CAIE M2 2011 November — Question 4 9 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2011
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeCone stability and toppling conditions
DifficultyStandard +0.8 This is a multi-part centre of mass problem requiring understanding of toppling conditions, geometric reasoning about centres of mass of standard solids (cone at h/4, cylinder at h/2), and critically, part (iii) involves a composite body in an unusual orientation (cone on curved surface) requiring careful geometric analysis of the combined centre of mass position. The conceptual demand and multi-step reasoning elevate this above routine exercises.
Spec6.04a Centre of mass: gravitational effect6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces

A uniform solid cylinder has radius 0.7 m and height \(h\) m. A uniform solid cone has base radius 0.7 m and height 2.4 m. The cylinder and the cone both rest in equilibrium each with a circular face in contact with a horizontal plane. The plane is now tilted so that its inclination to the horizontal, \(θ°\), is increased gradually until the cone is about to topple.
  1. Find the value of \(θ\) at which the cone is about to topple. [2]
  2. Given that the cylinder does not topple, find the greatest possible value of \(h\). [2]
The plane is returned to a horizontal position, and the cone is fixed to one end of the cylinder so that the plane faces coincide. It is given that the weight of the cylinder is three times the weight of the cone. The curved surface of the cone is placed on the horizontal plane (see diagram). \includegraphics{figure_4}
  1. Given that the solid immediately topples, find the least possible value of \(h\). [5]

AnswerMarks Guidance
(i) \(\tan\theta = 0.7/(2.4/4)\)M1
\(\theta = 49.4°\)A1 [2]
(ii) \(h/2 = 2.4/4\)M1
\(h = 1.2\)A1 [2]
(iii) \(4wVG = w \times 2.4 \times 3/4 + 3w(2.4 + h/2)\)M1 Table of values idea, accept \(w = 1\)
A1
M1Centre of mass above common circumference
\(VG = [\sqrt{(0.7^2 + 2.4^2)}]/\cos\alpha\)A1 \(\cos\alpha = 2.4/2.5 = 0.96\)
\(h = 0.944\)A1 [5]
**(i)** $\tan\theta = 0.7/(2.4/4)$ | M1 | 
$\theta = 49.4°$ | A1 | [2]

**(ii)** $h/2 = 2.4/4$ | M1 | 
$h = 1.2$ | A1 | [2]

**(iii)** $4wVG = w \times 2.4 \times 3/4 + 3w(2.4 + h/2)$ | M1 | Table of values idea, accept $w = 1$
| | A1 | 
| | M1 | Centre of mass above common circumference
$VG = [\sqrt{(0.7^2 + 2.4^2)}]/\cos\alpha$ | A1 | $\cos\alpha = 2.4/2.5 = 0.96$
$h = 0.944$ | A1 | [5]
A uniform solid cylinder has radius 0.7 m and height $h$ m. A uniform solid cone has base radius 0.7 m and height 2.4 m. The cylinder and the cone both rest in equilibrium each with a circular face in contact with a horizontal plane. The plane is now tilted so that its inclination to the horizontal, $θ°$, is increased gradually until the cone is about to topple.

\begin{enumerate}[label=(\roman*)]
\item Find the value of $θ$ at which the cone is about to topple. [2]
\item Given that the cylinder does not topple, find the greatest possible value of $h$. [2]
\end{enumerate}

The plane is returned to a horizontal position, and the cone is fixed to one end of the cylinder so that the plane faces coincide. It is given that the weight of the cylinder is three times the weight of the cone. The curved surface of the cone is placed on the horizontal plane (see diagram).

\includegraphics{figure_4}

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Given that the solid immediately topples, find the least possible value of $h$. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE M2 2011 Q4 [9]}}