| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Lamina with removed circle/semicircle |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass problem requiring composite shapes (difference of semicircles), followed by routine equilibrium with moments and force resolution. The centre of mass calculation is bookwork with given answer, and the equilibrium parts use standard techniques. Slightly above average due to the composite shape and three-part structure, but all methods are textbook standard. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| CoM of one semicircular lamina is \(\frac{4a}{\pi}\), \(\frac{4a}{3\pi}\) | B1 | oe, may be unsimplified, allow \(^{4r}/_{3\pi}\) |
| \(-\left(\frac{1}{2}\pi a^2\right)\left(\frac{4a}{3\pi}\right) + \left(\frac{1}{2}\pi(3a)^2\right)\left(\frac{12a}{3\pi}\right)\) | M1 | Table of values idea |
| A1 | ||
| \(= \left(\frac{1}{2}\pi(3a)^2 - \frac{1}{2}\pi a^2\right)x_G\) | A1 | |
| \(x_G = \frac{13a}{3\pi}\) | A1 [5] | AG Correctly shown |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | Moments equation in terms of \(a\) and \(T\), with correct number of terms, dimensionally correct, resolving on \(T\) side of equation; if conflicting evidence about point taking moments mark to benefit of candidate. | |
| \(3g\left(\frac{13a}{3\pi}\right) = (T \cos 40)(6a)\) | A1 | |
| \(T = 8.82 \text{ N}\) | A1 [3] | \(T = 8.822960...\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(X = T \cos 40\) | B1 | ft candidates value of \(T\) if substituted (\(X = 6.758779...\)) |
| \(Y = 3g + T \sin 40\) | B1 | ft candidates value of \(T\) if substituted (\(Y = 35.071289...\)) |
| \(\tan \theta = \frac{35.0712...}{6.7587...}\) | M1 | Any relevant angle |
| \(\theta = 79.1\) above horizontal (to the left) | A1 | \(\theta = 79.0919...\) (10.9 to the upward vertical); above or upwards may be implied by a correct diagram. |
| OR | ||
| \(x^2 = (3g)^2 + T^2 - 2 \times 3g \times T \times \cos 130\) | [4] | |
| Use of sine rule to find either of the missing angles | M1 A1 ft | |
| \(\theta = 79.1\) above horizontal (to the left) | M1 A1 | \(\theta = 79.0919...\) (10.9 to the upward vertical); above or upwards may be implied by a correct diagram. |
## (i)
CoM of one semicircular lamina is $\frac{4a}{\pi}$, $\frac{4a}{3\pi}$ | B1 | oe, may be unsimplified, allow $^{4r}/_{3\pi}$
$-\left(\frac{1}{2}\pi a^2\right)\left(\frac{4a}{3\pi}\right) + \left(\frac{1}{2}\pi(3a)^2\right)\left(\frac{12a}{3\pi}\right)$ | M1 | Table of values idea
| A1 |
$= \left(\frac{1}{2}\pi(3a)^2 - \frac{1}{2}\pi a^2\right)x_G$ | A1 |
$x_G = \frac{13a}{3\pi}$ | A1 [5] | AG Correctly shown
## (ii)
| M1 | Moments equation in terms of $a$ and $T$, with correct number of terms, dimensionally correct, resolving on $T$ side of equation; if conflicting evidence about point taking moments mark to benefit of candidate.
$3g\left(\frac{13a}{3\pi}\right) = (T \cos 40)(6a)$ | A1 |
$T = 8.82 \text{ N}$ | A1 [3] | $T = 8.822960...$
## (iii)
$X = T \cos 40$ | B1 | ft candidates value of $T$ if substituted ($X = 6.758779...$)
$Y = 3g + T \sin 40$ | B1 | ft candidates value of $T$ if substituted ($Y = 35.071289...$)
$\tan \theta = \frac{35.0712...}{6.7587...}$ | M1 | Any relevant angle
$\theta = 79.1$ above horizontal (to the left) | A1 | $\theta = 79.0919...$ (10.9 to the **upward vertical**); above or upwards may be implied by a correct diagram.
OR | |
$x^2 = (3g)^2 + T^2 - 2 \times 3g \times T \times \cos 130$ | [4] |
Use of sine rule to find either of the missing angles | M1 A1 ft |
$\theta = 79.1$ above horizontal (to the left) | M1 A1 | $\theta = 79.0919...$ (10.9 to the **upward vertical**); above or upwards may be implied by a correct diagram.
---\includegraphics{figure_1}
A uniform lamina $ABDC$ is bounded by two semicircular arcs $AB$ and $CD$, each with centre $O$ and of radii $3a$ and $a$ respectively, and two straight edges, $AC$ and $DB$, which lie on the line $AOB$ (see Fig. 1).
\begin{enumerate}[label=(\roman*)]
\item Show that the distance of the centre of mass of the lamina from $O$ is $\frac{13a}{3\pi}$. [5]
\end{enumerate}
\includegraphics{figure_2}
The lamina has mass 3 kg and is freely pivoted to a fixed point at $A$. The lamina is held in equilibrium with $AB$ vertical by means of a light string attached to $B$. The string lies in the same plane as the lamina and is at an angle of $40°$ below the horizontal (see Fig. 2).
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Calculate the tension in the string. [3]
\item Find the direction of the force acting on the lamina at $A$. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR M2 2016 Q3 [12]}}