ for M mark only |
| $Y = 4ma\left\{\dfrac{6g}{5a}(1-\cos\theta)\right\} - 3mg\cos\theta$ | A1 | $Y = \pm\dfrac{9}{10}mg$ |
| $3mg\sin\theta - X = 3mr\alpha$ | *M1 | For transverse acceleration $r\alpha$ - must sub their $\alpha$ - allow incorrect $m$ and $r$ for M mark only |
| $X = 3mg\sin\theta - 4ma\left(\dfrac{3g}{5a}\sin\theta\right)$ | A1 | $X = \pm\dfrac{3\sqrt{3}}{10}mg\ (\pm 0.519615\ldots mg)$ |
| $F = \sqrt{X^2 + Y^2} = mg\sqrt{\dfrac{81}{100} + \dfrac{27}{100}}$ | M1 dep* | Substituting $\theta = \dfrac{\pi}{3}$ into $X$ and $Y$ and applying formula for $F$. Must be using correct $m$ and $r$ |
| $F = \frac{3}{5}mg\sqrt{3}$ | A1 **[7]** | $1.039230\ldots mg$ |

## Part (v):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\omega^2 = \dfrac{6g}{5a}(1-\cos\theta) \Rightarrow \dfrac{d\theta}{dt} = 4(1-\cos\theta)^{\frac{1}{2}} \Rightarrow \ldots$ | M1 | Re-writing $\omega$ as $\dfrac{d\theta}{dt}$ (maybe implied) and attempt to set up integral by separating variables |
| $4t = \displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \dfrac{d\theta}{(1-\cos\theta)^{\frac{1}{2}}}$ | A1 | Condone lack of limits on integral (may still be in terms of $a$) |
| Re-write $(1-\cos\theta)^{\frac{1}{2}}$ as $\sqrt{2}\sin\left(\dfrac{\theta}{2}\right)$ | B1 | Applying the trigonometric identity $\sin^2 X = \frac{1}{2}(1 - \cos 2X)$ |
| Leading to $\dfrac{1}{4\sqrt{2}}\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \text{cosec}\left(\dfrac{\theta}{2}\right)\,d\theta$ | A1 | $k = \dfrac{1}{4\sqrt{2}}$ (= 0.1767766…) – condone lack of limits on integral; oe for $k$ e.g. $\sqrt{\dfrac{5a}{12g}}$ |
| $= \dfrac{1}{4\sqrt{2}}\left[2\ln\left\lvert\tan\left(\dfrac{\theta}{4}\right)\right\rvert\right]_{\frac{\pi}{6}}^{\frac{\pi}{3}}$ | B1 | Using the result $\int \text{cosec}\left(\dfrac{x}{2}\right)dx = 2\ln\left\lvert\tan\dfrac{1}{4}x\right\rvert (+c)$ or $-2\ln\left\lvert\text{cosec}\dfrac{1}{2}x + \cot\dfrac{1}{2}x\right\rvert (+c)$ |
| $= 0.251$ (3sf) | A1 **[6]** | 0.2512461… |

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