OCR MEI Further Mechanics Major 2022 June — Question 9 11 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Major (Further Mechanics Major)
Year2022
SessionJune
Marks11
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Mark schemeDownload PDF ↗
TopicCentre of Mass 2
TypeCentre of mass of solid of revolution
DifficultyChallenging +1.8 This is a challenging Further Maths mechanics problem requiring: (1) decomposition of a solid of revolution into cone minus curved solid, (2) application of Pappus's theorem or direct integration for centre of mass with exponential functions, (3) geometric reasoning about equilibrium when suspended. The 8-mark part (a) involves substantial algebraic manipulation to reach the given answer, and part (b) requires understanding of suspended equilibrium. While the techniques are standard for Further Maths, the execution demands careful multi-step work with non-trivial algebra.
Spec6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces
[In this question you may use the facts that for a uniform solid right circular cone of height \(h\) and base radius \(r\) the volume is \(\frac{1}{3}\pi r^2 h\) and the centre of mass is \(\frac{1}{4}h\) above the base on the line from the centre of the base to the vertex.] \includegraphics{figure_9} The diagram shows the shaded region S bounded by the curve \(y = e^{\frac{1}{4}x}\) for \(0 \leq x \leq 2\), the \(x\)-axis, the \(y\)-axis, and the line \(y = \frac{1}{4}e(6-x)\). The line \(y = \frac{1}{4}e(6-x)\) meets the curve \(y = e^{\frac{1}{4}x}\) at the point A with coordinates \((2, e)\). The region S is rotated through \(2\pi\) radians about the \(x\)-axis to form a uniform solid of revolution T.
  1. Show that the \(x\)-coordinate of the centre of mass of T is \(\frac{3(5e^2 + 1)}{7e^2 - 3}\). [8]
Solid T is freely suspended from A and hangs in equilibrium.
  1. Determine the angle between AO, where O is the origin, and the vertical. [3]
Question 9:
AnswerMarks Guidance
9(a) V =π ⌠  2( e 1 2 x )2 dx
1 ⌡
0
2 ( )
=πex =π e2 −1
 
0
V x =π ⌠  2 x ( e 1 2 x )2 dx=...
1 ⌡
0
2 2 2
V x =πxex −π∫ exdx=πxex −ex
1    
0 0 0
( )
V x =π e2 +1
1
V x +3   1 π(e)2(4)   =x   1 π(e)2(4)+π ( e2 −1 ) 
1 T
3  3 
( ) ( )
e2 +1 π+4πe2 =x 4πe2 +πe2 −π
T 3
( )
3 5e2 +1
( )
x 7e2 −1 =5e2 +1⇒x =
AnswerMarks
T 3 T 7e2 −3M1*
A1
M1*
A1
A1
M1dep*
A1ft
A1
AnswerMarks
[8]2.1
1.1
1.1
1.1
1.1
3.1a
1.1
AnswerMarks
2.2aCorrect integral representation for the volume V of
1
the curve y=e 1 2 x between 0 and 2
Correct integral representation for V x and attempt to
1
integrate by parts
Correct integration – limits not required for this mark
e2 +1
Implied by x =
e2 −1
Table of values idea – correct number of terms
Correct equation for the centre of mass of T with
their V x and correct V
1 1
AG – sufficient working must be shown – condone
absence of πthroughout for full marks
AnswerMarks
(b) 3 ( 5e2 +1 ) 
 −2 
 7e2 −3 
 
tanθ = (tanθ =0.1237...⇒θ =7.0541...)
1 1 1
e
2
tanθ = (tanθ =0.7357...⇒θ =36.3441...)
2 2 1
e
θ=θ +θ ⇒θ=43.4 
AnswerMarks
1 2B1
B1
B1
AnswerMarks
[3]3.1b
1.1
AnswerMarks
2.2aCorrect expression for either θor tanθwhere θis
1 1 1
the angle between the line through (2, 0) and A and
the line through A and (x ,0)
T
Correct expression for either θ or tanθ where θ is
2 2 2
the angle between the line through (2, 0) and A and
the line OA
43.3982198… or 0.75744182… (in radians) – where
θis the angle between OA and the vertical
Alternative method 1 (cosine rule)
( 4+e2 ) + ( e2 +(2−x)2 ) −x2
cosθ=
AnswerMarks Guidance
2 4+e2 e2 +(2−x)2B2 B2 for correct application of cosine rule with correct
x – for B1 allow one incorrect length
(3.37476...)2 +(2.7390...)2 −(2.33636...)2
cosθ=
2(3.37476...)(2.7390...)
AnswerMarks Guidance
θ=43.4 B1 43.3982198… or 0.75744182… (in radians) – where
θis the angle between OA and the vertical
AnswerMarks Guidance
Alternative method 2 (scalar product)[3]
−2(x −2)+e2 = 4+e2 (x −2)2 +e2 cosθ−2(x −2)+e2 = 4+e2 (x −2)2 +e2 cosθ B2
Or B1 for   =    cosθ
−e  −e  −e  −e 
AnswerMarks Guidance
θ=43.4 B1 If B0 then SC B2 for 136.601780…
[3]
Question 9:
9 | (a) | V =π ⌠  2( e 1 2 x )2 dx
1 ⌡
0
2 ( )
=πex =π e2 −1
 
0
V x =π ⌠  2 x ( e 1 2 x )2 dx=...
1 ⌡
0
2 2 2
V x =πxex −π∫ exdx=πxex −ex
1    
0 0 0
( )
V x =π e2 +1
1
V x +3   1 π(e)2(4)   =x   1 π(e)2(4)+π ( e2 −1 ) 
1 T
3  3 
( ) ( )
e2 +1 π+4πe2 =x 4πe2 +πe2 −π
T 3
( )
3 5e2 +1
( )
x 7e2 −1 =5e2 +1⇒x =
T 3 T 7e2 −3 | M1*
A1
M1*
A1
A1
M1dep*
A1ft
A1
[8] | 2.1
1.1
1.1
1.1
1.1
3.1a
1.1
2.2a | Correct integral representation for the volume V of
1
the curve y=e 1 2 x between 0 and 2
Correct integral representation for V x and attempt to
1
integrate by parts
Correct integration – limits not required for this mark
e2 +1
Implied by x =
e2 −1
Table of values idea – correct number of terms
Correct equation for the centre of mass of T with
their V x and correct V
1 1
AG – sufficient working must be shown – condone
absence of πthroughout for full marks
(b) |  3 ( 5e2 +1 ) 
 −2 
 7e2 −3 
 
tanθ = (tanθ =0.1237...⇒θ =7.0541...)
1 1 1
e
2
tanθ = (tanθ =0.7357...⇒θ =36.3441...)
2 2 1
e
θ=θ +θ ⇒θ=43.4 
1 2 | B1
B1
B1
[3] | 3.1b
1.1
2.2a | Correct expression for either θor tanθwhere θis
1 1 1
the angle between the line through (2, 0) and A and
the line through A and (x ,0)
T
Correct expression for either θ or tanθ where θ is
2 2 2
the angle between the line through (2, 0) and A and
the line OA
43.3982198… or 0.75744182… (in radians) – where
θis the angle between OA and the vertical
Alternative method 1 (cosine rule)
( 4+e2 ) + ( e2 +(2−x)2 ) −x2
cosθ=
2 4+e2 e2 +(2−x)2 | B2 | B2 for correct application of cosine rule with correct
x – for B1 allow one incorrect length
(3.37476...)2 +(2.7390...)2 −(2.33636...)2
cosθ=
2(3.37476...)(2.7390...)
θ=43.4  | B1 | 43.3982198… or 0.75744182… (in radians) – where
θis the angle between OA and the vertical
Alternative method 2 (scalar product) | [3]
−2(x −2)+e2 = 4+e2 (x −2)2 +e2 cosθ | −2(x −2)+e2 = 4+e2 (x −2)2 +e2 cosθ | B2 | B2 | −2 x −2 −2 x −2
Or B1 for   =    cosθ
−e  −e  −e  −e 
θ=43.4  | B1 | If B0 then SC B2 for 136.601780… | If B0 then SC B2 for 136.601780…
[3]
[In this question you may use the facts that for a uniform solid right circular cone of height $h$ and base radius $r$ the volume is $\frac{1}{3}\pi r^2 h$ and the centre of mass is $\frac{1}{4}h$ above the base on the line from the centre of the base to the vertex.]

\includegraphics{figure_9}

The diagram shows the shaded region S bounded by the curve $y = e^{\frac{1}{4}x}$ for $0 \leq x \leq 2$, the $x$-axis, the $y$-axis, and the line $y = \frac{1}{4}e(6-x)$.

The line $y = \frac{1}{4}e(6-x)$ meets the curve $y = e^{\frac{1}{4}x}$ at the point A with coordinates $(2, e)$.

The region S is rotated through $2\pi$ radians about the $x$-axis to form a uniform solid of revolution T.

\begin{enumerate}[label=(\alph*)]
\item Show that the $x$-coordinate of the centre of mass of T is $\frac{3(5e^2 + 1)}{7e^2 - 3}$. [8]
\end{enumerate}

Solid T is freely suspended from A and hangs in equilibrium.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Determine the angle between AO, where O is the origin, and the vertical. [3]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2022 Q9 [11]}}
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