OCR Further Mechanics 2019 June — Question 1 8 marks

Exam BoardOCR
ModuleFurther Mechanics (Further Mechanics)
Year2019
SessionJune
Marks8
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Mark schemeDownload PDF ↗
TopicCentre of Mass 2
TypeCentre of mass of solid of revolution
DifficultyChallenging +1.2 This is a standard Further Maths mechanics question on centre of mass of solids of revolution. Part (a) requires recognizing symmetry (routine), part (b) involves applying the standard formula for x-coordinate of COM (∫x·πy²dx / ∫πy²dx), and part (c) uses equilibrium geometry. The cubic under the square root makes integration slightly tedious but the question follows a predictable template with no novel insight required. Slightly above average difficulty due to being Further Maths content and requiring careful calculation.
Spec6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces
1 The region bounded by the \(x\)-axis, the curve \(\mathrm { y } = \sqrt { 2 x ^ { 3 } - 15 x ^ { 2 } + 36 x - 20 }\) and the lines \(x = 1\) and \(x = 4\) is rotated through \(2 \pi\) radians about the \(x\)-axis to form a uniform solid of revolution \(R\). The centre of mass of \(R\) is the point \(G\) (see diagram). \includegraphics[max width=\textwidth, alt={}, center]{9bc86277-9e6b-41f6-a2c3-94c85e7b1360-2_569_463_507_280}
  1. Explain why the \(y\)-coordinate of \(G\) is 0 .
  2. Find the \(x\)-coordinate of \(G\). \(P\) is a point on the edge of the curved surface of \(R\) where \(x = 4 . R\) is freely suspended from \(P\) and hangs in equilibrium.
  3. Find the angle between the axis of symmetry of \(R\) and the vertical.
Question 1:
AnswerMarks Guidance
1(a) (G lies on the axis of) symmetry (of R)
[1]1.2 Accept ‘symmetrical’
around the axis’
AnswerMarks Guidance
1(b) 4 45
 (2x315x236x20)dx
1 2
4 1179
 (2x415x336x220x)dx
1 20
4
 (2x415x336x220x)dx
x  1
4
 (2x315x236x20)dx
1
1179
131
x  20  or 2.62
45 50
AnswerMarks
2B1
B1
M1
A1
AnswerMarks
[4]1.1a
1.1
1.1
AnswerMarks
1.1BC With or without  soi
BC With or without  soi
Either no  or seen on both top and
AnswerMarks Guidance
bottomIf treats as lamina award 0
1(c) PC 12= 2√3 =3.46…
"√12"
tan𝜃 = = (2.51…)
4−"2.62"
AnswerMarks
1.19 rad or 68.3B1
M1
A1
AnswerMarks
[3]1.1
1.1
AnswerMarks
1.1soi
42.62
Allow (0.398...)
AnswerMarks
12C is the centre of the larger
circle
Can use a clearly stated
value of PC
1< “2.62” <4
Question 1:
1 | (a) | (G lies on the axis of) symmetry (of R) | B1
[1] | 1.2 | Accept ‘symmetrical’ | Not ‘distributed uniformly
around the axis’
1 | (b) | 4 45
 (2x315x236x20)dx
1 2
4 1179
 (2x415x336x220x)dx
1 20
4
 (2x415x336x220x)dx
x  1
4
 (2x315x236x20)dx
1
1179
131
x  20  or 2.62
45 50
2 | B1
B1
M1
A1
[4] | 1.1a
1.1
1.1
1.1 | BC With or without  soi
BC With or without  soi
Either no  or seen on both top and
bottom | If treats as lamina award 0
1 | (c) | PC 12= 2√3 =3.46…
"√12"
tan𝜃 = = (2.51…)
4−"2.62"
1.19 rad or 68.3 | B1
M1
A1
[3] | 1.1
1.1
1.1 | soi
42.62
Allow (0.398...)
12 | C is the centre of the larger
circle
Can use a clearly stated
value of PC
1< “2.62” <4
1 The region bounded by the $x$-axis, the curve $\mathrm { y } = \sqrt { 2 x ^ { 3 } - 15 x ^ { 2 } + 36 x - 20 }$ and the lines $x = 1$ and $x = 4$ is rotated through $2 \pi$ radians about the $x$-axis to form a uniform solid of revolution $R$. The centre of mass of $R$ is the point $G$ (see diagram).\\
\includegraphics[max width=\textwidth, alt={}, center]{9bc86277-9e6b-41f6-a2c3-94c85e7b1360-2_569_463_507_280}
\begin{enumerate}[label=(\alph*)]
\item Explain why the $y$-coordinate of $G$ is 0 .
\item Find the $x$-coordinate of $G$.\\
$P$ is a point on the edge of the curved surface of $R$ where $x = 4 . R$ is freely suspended from $P$ and hangs in equilibrium.
\item Find the angle between the axis of symmetry of $R$ and the vertical.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Mechanics 2019 Q1 [8]}}
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