| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2020 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Lamina with hole removed |
| Difficulty | Challenging +1.8 This is a challenging Further Maths mechanics question requiring: (a) symmetry argument, (b) proof using integration of a segment's centre of mass (non-standard formula), (c) application using composite bodies with hole removal, and (d) moments equilibrium. The proof in part (b) and the geometric setup in parts (c-d) require significant problem-solving beyond routine exercises, though the techniques are standard for FM students. |
| Spec | 6.04a Centre of mass: gravitational effect6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (a) | The central radius is a line of symmetry of the shape. |
| [1] | 2.4 | Allow equal area each side |
| 7 | (b) | Area/mass of sector 1r2×2θ and CoM of sector at |
| Answer | Marks |
|---|---|
| 3(θ−sinθcosθ) | B1 |
| Answer | Marks |
|---|---|
| [4] | 1.2 |
| Answer | Marks |
|---|---|
| 1.1 | Must be used |
| Answer | Marks |
|---|---|
| must be seen. | Must see change from double |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (c) | Sector angle is 2cos−12 or θ=cos−12 |
| Answer | Marks |
|---|---|
| = –0.0988 (3 sf) | B1 |
| Answer | Marks |
|---|---|
| [5] | 2.2a |
| Answer | Marks |
|---|---|
| 1.1 | 1.682... |
| Answer | Marks |
|---|---|
| Their 3.097, 2.406 and 75.44 | 0.841… |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (d) | –0.0988M + 5m = 0 |
| So the mass of the component is 50.6m kg | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | May see Mg and mg |
Question 7:
7 | (a) | The central radius is a line of symmetry of the shape. | B1
[1] | 2.4 | Allow equal area each side
7 | (b) | Area/mass of sector 1r2×2θ and CoM of sector at
2
2rsinθ
from centre used
3θ
Area/mass of triangle (−)1r2sin(2θ) and CoM of
2
triangle at 2rcosθ from centre used
3
2rsinθ
1r2×2θ× −1r2sin2θ×2rcosθ
2 3θ 2 3
1r2×2θ−1r2sin2θ
2 2
4rsinθ−2rsin2θcosθ
=
3(2θ−sin2θ)
2r(sinθ−sinθcos2θ) 2rsinθ(1−cos2θ)
= =
3(θ−sinθcosθ) 3(θ−sinθcosθ)
2rsin3θ
=
3(θ−sinθcosθ) | B1
B1
M1
A1
[4] | 1.2
1.2
3.1b
1.1 | Must be used
Σmx
Using x = i i with sector
Σm
i
and triangle of –ve mass, oe
AG
At least one intermediate step
must be seen. | Must see change from double
angle
7 | (c) | Sector angle is 2cos−12 or θ=cos−12
3 3
Area/Mass of component:
π×52−1×32(1.682−sin1.682)
2
= 75.44
(−3.097...×2.406...)
y =
75.44...
= –0.0988 (3 sf) | B1
M1
A1
M1
A1
[5] | 2.2a
1.1
1.1
3.1b
1.1 | 1.682...
Attempting to find mass of
component using ‘negative’
mass
Their 3.097, 2.406 and 75.44 | 0.841…
Allow use of 0.841
If θ=2cos−12 used this gives
3
1.095... instead of 2.406...
7 | (d) | –0.0988M + 5m = 0
So the mass of the component is 50.6m kg | M1
A1
[2] | 1.1
1.1 | May see Mg and mg7 Fig. 7.1 shows a uniform lamina in the shape of a sector of a circle of radius $r$ and angle $2 \theta$ where $\theta$ is in radians. The sector consists of a triangle $O A B$ and a segment bounded by the chord $A B$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{831ba5da-df19-43bb-b163-02bbddb4e2b8-6_364_556_342_246}
\captionsetup{labelformat=empty}
\caption{Fig. 7.1}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Explain why the centre of mass of the segment lies on the radius through the midpoint of $A B$.
\item Show that the distance of the centre of mass of the segment from $O$ is $\frac { 2 r \sin ^ { 3 } \theta } { 3 ( \theta - \sin \theta \cos \theta ) }$.
A uniform circular lamina of radius 5 units is placed with its centre at the origin, $O$, of an $x - y$ coordinate system. A component for a machine is made by removing and discarding a segment from the lamina. The radius of the circle from which the segment is formed is 3 units and the centre of this circle is $O$. The centre of the straight edge of the segment has coordinates ( 0,2 ) and this edge is perpendicular to the $y$-axis (see Fig. 7.2).
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{831ba5da-df19-43bb-b163-02bbddb4e2b8-6_766_757_1400_244}
\captionsetup{labelformat=empty}
\caption{Fig. 7.2}
\end{center}
\end{figure}
\item Find the $y$-coordinate of the centre of mass of the component, giving your answer correct to 3 significant figures.\\
$C$ is the point on the component with coordinates ( 0,5 ). The component is now placed horizontally and supported only at $O$. A particle of mass $m \mathrm {~kg}$ is placed on the component at $C$ and the component and particle are in equilibrium.
\item Find the mass of the component in terms of $m$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2020 Q7 [12]}}