Practise › Questions › Centres of mass of plane figures and frameworks
Centres of mass of plane figures and frameworks questions
Cut a shape into pieces whose centres you already know, weight each by its area, and add. Removing a piece is the same sum with a minus sign.
6 original questions · 22 marks · the centres of mass of plane figures and frameworks notes · Further Mechanics 2
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
State the centre of mass of a uniform triangular lamina and of a uniform rectangular one, and say how to use symmetry to save work.
Worked answer
A rectangle's centre of mass is at its centre, where the diagonals cross. A triangle's is at the centroid, a third of the way from each side towards the opposite vertex. Any axis of symmetry must contain the centre of mass, so a symmetric shape needs only one coordinate calculating. B1 for the two standard positions, B1 for the symmetry argument.A uniform lamina is made of a 6 by 2 rectangle with corners (0, 0) and (6, 2), and a 2 by 4 rectangle with corners (0, 2) and (2, 6). Find its centre of mass.
Worked answer
Split the shape into the two rectangles and treat each as a particle at its own centre. Areas 12 and 8, total 20, with centres (3, 1) and (1, 4).
x(G) = (12 × 3 + 8 × 1)/20 = 44/20 = 2.2.
y(G) = (12 × 1 + 8 × 4)/20 = 44/20 = 2.2.
M1 for the areas acting at their own centres, A1 for the x coordinate, A1 for the y coordinate.
The L is symmetric about the line y = x, so the two coordinates were bound to agree. That makes a quick check on the arithmetic.A uniform disc of radius 4 centred at the origin has a disc of radius 2 centred at (2, 0) removed. Find the centre of mass of what is left.
Worked answer
Areas 16π and 4π, leaving 12π. By symmetry the centre of mass lies on the x-axis. Treating the hole as a negative area: x(G) = (16π × 0 − 4π × 2)/(12π) = −8/12 = −2/3. B1 for the symmetry, M1 for treating the hole as a negative area, A1 for the moment equation, A1 for the position. It has moved away from the hole, as it must, and the π has cancelled throughout.A uniform wire framework consists of three rods forming three sides of a rectangle: from (0, 0) to (6, 0), from (6, 0) to (6, 4), and from (0, 0) to (0, 4). Find its centre of mass.
Worked answer
Lengths 6, 4 and 4, total 14, acting at (3, 0), (6, 2) and (0, 2). x(G) = (18 + 24 + 0)/14 = 42/14 = 3, which symmetry about x = 3 confirms. y(G) = (0 + 8 + 8)/14 = 16/14 = 8/7 ≈ 1.14. M1 for the lengths acting at their midpoints, A1 for the x coordinate, M1 A1 for the y coordinate. It sits below the middle because the long rod lies along the bottom.Explain why a triangular lamina and a triangular wire framework of the same shape have different centres of mass.
Worked answer
The lamina has its mass spread over the area, so each element is weighted by area and the answer is the centroid of the region. The framework has its mass along the perimeter, so each rod is weighted by its length and acts at its own midpoint. Different distributions of the same total mass give different balance points. B1 for the lamina weighted by area, B1 for the framework weighted by rod length, B1 for each rod acting at its own midpoint.A uniform wire of total length 12 cm is bent to form the triangle with vertices (0, 0), (4, 0) and (0, 3). Find the centre of mass of the framework, and compare it with the centre of mass of the uniform triangular lamina having the same three vertices.
Worked answer
For a framework the mass of each rod is proportional to its length, and each rod acts at its own midpoint. Setting the work out as a table of length against midpoint is what earns the method marks.
The three rods have lengths 4, 3 and 5, the last by Pythagoras, giving a total of 12. Their midpoints are (2, 0), (0, 1.5) and (2, 1.5).
x(G) = [4(2) + 3(0) + 5(2)]/12 = 18/12 = 1.5.
y(G) = [4(0) + 3(1.5) + 5(1.5)]/12 = 12/12 = 1.
For the lamina the mass is spread over the area, and the centre of mass is the centroid, the average of the three vertices: x(G) = (0 + 4 + 0)/3 = 4/3 and y(G) = (0 + 0 + 3)/3 = 1.
The two heights agree at y = 1, which is a coincidence of these particular numbers rather than a rule. The x coordinates differ, 1.5 against 4/3, because the framework's heaviest rod is the hypotenuse, whose midpoint is at x = 2, and that pulls the framework to the right.
M1 for the lengths and their midpoints, A1 for the x coordinate, A1 for the y coordinate, M1 A1 for the centroid of the lamina, B1 for the comparison.
Using the centroid formula on a framework is the mistake this question exists to catch. It is only valid for a lamina.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise centres of mass of plane figures and frameworks one question at a time
The player marks nothing for you. It shows one question, waits, then shows the worked answer so you can mark yourself, and brings a question back sooner when it went badly.