Centres of mass

[The centre of mass of a semicircular lamina of radius r is from the centre.]

Figure 2 shows the uniform lamina ABCDE, such that ABDE is a square with sides of length 2 a and BCD is a semicircle with diameter BD.

Show that the distance of the centre of mass of the lamina from BD is

The lamina is freely suspended from D and hangs in equilibrium.

Find, to the nearest degree, the angle that DE makes with the downward vertical.

The uniform rectangular lamina ABDE, shown in Figure 1, has side AB of length 2 a and side BD of length 6 a. The point C divides BD in the ratio 1:2 and the point F divides EA in the ratio 1:2. The rectangular lamina is folded along FC to produce the folded lamina L, shown in Figure 2.

Show that the centre of mass of L is a from EF.

The folded lamina, L, is freely suspended from C and hangs in equilibrium.

Find the size of the angle between CF and the downward vertical.

The uniform lamina OBC is one quarter of a circular disc with centre O and radius 4m. The points A and D, on OB and OC respectively, are 3m from O. The uniform lamina ABCD, shown shaded in Figure 1, is formed by removing the triangle OAD from OBC. Given that the centre of mass of one quarter of a uniform circular disc of radius r is at a

find the distance of the centre of mass of the lamina ABCD from AD.

The lamina is freely suspended from D and hangs in equilibrium.

Find, to the nearest degree, the angle between DC and the downward vertical.

Three particles of mass 3 m, 2 m and km are placed at the points whose coordinates are (1, 5), (6, 4) and (a, 1) respectively. The centre of mass of the three particles is at the point with coordinates (3, 3). Find

the value of k,

the value of a.

The uniform lamina ABCDEF, shown shaded in Figure 1, is symmetrical about the line through B and E. It is formed by removing the isosceles triangle FED, of height 6 a and base 8 a, from the isosceles triangle ABC of height 9 a and base 12 a.

Find, in terms of a, the distance of the centre of mass of the lamina from AC.

The lamina is freely suspended from A and hangs in equilibrium.

Find, to the nearest degree, the size of the angle between AB and the downward vertical.

Figure 2 shows a lamina L. It is formed by removing a square PQRS from a uniform triangle ABC. The triangle ABC is isosceles with AC = BC and AB = 12 cm. The midpoint of AB is D and DC = 8 cm. The vertices P and Q of the square lie on AB and PQ = 4 cm. The centre of the square is O. The centre of mass of L is at G.

Find the distance of G from AB.

When L is freely suspended from A and hangs in equilibrium, the line AB is inclined at 25° to the vertical.

Find the distance of O from DC.

The uniform lamina ABCDEF is a regular hexagon with centre O and sides of length 2 m, as shown in Figure 1.

The triangles OAF and OEF are removed to form the uniform lamina OABCDE, shown in Figure 2.

Find the distance of the centre of mass of OABCDE from O.

The lamina OABCDE is freely suspended from E and hangs in equilibrium.

Find the size of the angle between EO and the downward vertical.

A uniform triangular lamina ABC of mass M is such that AB = AC, BC = 2a and the distance of A from BC is h. A line, parallel to BC and at a distance from A, cuts AB at D and cuts AC at E, as shown in Figure 2. It is given that the mass of the trapezium BCED is

Show that the centre of mass of the trapezium BCED is from BC.

The portion ADE of the lamina is folded through 180° about DE to form the folded lamina shown in Figure 3.

Find the distance of the centre of mass of the folded lamina from BC.

The folded lamina is freely suspended from D and hangs in equilibrium. The angle between DE and the downward vertical is .

Find tan in terms of a and h.

Two uniform rods AB and BC are rigidly joined at B so that ABC = 90°. Rod AB has length 0.5 m and mass 2 kg. Rod BC has length 2 m and mass 3 kg. The centre of mass of the framework of the two rods is at G.

Find the distance of G from BC.

The distance of G from AB is 0.6 m. The framework is suspended from A and hangs freely in equilibrium.

Find the angle between AB and the downward vertical at A.

A uniform circular disc has centre O and radius 4 a. The lines PQ and ST are perpendicular diameters of the disc. A circular hole of radius 2 a is made in the disc, with the centre of the hole at the point R on OP where OR = 2 a, to form the lamina L, shown shaded in Figure 2.

The mass of L is m and a particle of mass km is now fixed to L at the point P. The system is now suspended from the point S and hangs freely in equilibrium. The diameter ST makes an angle with the downward vertical through S, where tan

Find the value of k.

Figure 1 shows a uniform lamina ABCDE such that ABDE is a rectangle, BC = CD, AB = 4 a and AE = 2 a. The point F is the midpoint of BD and FC = a.

Find, in terms of a, the distance of the centre of mass of the lamina from AE.

The lamina is freely suspended from A and hangs in equilibrium.

Find the angle between AB and the downward vertical.

The uniform L-shaped lamina ABCDEF, shown in Figure 2, has sides AB and FE parallel, and sides BC and ED parallel. The pairs of parallel sides are 9 cm apart. The points A, F, D and C lie on a straight line. AB = BC = 36 cm, FE = ED = 18 cm. ABC = FED = 90°, and BCD = EDF = EFD = BAC = 45°.

Find the distance of the centre of mass of the lamina from

side AB,

side BC.

The lamina is freely suspended from A and hangs in equilibrium.

Find, to the nearest degree, the size of the angle between AB and the vertical.

A triangular frame is formed by cutting a uniform rod into 3 pieces which are then joined to form a triangle ABC, where AB = AC = 10 cm and BC = 12 cm, as shown in Figure 1.

Find the distance of the centre of mass of the frame from BC.

The frame has total mass M. A particle of mass M is attached to the frame at the mid-point of BC. The frame is then freely suspended from B and hangs in equilibrium.

Find the size of the angle between BC and the vertical.

[The centre of mass of a semi-circular lamina of radius r is from the centre]

A template T consists of a uniform plane lamina PQROS, as shown in Figure 3. The lamina is bounded by two semicircles, with diameters SO and OR, and by the sides SP, PQ and QR of the rectangle PQRS. The point O is the mid-point of SR, PQ = 12 cm and QR = 2 x cm.

The template T is freely suspended from the point P and hangs in equilibrium. Given that x = 2 and that θ is the angle that PQ makes with the horizontal,

A shop sign ABCDEFG is modelled as a uniform lamina, as illustrated in Figure 2. ABCD is a rectangle with BC = 120 cm and DC = 90 cm. The shape EFG is an isosceles triangle with EG = 60 cm and height 60 cm. The mid-point of AD and the mid-point of EG coincide.

Find the distance of the centre of mass of the sign from the side AD.

The sign is freely suspended from A and hangs at rest.

Find the size of the angle between AB and the vertical.

Figure 3

Figure 3 shows a rectangular lamina OABC. The coordinates of O, A, B and C are (0, 0), (8, 0), (8, 5) and (0, 5) respectively. Particles of mass km, 5 m and 3 m are attached to the lamina at A, B and C respectively. The x-coordinate of the centre of mass of the three particles without the lamina is 6.4.

Show that k = 7.

The lamina OABC is uniform and has mass 12 m.

Find the coordinates of the centre of mass of the combined system consisting of the three particles and the lamina.

The combined system is freely suspended from O and hangs at rest.

Find the angle between OC and the horizontal.

A set square S is made by removing a circle of centre O and radius 3 cm from a triangular piece of wood. The piece of wood is modelled as a uniform triangular lamina ABC, with ABC = 90°, AB = 12 cm and BC = 21 cm. The point O is 5 cm from AB and 5 cm from BC, as shown in Figure 1.

Find the distance of the centre of mass of S from

AB,

BC.

The set square is freely suspended from C and hangs in equilibrium.

Find, to the nearest degree, the angle between CB and the vertical.