# Centre of Mass Often times, we want to generalise and model the motion of a larger system, such as a bottle, or a stapler. However, it might not be possible to model a system as a single particle, requiring us to treat an arbitrary point within the object as its **centre of mass**. The centre of mass of an object is defined as: >***The point that moves as though all of the system's mass were concentrated there or if all external forces were applied there.*** Take the example of a system with two particles, each particle with a distinct mass $m_1$ and $m_2$. The centre of mass can be found as a function of the distance $d$ between the two particles: $x_{com} = \frac{m_{2}}{m_{1}+m_{2}}d$ So it'll allow the position of the centre of mass within a system to change as the masses of the particle change. Even then, this equation only holds if one of the particles exists on an arbitrary standard axis (like the x-axis!), so we've a more generalised equation to help us with systems that are not on the standard coordinate axes. $x_{com} = \frac{m_{1}x_{1} + m_{2}{x_{2}}}{m_{1}+m_{2}}$ Where $x_1$ and $x_2$ are the initial positions of $m_1$ and $m_2$. So if there are more particles, all we have to do is add $m_nx_n$ to the equation divided by the mass of the particle itself. So the denominator can just be written as $M$, the total mass of the object. The centre of mass can therefore be defined as a function of the sum of the positions of all the particles: $x_{com} = \frac{1}{M}\sum_{i=1}^nm_{i}x_{i}$ Where $x$ allows for the average displacements from the centre of mass of all particles in the system to be accounted for, up to a certain number of particles $n$. Finally, if need be (although cases are very rare!), we can find the centre of mass of a three-dimensional object! It's quite a bit of work though, and for the context of iGCSEs and IB may be obsolete. Very, very fun regardless! >[!Abstract]- Three-Dimensional Centre of Masses >If the particles are distributed in three dimensions (e.g in a stapler), the centre of mass becomes a coordinate where: >$x_{com} = \frac{1}{M}\sum_{i=1}^nm_{i}x_{i}$ >$y_{com} = \frac{1}{M}\sum_{i=1}^nm_{i}y_{i}$ >$z_{com} = \frac{1}{M}\sum_{i=1}^nm_{i}z_{i}$ >Where $M$ is the collective mass of the system of particles. >So the individual $x$, $y$ and $z$ components of each particle have to be evaluated! Lots of work, right? >We can create a displacement vector $\overrightarrow{r}$ that defines the coordinate of the centre of mass in relation to each other by adding up the position vectors of the object relative to the position of the centre of mass, like so: >$\overrightarrow{r}_{com} = x_{com}\hat{i} + y_{com}\hat{j} + z_{com}\hat{k}$ >So finding the centre of mass just becomes: >$\overrightarrow{r}_{com} = \frac{1}{M}\sum_{i=1}^nm_{i}\overrightarrow{r}_{i}$ In the event that there is a large mass (or an amount of particles which is too large to parse), we need to use another method: >[!Abstract]- The Centre of Mass of a large object >As there are too many particles, we need to rewrite the centre of mass equations as a function of the total mass $M$. We can thus substitute each of the masses of the particles with a value for the coordinate component, given by the coordinate variable ($x$, $y$, $z$). The integral acts as a sort of sum for all the coordinate components for the discrete particles, allowing us to find the COM. >We can rewrite the equations defining each of the COM coordinates as: >$x_{com} = \frac{1}{M} \int x\, dm $ >$y_{com} = \frac{1}{M} \int y\, dm $ >$z_{com} = \frac{1}{M} \int z\, dm $ >Integrating along the mass elements across each axis, given $dm$ is the mass element along the axis. > >Since it is difficult to find the centre of mass for an irregular object (like a TV) like this, we can use the substitution $\rho = M/V$ for density to create: >$x_{com} = \frac{1}{V} \int x\, dV $ >$y_{com} = \frac{1}{V} \int y\, dV $ >$z_{com} = \frac{1}{V} \int z\, dV $ >Where $dV$ is the volume occupied by the mass elements $dm$. Don't worry! Most objects in GCSEs have an implied centre of mass in the dead centre of the object (either that or it's explicitly stated with a complimentary free-body diagram). It's only in IB that you begin to see lopsided objects. # Example Problems >[!Example]- Module 9-1 Problem 1 >A 2.00 kg particle has the xy coordinates (-1.20 m, 0.500 m), and a 4.00 kg particle has the xy coordinates (0.600 m, -0.750 m). Both lie on a horizontal plane. At what (a) x and (b) y coordinates must you place a 3.00 kg particle such that the centre of mass of the three-particle system has the coordinates (-0.500 m, -0.700 m)? > >**Solution a)** >Let's find the total x position vector of the two particles and the unknown third relative to our target centre of mass coordinate, taking care of signs. We can assign weights to the variables, giving us: >$\overrightarrow{r}_{x} = 2 \times -0.7 + 4 \times 1.1 + 3x_{x} = 3 + 3x_{x}$ >The x coordinate of the third particle can thus be easily found: >$3x_{x} = -3$ >Meaning relative to our target centre of mass the third particle is 1 unit behind it. This gives us: >$x_{3} = -1$ > >**Solution b)** >Just do the same for the $y$ coordinates! >$-0.7 = 2 \times 1.2 + 2 \times -0.05 + 3x_{y} = 2.3$ >$3x_{y} = -2.3 - 0.7 = -3$ >$x_{y} = -1$ >The total coordinate of particle 3 is therefore $(-1, -1)$. # What's Next? To navigate back to the physics menu, click here: [[The (Incomplete) Physics Almanac]] For more mechanics, click here: [[Mechanics - A Contents Page]] To check out some of the prerequisite maths, click here: [[Maths Contents Page]]