I am currently studying moments. The concepts are fairly obvious, but there is one thing I don't understand: the concept of center of mass. Why do we consider that the weight of a uniform object is concentrated on one point.
Let's say that we have a uniform plank of weight 100 N and length 5m, which rests on a pivot which is 2 meters from the left.
Now why do we consider that the weight acts at the center (2.5m) and apply this concept when calculating the clockwise and anticlockwise moment?
Imagine a thin stick that you place on the floor and you want it stand. If you succeed to put the stick so as its weight, considered as acting on the center of mass, pass through the point $O$ of contact with the floor, the stick will stand. Otherwise it will fall.
Let's see why. Let's decompose the weight of the stick into a component along the stick, and one perpendicular to it. The latter component creates a torque around the point $O$. Let's calculate this torque.
$$\tau = g\int_A^B R \ \rho (R) \ \text dR \tag{1},$$
where $R$ is the distance from the point $O$, $\rho$ is the density of the stick per unit length, $A$ and $B$ are the extremities of the stick.
Now let me express $R = R_0 + r$ where $R_0$ is a point whose meaning will appear below. We get
$$\tau = g R_0\int_{R_A - R_0}^{R_B - R_0} \rho (R_0 + r) \ \text dr \ + g \int_{R_A - R_0}^{R_B - R_0}r \ \rho (R_0 + r) \ \text dr. \tag{2}$$
Well, the 1st integral gives
$$\tau = g R_0 M = GR_0 \tag{3}$$
where $G$ is the weight of the stick, and this is the final result if we choose the point $R_0$ in such a way that the 2nd integral in (2) be zero. The point $R_0$ for which the 2nd integral in (2) vanishes is the center-of-mass. As to the result (3), it says that the torque imposed by the weight of the stick is equal to the distance to the center-of-mass, $R_0$, times the weight of the stick as if it were concentrated, all of it, in the center-of-mass.
Two simple examples If the stick has uniform density, then the 2nd integral yields the expression $g\ \rho \frac { (R_B - R_0)^2 - (R_A - R_0)^2}{2}$ which is clearly zero for $R_0$ chosen in the middle of the stick. But if, say, the lower third of the stick is has four times the linear density of the rest, the 2nd integral is zero for $R_0$ chosen at 1/3 height, i.e. $R_A - R_0 = -L/3$, where L is the stick length, and $R_B - R_0 = 2L/3$. Indeed, $g\frac {1}{2}[-4\rho \frac {L^2}{9} + \rho \frac {4L^2}{9}]$.
The center of mass is defined as it is because the center of mass obeys Newton's second law. Consider a collection of particles, labeled by an index $i$. Then each particle obeys $$\mathbf{F}_i=\dot{\mathbf{p}}_i$$ We decompose the force into the interaction between particles and an external force $$\mathbf{F}_i=\sum_{j\ne i}\mathbf{F}_{ij}+\mathbf{F}^\text{ext}_i$$ Consider the sum of all forces $$\tag{1}\sum_i \mathbf{F}_i=\sum_{i,j\text{ with }j\ne i}\mathbf{F}_{ij}+\sum_i \mathbf{F}^\text{ext}_i=\sum_{i<j}(\mathbf{F}_{ij}+\mathbf{F}_{ji})+\sum_i \mathbf{F}^\text{ext}_i$$ The first term vanishes by Newton's third law. We define $$\sum_i \mathbf{F}^\text{ext}_i=\mathbf{F}^\text{ext},\quad M=\sum_i m_i$$ The center of mass is $$\mathbf{R}=\frac{1}{M}\sum_i m_i\mathbf{r}_i$$ Taking the second derivative of this and using (1), we see that the center of mass obeys Newton's second law $$\mathbf{F}^\text{ext}=M\ddot{\mathbf{R}}$$ This can be generalized to continua by taking limits and integrals where appropriate.
Now if the object mass is uniform throughout the object then :
let masses of a single point on the object be: m_1 , m_2, m_3... m_n
let the displacements of that point from the centre be: x_1 ,x_2, x_3...m_n
Remember : The sum of individual moments equals the total moment.
Let M (total mass of the object) be : m_1 + m_2 +...+ m_n
And let X (net displacement from the centre point) of the object be : x_1 + x_2+...+ x_n
[which must equal to zero as the displacement taken from either side of the object will cancel out to equate to zero= centre point.] Hence X = centre point.
And as sum of individual moments = total moment, therefore MX = m_1x_1 + m_2x_2... + m_nx_n
Hence in a torque problem, the mass of a uniform object is always taken to be at the centre because all the moments are 'Concentrated' in the middle.
