CAIE Further Math AS 3.2 Rigid Body Equilibrium

[Maximum number: 4]

A child's toy consists of a uniform solid circular cone, of vertical height 3 r and radius r, and a uniform solid hemisphere of radius r. The circular bases of the cone and the hemisphere are joined together so that they coincide. The cone and the hemisphere are made of the same material.

Show that the centre of mass of the toy is at a distance $\frac{27}{10} r$ from the vertex of the cone.

COM from vertex
$3 r+\frac{3}{8} r$ $\frac{3}{4} \times 3 r$ $\bar{x}$ $\frac{1}{3} \pi r^{2} \times 3 r$
Volume
$\frac{5}{3} \pi r^{3}$ $\frac{2}{3} \pi r^{3}$
Hemisphere
Combined
Density $\rho$
Cone

1
For COM (centre of mass) $\frac{3}{4} \cdot 3 r$ or $3 r+\frac{3}{8} r$

Take moments about vertex: (allow $\rho$ omitted)
$\frac{5}{3} \pi r^{3} \times \bar{x}=\pi r^{3} \times \frac{9}{4} r+\frac{2}{3} \pi r^{3} \times \frac{27}{8} r$

1
Taking moments equation

leading to $\bar{x}=\frac{27}{10} r$

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distance of COM from vertex $=\frac{27}{10} r$

1
AG Correct answer, with convincing working

4

[Maximum number: 8]

Figure for Question 2 — CAIE A-Level Further Math AS

A uniform square lamina of side 2 a and weight W is suspended from a light inextensible string attached to the midpoint E of the side AB. The other end of the string is attached to a fixed point P on a rough vertical wall. The vertex B of the lamina is in contact with the wall. The string EP is perpendicular to the side AB and makes an angle $\theta$ with the wall (see diagram). The string and the lamina are in a vertical plane perpendicular to the wall. The coefficient of friction between the wall and the lamina is $\frac{1}{2}$ .

Given that the vertex B is about to slip up the wall, find the value of $\tan \theta$ .

Let N be normal reaction at B and F the frictional force acting downwards
$\uparrow T \cos \theta=F+W$ $\rightarrow T \sin \theta=N$
Moments about $B: T a=W \sin \theta \times a+W \cos \theta \times a$
A moments equation with all relevant forces.

$F=\frac{1}{2} N$ used
$(\cos \theta+\sin \theta)\left(\cos \theta-\frac{1}{2} \sin \theta\right)=1$ ое
Combine to obtain equation in $\theta$ .
Equation in trigonometric functions only.

$\begin{aligned} \frac{1}{2} \cos \theta \sin \theta=\frac{3}{2}(\sin \theta)^{2} \sin \theta(\cos \theta-3 \sin \theta)=0 \end{aligned}$
Solve trigonometric equation.

$\tan \theta=\frac{1}{3}$
8

[Maximum number: 8]

Figure for Question 3 — CAIE A-Level Further Math AS

A uniform square lamina of side 2 a and weight W is suspended from a light inextensible string attached to the midpoint E of the side AB. The other end of the string is attached to a fixed point P on a rough vertical wall. The vertex B of the lamina is in contact with the wall. The string EP is perpendicular to the side AB and makes an angle $\theta$ with the wall (see diagram). The string and the lamina are in a vertical plane perpendicular to the wall. The coefficient of friction between the wall and the lamina is $\frac{1}{2}$ .

Given that the vertex B is about to slip up the wall, find the value of $\tan \theta$ .

Let N be normal reaction at B and F the frictional force acting downwards
$\uparrow T \cos \theta=F+W$ $\rightarrow T \sin \theta=N$
Moments about $B: T a=W \sin \theta \times a+W \cos \theta \times a$
A moments equation with all relevant forces.

$F=\frac{1}{2} N$ used
$(\cos \theta+\sin \theta)\left(\cos \theta-\frac{1}{2} \sin \theta\right)=1$ ое
Combine to obtain equation in $\theta$ .
Equation in trigonometric functions only.

$\frac{1}{2} \cos \theta \sin \theta=\frac{3}{2}(\sin \theta)^{2}$ $\sin \theta(\cos \theta-3 \sin \theta)=0$
Solve trigonometric equation.

$\tan \theta=\frac{1}{3}$
8

CAIE A-Level Further Math AS 3.2 Equilibrium of a Rigid Body Questions

Exam points

Covered syllabus points

Covered syllabus points

Question 1

Question 2

Question 3