Edexcel M3 Specimen — Question 4 10 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
SessionSpecimen
Marks10
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Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeSolid with removed cone from cone or cylinder
DifficultyStandard +0.3 This is a standard M3 centre of mass question requiring systematic application of the composite body formula (cylinder minus cone) and then a toppling condition. The geometry is clearly defined, the calculation is methodical with no conceptual surprises, and toppling problems are routine at this level. Slightly easier than average due to the straightforward setup and standard techniques.
Spec6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces

\includegraphics{figure_3} A container is formed by removing a right circular solid cone of height \(4l\) from a uniform solid right circular cylinder of height \(6l\). The centre \(O\) of the plane face of the cone coincides with the centre of a plane face of the cylinder and the axis of the cone coincides with the axis of the cylinder, as shown in Figure 3. The cylinder has radius \(2l\) and the base of the cone has radius \(l\).
  1. Find the distance of the centre of mass of the container from \(O\). [6]
\includegraphics{figure_4} The container is placed on a plane which is inclined at an angle \(\theta°\) to the horizontal. The open face is uppermost, as shown in Figure 4. The plane is sufficiently rough to prevent the container from sliding. The container is on the point of toppling.
  1. Find the value of \(\theta\). [4]

Part (a)
AnswerMarks Guidance
conecontainer cylinder
mass ratio\(\frac{4\pi l^3}{3}\) \(\frac{68\pi l^3}{3}\)
468 72
dist from O\(l\) \(\frac{68}{x}\)
Moments: \(4l + 68\bar{x} = 72 \times 3l\)M1 A1ft
\(\bar{x} = \frac{212l}{68} = \frac{53}{17}\) accept \(3.12/\)A1 (6 marks total)
Part (b)
AnswerMarks Guidance
\(GX = 6l - \bar{x}\) seenM1
\(\tan\theta = \frac{2l}{6l - \bar{x}} = \frac{2 \times 17}{49} = \frac{34}{49}\)M1 A1
\(\theta = 34.75\ldots = 34.8\) or \(35\)A1 (4 marks total)
[10 marks total]
## Part (a)

| | cone | container | cylinder |
|---|---|---|---|
| mass ratio | $\frac{4\pi l^3}{3}$ | $\frac{68\pi l^3}{3}$ | $24\pi l^3$ |
| | 4 | 68 | 72 |
| dist from O | $l$ | $\frac{68}{x}$ | $3l$ |

Moments: $4l + 68\bar{x} = 72 \times 3l$ | M1 A1ft |
$\bar{x} = \frac{212l}{68} = \frac{53}{17}$ accept $3.12/$ | A1 | (6 marks total)

## Part (b)

$GX = 6l - \bar{x}$ seen | M1 |
$\tan\theta = \frac{2l}{6l - \bar{x}} = \frac{2 \times 17}{49} = \frac{34}{49}$ | M1 A1 |
$\theta = 34.75\ldots = 34.8$ or $35$ | A1 | (4 marks total)

**[10 marks total]**
\includegraphics{figure_3}

A container is formed by removing a right circular solid cone of height $4l$ from a uniform solid right circular cylinder of height $6l$. The centre $O$ of the plane face of the cone coincides with the centre of a plane face of the cylinder and the axis of the cone coincides with the axis of the cylinder, as shown in Figure 3. The cylinder has radius $2l$ and the base of the cone has radius $l$.

\begin{enumerate}[label=(\alph*)]
\item Find the distance of the centre of mass of the container from $O$. [6]
\end{enumerate}

\includegraphics{figure_4}

The container is placed on a plane which is inclined at an angle $\theta°$ to the horizontal. The open face is uppermost, as shown in Figure 4. The plane is sufficiently rough to prevent the container from sliding. The container is on the point of toppling.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the value of $\theta$. [4]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M3  Q4 [10]}}