| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | November |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Coplanar forces in equilibrium |
| Difficulty | Challenging +1.8 This is a challenging mechanics problem requiring knowledge of the center of mass of a hemisphere (3a/8 from base), moments about a point, friction conditions, and rotational dynamics. Part (i) requires setting up moment and force equations; part (ii) is a proof comparing friction coefficients; part (iii) involves work-energy principles with limiting friction; part (iv) requires calculating angular acceleration using torque. The multi-part structure, need for the non-trivial COM location, and integration of statics with dynamics concepts places this well above average difficulty, though it follows standard M2 problem-solving patterns. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks |
|---|---|
| 7 (i) | π × 0.5 2 × 0.4 × 0.2+ π × 0.5 2 × |
| Answer | Marks |
|---|---|
| d = 0.275 m | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| A1 | [3] | Uses table of moments idea |
| (ii) | (0.4 + 0.4)F = 0.5 × 60 | |
| F = 37.5 | M1 | |
| A1 | [2] | Takes moments |
| (iii) | µ (= 37.5/60) = 0.625 | B1 |
| ft | [1] | cv(F)/60 |
| (iv) | F/R=(60sin30)/(60cos30) (= 0.577..) |
| Answer | Marks |
|---|---|
| = 14 AG | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| A1 | [4] | p |
| Answer | Marks |
|---|---|
| 0.4 – 0.29 0.275, topples | 10 |
Question 7:
--- 7 (i) ---
7 (i) | π × 0.5 2 × 0.4 × 0.2+ π × 0.5 2 ×
0.4 × 0.5/3
= ( π × 0.5 2 × 0.4+ π × 0.5 2 ×
0.4/3)0G AG
d = 0.275 m | M1
A1
A1 | [3] | Uses table of moments idea
(ii) | (0.4 + 0.4)F = 0.5 × 60
F = 37.5 | M1
A1 | [2] | Takes moments
(iii) | µ (= 37.5/60) = 0.625 | B1
ft | [1] | cv(F)/60
(iv) | F/R=(60sin30)/(60cos30) (= 0.577..)
0.577 p 0.625 (or µ ), no sliding AG
θ
tan = (0.4 – 0.275)/0.5
θ o
= 14 AG | M1
A1
M1
A1 | [4] | p
Or quotes tan30 0.625
Or 0.5tan30 = 0.288..
p
0.4 – 0.29 0.275, topples | 10\includegraphics{figure_7}
A uniform solid hemisphere of mass $M$ and radius $a$ is placed with its curved surface on rough horizontal ground. A horizontal force $P$ is applied to the hemisphere at the centre of its flat circular face.
\begin{enumerate}[label=(\roman*)]
\item Find the minimum value of the coefficient of friction $\mu$ between the hemisphere and the ground for the hemisphere to slide without toppling.
\item Show that if $\mu < \frac{3}{8}$, the hemisphere will topple.
\item Find the maximum horizontal distance that the centre of mass of the hemisphere moves before toppling begins, given that $\mu = \frac{1}{4}$ and the hemisphere starts from rest.
\item Find the angular acceleration of the hemisphere about its point of contact with the ground at the instant when toppling begins.
\end{enumerate}
[16]
\hfill \mbox{\textit{CAIE M2 2013 Q7 [16]}}