| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Hemisphere or sphere resting on plane or wall |
| Difficulty | Challenging +1.8 This is a challenging Further Maths mechanics problem requiring: (1) knowledge that the center of mass of a hemisphere is at 3a/8 from the base, (2) careful resolution of forces and friction on a curved surface, (3) taking moments about the contact point with non-trivial geometry, and (4) applying the limiting friction condition. The multi-part nature, geometric complexity, and need to manipulate inequalities involving the friction coefficient make this significantly harder than standard A-level mechanics, though the techniques are systematic once the setup is understood. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03t Coefficient of friction: F <= mu*R model6.04e Rigid body equilibrium: coplanar forces |
**Question 3 (i)**
Resolve horizontally: $F \cos \alpha = R \sin \alpha$ B2
**Question 3 (ii)**
EITHER: Resolve horizontally: $F \cos \alpha = R \sin \alpha$ B2
OR: Resolve along plane to find friction $F$: $F = (M + m)g \sin \alpha$ (B1)
Resolve normal to plane for reaction $R$: $R = (M + m)g \cos \alpha$ (B1)
Use $F/R \leq \frac{1}{2}$: $\tan \alpha \leq \frac{1}{2}$ A.G. M1 A1
EITHER: Take moments about pt of contact: $mg (a - a \sin \alpha) = Mga \sin \alpha$
OR: Take moments about centre: $mg = F = (M + m)g \sin \alpha$ M1 A1
Find inequality for $\sin \alpha$: $\sin \alpha \leq \frac{1}{\sqrt{5}}$ B1
Combine: $m = \frac{M}{\frac{1}{\sin \alpha} - 1} \leq \frac{M}{\sqrt{5} - 1}$
$m \leq \frac{M}{1 + \sqrt{5}} / 4$ A.G. M1 A13\\
\includegraphics[max width=\textwidth, alt={}, center]{020ebd88-b920-40ce-84cf-5c26d45e2935-2_355_695_1073_726}
A uniform solid hemisphere, of radius $a$ and mass $M$, is placed with its curved surface in contact with a rough plane that is inclined at an angle $\alpha$ to the horizontal. A particle $P$ of mass $m$ is attached to the rim of the hemisphere. The system rests in equilibrium with the rim of the hemisphere horizontal and $P$ at the point on the rim that is closest to the inclined plane (see diagram). Given that the coefficient of friction between the plane and the hemisphere is $\frac { 1 } { 2 }$, show that\\
(i) $\tan \alpha \leqslant \frac { 1 } { 2 }$,\\
(ii) $m \leqslant \frac { M ( 1 + \sqrt { } 5 ) } { 4 }$.
\hfill \mbox{\textit{CAIE FP2 2011 Q3 [9]}}