CAIE M2 2017 June — Question 5 7 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2017
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeRod on smooth peg or cylinder
DifficultyStandard +0.8 This is a non-trivial statics problem requiring knowledge that the center of mass of a semicircular lamina is at distance 4r/(3π) from the diameter, setting up moment equilibrium about point A with careful geometry (30° angle, forces at different points), and resolving forces in two directions. The multi-step nature, geometric complexity, and need to recall/apply a non-standard center of mass formula place it above average difficulty.
Spec3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces

\includegraphics{figure_3} A uniform semicircular lamina of radius \(0.7\) m and weight \(14\) N has diameter \(AB\). The lamina is in a vertical plane with \(A\) freely pivoted at a fixed point. The straight edge \(AB\) rests against a small smooth peg \(P\) above the level of \(A\). The angle between \(AB\) and the horizontal is \(30°\) and \(AP = 0.9\) m (see diagram).
  1. Show that the magnitude of the force exerted by the peg on the lamina is \(7.12\) N, correct to 3 significant figures. [4]
  2. Find the angle with the horizontal of the force exerted by the pivot on the lamina at \(A\). [3]

Question 5:

AnswerMarks Guidance
5(i)OG = 2 × 0.7sin(π / 2) / (3π / 2) (= 0.297) B1
0.9R = 14(0.7cos30 – 0.297sin30)M1A1 Attempts to take moments about A
R = 7.12 NA1
Total:4

AnswerMarks Guidance
5(ii)H = 7.12sin30 and V = 14 − Rcos30 M1
tanθ = (14 – 7.12cos30) / (7.12sin30)M1 Uses tanθ = V / H, where θ is the required angle
θ = 65.6A1
Total:3
QuestionAnswer Marks
Question 5:
--- 5(i) ---
5(i) | OG = 2 × 0.7sin(π / 2) / (3π / 2) (= 0.297) | B1
0.9R = 14(0.7cos30 – 0.297sin30) | M1A1 | Attempts to take moments about A
R = 7.12 N | A1
Total: | 4
--- 5(ii) ---
5(ii) | H = 7.12sin30 and V = 14 − Rcos30 | M1 | Resolves horizontally and vertically
tanθ = (14 – 7.12cos30) / (7.12sin30) | M1 | Uses tanθ = V / H, where θ is the required angle
θ = 65.6 | A1
Total: | 3
Question | Answer | Marks | Guidance
\includegraphics{figure_3}

A uniform semicircular lamina of radius $0.7$ m and weight $14$ N has diameter $AB$. The lamina is in a vertical plane with $A$ freely pivoted at a fixed point. The straight edge $AB$ rests against a small smooth peg $P$ above the level of $A$. The angle between $AB$ and the horizontal is $30°$ and $AP = 0.9$ m (see diagram).

\begin{enumerate}[label=(\roman*)]
\item Show that the magnitude of the force exerted by the peg on the lamina is $7.12$ N, correct to 3 significant figures. [4]
\item Find the angle with the horizontal of the force exerted by the pivot on the lamina at $A$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE M2 2017 Q5 [7]}}