Edexcel M3 2009 June — Question 2 8 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Year2009
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeConical or hemispherical shell composite
DifficultyStandard +0.3 This is a standard M3 centre of mass question requiring composite body calculations and toppling conditions. Part (a) involves straightforward application of the centre of mass formula for composite bodies using given information. Part (b) applies the standard toppling criterion (vertical line through COM passes through pivot point) with basic trigonometry. While it requires careful setup and multiple steps (8 marks total), it follows predictable M3 patterns without requiring novel insight or complex problem-solving.
Spec6.04c Composite bodies: centre of mass

[The centre of mass of a uniform hollow cone of height \(h\) is \(\frac{1}{3}h\) above the base on the line from the centre of the base to the vertex.] \includegraphics{figure_1} A marker for the route of a charity walk consists of a uniform hollow cone fixed on to a uniform solid cylindrical ring, as shown in Figure 1. The hollow cone has base radius \(r\), height \(9h\) and mass \(m\). The solid cylindrical ring has outer radius \(r\), height \(2h\) and mass \(3m\). The marker stands with its base on a horizontal surface.
  1. Find, in terms of \(h\), the distance of the centre of mass of the marker from the horizontal surface. [5]
When the marker stands on a plane inclined at arctan \(\frac{1}{12}\) to the horizontal it is on the point of toppling over. The coefficient of friction between the marker and the plane is large enough to be certain that the marker will not slip.
  1. Find \(h\) in terms of \(r\). [3]

Part (a)
AnswerMarks Guidance
ObjectMass c of m above base
Cone\(m\) \(2h + 3h\)
Base\(3m\) \(h\)
Marker\(4m\) \(d\)
\(m \times 5h + 3m \times h = 4m \times d\)B1(ratio masses), B1(distances)
\(d = 2h\)M1A1ft, A1
Part (b)
AnswerMarks
\(\frac{r}{d} = \frac{1}{12}\)M1A1ft
\(6r = h\)A1
## Part (a)

| Object | Mass | c of m above base |
|--------|------|-------------------|
| Cone | $m$ | $2h + 3h$ |
| Base | $3m$ | $h$ |
| Marker | $4m$ | $d$ |

$m \times 5h + 3m \times h = 4m \times d$ | B1(ratio masses), B1(distances) |
$d = 2h$ | M1A1ft, A1 |

## Part (b)

$\frac{r}{d} = \frac{1}{12}$ | M1A1ft |

$6r = h$ | A1 |

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[The centre of mass of a uniform hollow cone of height $h$ is $\frac{1}{3}h$ above the base on the line from the centre of the base to the vertex.]

\includegraphics{figure_1}

A marker for the route of a charity walk consists of a uniform hollow cone fixed on to a uniform solid cylindrical ring, as shown in Figure 1. The hollow cone has base radius $r$, height $9h$ and mass $m$. The solid cylindrical ring has outer radius $r$, height $2h$ and mass $3m$. The marker stands with its base on a horizontal surface.

\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $h$, the distance of the centre of mass of the marker from the horizontal surface. [5]
\end{enumerate}

When the marker stands on a plane inclined at arctan $\frac{1}{12}$ to the horizontal it is on the point of toppling over. The coefficient of friction between the marker and the plane is large enough to be certain that the marker will not slip.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find $h$ in terms of $r$. [3]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M3 2009 Q2 [8]}}