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C entres of Mass

C entres of Mass. Chapter objectives. Find the centre of mass of a system of particles distributed in one dimension Find the centre of mass of a system of particles distributed in two dimensions Use knowledge of standard results to find the centre of mass of plane figures

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C entres of Mass

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  1. Centres of Mass

  2. Chapter objectives • Find the centre of mass of a system of particles distributed in one dimension • Find the centre of mass of a system of particles distributed in two dimensions • Use knowledge of standard results to find the centre of mass of plane figures • Consider the equilibrium of a lamina which is suspended from a fixed point or placed on an inclined plane

  3. Keywords centre of mass – the point at which the whole mass of the body can be considered to be concentrated Lamina – an object which is very thin compared to it’s other two dimensions is modelled as a lamina e.g. a sheet of paper Uniform – mass is evenly distributed Find the centre of mass of a system of particles distributed in one dimension

  4. Find the centre of mass of a system of particles distributed in one dimension This means your particles are arranged along a straight line For example Three particles of mass 6kg, 3kg and 2.5kg are attached to a light rod PQ of length 3m at the points P, Q and R, where PR = 0.9m. Find the position of the centreof mass of the system. Draw a diagram or 2!

  5. P P Q Q R R G 0.9m 0.9m 2.5g 2.5g 3g 3g 11.5g 6g 6g Start by adding the centreof mass to the diagram and let the distance PG be x.

  6. P Q R G 0.9m 2.5g 3g 11.5g 6g Taking moments about P gives: 11.5g × x = 3g × 0.9 + 3 × 2.5g x = 0.89m

  7. There is a general rule you can apply to this And our text book says you should learn it!

  8. Find the centre of mass of a system of particles distributed in two dimensions Now our particles are arranged in a plane For example Particles of mass 2kg, 4kg, 5kg and 6kg are attached to the corners of a light rectangular plate PQRS. Given that PQ = 5cm and QR = 12cm calculate the distance of the centre of mass of the system from a) PQ b) PS

  9. Draw a diagram The question asks us to: calculate the distance of the centre of mass of the system from a) PQ b) PS

  10. calculate the distance of the centre of mass of the system from a) PQ PS There are 3 different methods to solve this question Method 1: taking P as (0, 0) put the information in a table

  11. calculate the distance of the centre of mass of the system from a) PQ PS To find the distance of the centre of mass from PQ we use the formula: To find the distance from PS we use the formula: centre of mass is at (7.66, 2.65)

  12. Method 2 Take moments about the Y axis PQ: 5 x 12 + 6 x 12 = 17n.b. g has been cancelled = 7.66 Take moments about the X axis PS: 4 x 5 + 5 x 5 = 17 = 2.65 centre of mass is at (7.66, 2.65)

  13. Method 3 Use position vectors taking P as (0, 0) 2 + 6 + 5 + 4 = 17 + + + = 17 = centre of mass is at (7.66, 2.65)

  14. Pick a method and try this example The diagram below shows a series of particles that make up a system. The centreof mass of the system is at the point (x,y). Find the coordinates of the centreof mass of the system

  15. 2 + 3 + 4 + 2 = 1 + + + = 1 = 1 = centre of mass is at (2.22, -0.04)

  16. There is a general rule that goes with all this. It can be found on page 35. There is an error in question 9 on page 39 The third side is CD not CA

  17. Use knowledge of standard results to find the centre of mass of plane figures • There are some standard results to be taken for granted when it comes to the centre of mass. • Uniform rectangular lamina:- at centre of the shape • Uniform circular disc:- at centre of disc • Uniform triangular lamina:- • Equilateral:- at centre • Isosceles:- at the intersection of the medians • A median is a line that joins a vertex of a triangle to the centre of the side opposite to the vertex. The centre of mass of a scalene triangle is at a point one third of the way along the median (from the edge). The centre of mass of the triangle ABC is at the point G, where EG = ⅓EC

  18. If the coordinates of the three vertices of a uniform triangular lamina are (x1, y1), (x2, y2) and (x3, y3) then the coordinates of the centre of mass are found by taking the mean of the coordinates of the 3 vertices.

  19. B 12m 18m C A Calculate the coordinates of the centre of mass of the uniform triangular lamina ABC if the point A is placed at the origin. = centre of mass G is at the point

  20. This formulae is in the booklet

  21. C G A O B Find the centre of mass of the lamina with radius 5 cm The centre of mass lies on the line of symmetry OC Using the result for a sector with r = 5 and α = Then OG = = • The centre of mass lies on the line OC a distance from O.

  22. Application to composite figures You can find the centre of mass for this shape by applying the ideas we have already come across: Work out the centre of mass for the rectangle Work out the centre of mass for the triangle Find the centre of mass of both ‘particles’ centre of mass for ABCE Take A as the origin then (6, 4.5) is the centre of mass centre of mass for CDE = = (14.5, 4.5)

  23. Because the shapes are uniform the area of each can be used to represent it’s mass Mass of rectangle = 108 cm² mass of triangle = 33.75cm² centre of mass of shape = 108 141.75 centre of mass is 8.02 cm from AB and 4.5 cm from AE

  24. 3m 6m The object below is formed by removing a uniform semi circular disc of radius 3m from a second uniform semi circular disc of radius 6m. Calculate the centre of mass of the object.

  25. Using the tabular approach the disc we are interested in can be found by subtracting the small semi circular disc from the larger one.

  26. So the centre of mass is 2.97cm from the centre along the axis of symmetry.

  27. 16cm B C 1.5cm 10cm 3cm 6cm A D 12cm The object below is formed by removing two uniform circular discs from a uniform rectangular lamina. Calculate the centre of mass of the plate.

  28. The same principle of adding and subtracting parts can be applied to this question (120-4.5∏) Plate = Rectangular Lamina - Circle 1 - Circle 2 = 720 – (7.75∏ + 27∏) = 5.77 cm

  29. B 20cm 12cm 16cm C A 2.5 You can find the centre of mass of a framework by using the centre of mass of each rod or wire that makes up the framework The framework ABC is made up of three uniform rods and a semi circular arc. Find the centre of mass of the framework assuming that A is at the origin

  30. 2.6 Laminas in equilibrium A suspended lamina will be in equilibrium when its centre of mass is directly below the point of suspension It can be suspended from a fixed point or a pivot R Fixed point string T pivot O Point of suspension O G G W W

  31. Lets look at the example in the book on page 54

  32. G θ θ G Equilibrium of a uniform lamina on an inclined plane. For an inclined lamina to remain in equilibrium on an inclined plane the line of action of the weight must fall within the side of the lamina that is in contact with the plane (as shown in the diagram below). In the second diagram the object will topple over

  33. 6cm 15cm G θ A uniform rectangular lamina is placed on a plane inclined at an angle θ. Given that the lamina is in limiting equilibrium find the angle θ. By simple trigonometry:

  34. Done

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