Chapter+2+-+CENTRE+OF+MASS

= Centre of Mass = It is the point through which the weight acts. It doesn’t have to be on the object. The moments clockwise about the centre of mass are equal to the moments anticlockwise about the centre of mass. = One Dimension = On a one dimensional system with two particles of equal mass the centre of mass is at the midpoint of the two particles, the weight acts through this point. This means on a uniform rod the weight acts through the centre. If you have more than one particle of different weights. The method for calculating the centre of mass is as follows: /*Diagram to go here*/ Pick a reference point to measure distances to each point from. This equation gives the centre of mass with regards to the reference point you picked.

Examples: A light inflexible plank AB has stones balanced on it x,y,z with masses 2 kg, 6kg and 4 kg respectively. The displacement xy is 2m and the displacement yz is 0.5m. Find the displacement to the centre of mass from y and the weight which acts through that point.


 * Solution: A1. **

=** In Two Dimensions **= media type="custom" key="11544086" Another note for triangles! For a triangle with points A, B and C the centre of mass is given by:
 * The technique for calculating the centre of mass in two dimensions is much the same as for calculating it in one however it must be done twice, once in the x plane and once in the y plane which gives a resulting set of co-ordinates. **
 * Table of two dimensional objects: **

= Life's not so simple! = Sadly most of the questions to do with centre of mass feature more complicated objects, which comprise multiple simpler objects, in order to find the mass of the whole object each simpler object's centre of mass must first be worked out which can then be used to calculate the whole in the ratio of the masses or areas of each lamina. Some people like to do this in a table! An example follows:

= Removing Pieces = If a piece is removed from a standard lamina the centre of mass of the remaining area can be calculated as follows.
 * || Removed piece || remaining area || Full object ||
 * X || Xp || Xr || Xt ||
 * Y || Yp || Yr || Yt ||

Xt = (Ap * Xp + Ar * Xr)/(Ap + Ar) -> rearrage to find Xr! Yt = (Ap * Yp+ Ar * Yr)/(Ap + Ar) -> rearrage to find Yr!

= Toppling & Hanging = If the lamina is freely suspended about a point and allowed to rotate, the lamina rotates until the line of action of gravity and line of action of the upwards force and in line, this allows the lamina to be in equilibrium, this means that the line will always pass through the centre of mass as gravity acts on the object through that point.

If the object is sat on a surface and the line of action of the force due to gravity is outside the base of the object the object will topple, unless there are other forces acting on it, in which case it may not. This allows you to calculate at what angle the surface on which the object rests needs to be at before the object will topple. Hanging

=** Real life examples **=


 * Tom is reclining on a desk-chair hybrid, find his centre of mass **


 * /* And picture */ **

solution B1

= An unlikely situation! =

Fortunate has decided to bring a cake into further maths and wants to balance it on the end of her finger, unfortunately she has already eaten a large bite out of the middle. find it’s centre of mass. Height: CAKE :5cm BITE: 2cm Width: CAKE 20cm BITE: 10cm The bite is perfectly centered **solution B2

Solutions A1. Taking y as the reference point and left as –ve. Distance x mass of each X: -2 x 2 = -4 kgm Y: 0 x 6 = 0 kgm Z: 0.5 x 4 = 2 kgm Total mass = 12kg Centre of mass = = -  m Weight: (total mass)*g = 12g = 117.6N B1. // tom’s chair //

// B2. // The area of the bite is 20 cm3, the area of the cake is 50 cm3.
 * || Cake-Bite || Bite || Cake ||
 * X ||  || 10 || 10 ||
 * Y ||  || 1 || 2.5 ||

x = 10 as the same 2.5 = (20y + x30)/(