OCR MEI Further Mechanics Major 2020 November — Question 8 13 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Major (Further Mechanics Major)
Year2020
SessionNovember
Marks13
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Mark schemeDownload PDF ↗
TopicCentre of Mass 2
TypeDeriving standard centre of mass formulae by integration
DifficultyStandard +0.8 This is a multi-part Further Maths mechanics question requiring integration to derive a centre of mass formula (non-trivial setup with volume elements), then applying it to a composite body problem with toppling analysis. The integration in part (a) requires careful setup of disk elements and is more demanding than standard A-level. Part (b) involves finding combined centre of mass and checking if it lies within the base of support—a moderately challenging application. Overall, this is harder than typical A-level but standard for Further Maths mechanics.
Spec6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces
[In this question, you may use the fact that the volume of a right circular cone of base radius \(r\) and height \(h\) is \(\frac{1}{3}\pi r^2 h\).]
  1. By using integration, show that the centre of mass of a uniform solid right circular cone of height \(h\) and base radius \(r\) is at a distance \(\frac{3}{4}h\) from the vertex. [5]
\includegraphics{figure_8} Fig. 8 shows the side view of a toy formed by joining a uniform solid circular cylinder of radius \(r\) and height \(2r\) to a uniform solid right circular cone, made of the same material as the cylinder, of radius \(r\) and height \(r\). The toy is placed on a horizontal floor with the curved surface of the cone in contact with the floor.
  1. Determine whether the toy will topple. [7]
  2. Explain why it is not necessary to know whether the floor is rough or smooth in answering part (b). [1]
Question 8:
AnswerMarks Guidance
8(a) r
y= x
AnswerMarks Guidance
hB1 1.2
could be rotated about the y-axis
2
h r 
Vx =π∫ x x dx
AnswerMarks Guidance
0 h M1 2.1
h
1 πr2 x4
πr2hx =  
3 h2  4 
AnswerMarks Guidance
0M1 1.1
Use of V = πr2h and attempt at
3
AnswerMarks
integrationLimits not required
1 πr2 h4 
πr2hx =  −0⇒x =...
AnswerMarks Guidance
3 h2   4  M1 1.1
for x
3h
x =
AnswerMarks Guidance
4A1 1.1
SC B0M1M0M1A0 for those who use
y = x
[5]
AnswerMarks Guidance
8(b) M1*
terms (dimensionally consistent)
(1πr3+πr2(2r) ) =1πr3(3r) ( πr2(2r))
x +2r
AnswerMarks
3 G 3 4A1
A11.1
1.1Correct LHS
Correct RHSx is the centre of
G
mass of the composite
body from the vertex
51
x = r
G
AnswerMarks Guidance
28A1 1.1
oe (e.g. r from the base)
28
x  51 
cos45=  ⇒x= 2r
51r  56 
 
AnswerMarks Guidance
 28 M1dep* 3.1b
G
horizontal distance distance of centre of
AnswerMarks
mass from vertex51r−r
28
Or tanθ=
r
AnswerMarks Guidance
Slant height of cone is r 2B1 1.1
(dependent on first M mark)
51
r 2<r 2⇒ does not topple
AnswerMarks Guidance
56A1 3.2a
[7]
AnswerMarks Guidance
8(c) Moment of frictional force about any point of contact
with the horizontal floor is zero and so has no effect on
AnswerMarks Guidance
the stability of the toyB1 2.4
[1]
Question 8:
8 | (a) | r
y= x
h | B1 | 1.2 | oe e.g. a correct equation for a line that
could be rotated about the y-axis
2
h r 
Vx =π∫ x x dx
0 h  | M1 | 2.1 | Use of Vx =π∫xy2dxwith their y | Limits not required
h
1 πr2 x4
πr2hx =  
3 h2  4 
0 | M1 | 1.1 | 1
Use of V = πr2h and attempt at
3
integration | Limits not required
1 πr2 h4 
πr2hx =  −0⇒x =...
3 h2   4   | M1 | 1.1 | Use of correct limits and attempt to solve
for x
3h
x =
4 | A1 | 1.1 | AG
SC B0M1M0M1A0 for those who use
y = x
[5]
8 | (b) | M1* | 2.1 | Table of values idea – correct number of
terms (dimensionally consistent)
(1πr3+πr2(2r) ) =1πr3(3r) ( πr2(2r))
x +2r
3 G 3 4 | A1
A1 | 1.1
1.1 | Correct LHS
Correct RHS | x is the centre of
G
mass of the composite
body from the vertex
51
x = r
G
28 | A1 | 1.1 | 33
oe (e.g. r from the base)
28
x  51 
cos45=  ⇒x= 2r
51r  56 
 
 28  | M1dep* | 3.1b | Use of correct angle and their x to find
G
horizontal distance distance of centre of
mass from vertex | 51r−r
28
Or tanθ=
r
Slant height of cone is r 2 | B1 | 1.1 | Or comparison with 45 or tanθ=1
(dependent on first M mark)
51
r 2<r 2⇒ does not topple
56 | A1 | 3.2a
[7]
8 | (c) | Moment of frictional force about any point of contact
with the horizontal floor is zero and so has no effect on
the stability of the toy | B1 | 2.4 | Or equivalent
[1]
[In this question, you may use the fact that the volume of a right circular cone of base radius $r$ and height $h$ is $\frac{1}{3}\pi r^2 h$.]

\begin{enumerate}[label=(\alph*)]
\item By using integration, show that the centre of mass of a uniform solid right circular cone of height $h$ and base radius $r$ is at a distance $\frac{3}{4}h$ from the vertex. [5]
\end{enumerate}

\includegraphics{figure_8}

Fig. 8 shows the side view of a toy formed by joining a uniform solid circular cylinder of radius $r$ and height $2r$ to a uniform solid right circular cone, made of the same material as the cylinder, of radius $r$ and height $r$.

The toy is placed on a horizontal floor with the curved surface of the cone in contact with the floor.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Determine whether the toy will topple. [7]
\item Explain why it is not necessary to know whether the floor is rough or smooth in answering part (b). [1]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2020 Q8 [13]}}
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