CAIE M2 2017 June — Question 3 7 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2017
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeSolid with removed cylinder or hemisphere from solid
DifficultyStandard +0.3 This is a straightforward centre of mass problem using standard formulas. Part (i) requires applying the composite body formula with given hemisphere volumes and the standard result for hemisphere COM position (3r/8 from base). Part (ii) is a simple application of the COM formula with known positions and masses. The question is slightly easier than average as it provides the volume formula, uses clean numbers, and follows a standard template with no geometric insight or proof required.
Spec6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids

\includegraphics{figure_3} An object is made from a uniform solid hemisphere of radius \(0.56\) m and centre \(O\) by removing a hemisphere of radius \(0.28\) m and centre \(O\). The diagram shows a cross-section through \(O\) of the object.
  1. Calculate the distance of the centre of mass of the object from \(O\). [4] [The volume of a hemisphere is \(\frac{2}{3}\pi r^3\).]
  2. The object has weight \(24\) N. A uniform hemisphere \(H\) of radius \(0.28\) m is placed in the hollow part of the object to create a new uniform hemisphere with centre \(O\). The centre of mass of the non-uniform hemisphere is \(0.15\) m from \(O\). Calculate the weight of \(H\). [3]

Question 3:

AnswerMarks
3(i)3 3
CofM of hemisphere = × 0.56 or × 0.28
AnswerMarks
8 8B1
2 2 2 3 2
[ π × 0.563– π × 0.283]X = π × 0.563 × × 0.56 – π × 0.283×
3 3 3 8 3
3
× 0.28
AnswerMarks Guidance
8M1A1 Take moments about O
X = 0.225 mA1
Total:4

AnswerMarks Guidance
3(ii)24 × 0.225 + W(3 × 0.28 / 8) = (24 + W) × 0.15 M1A1
W = weight of uniform hemi-sphere
AnswerMarks
W = 40 NA1
Total:3
Question 3:
--- 3(i) ---
3(i) | 3 3
CofM of hemisphere = × 0.56 or × 0.28
8 8 | B1
2 2 2 3 2
[ π × 0.563– π × 0.283]X = π × 0.563 × × 0.56 – π × 0.283×
3 3 3 8 3
3
× 0.28
8 | M1A1 | Take moments about O
X = 0.225 m | A1
Total: | 4
--- 3(ii) ---
3(ii) | 24 × 0.225 + W(3 × 0.28 / 8) = (24 + W) × 0.15 | M1A1 | Attempts to take moments about O
W = weight of uniform hemi-sphere
W = 40 N | A1
Total: | 3
\includegraphics{figure_3}

An object is made from a uniform solid hemisphere of radius $0.56$ m and centre $O$ by removing a hemisphere of radius $0.28$ m and centre $O$. The diagram shows a cross-section through $O$ of the object.

\begin{enumerate}[label=(\roman*)]
\item Calculate the distance of the centre of mass of the object from $O$. [4]

[The volume of a hemisphere is $\frac{2}{3}\pi r^3$.]

\item The object has weight $24$ N. A uniform hemisphere $H$ of radius $0.28$ m is placed in the hollow part of the object to create a new uniform hemisphere with centre $O$. The centre of mass of the non-uniform hemisphere is $0.15$ m from $O$.

Calculate the weight of $H$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE M2 2017 Q3 [7]}}