In summary, the centre of mass for a set of particles is located where the sum of the masses of all the particles is equal to the total mass of the set.
Is the last one supposed to be a 60 degree angle, or an arbitrary angle?
Anyway, first you need to choose a convenient coordinate system in each case (origin + two orthogonal axes), preferably making the most of the symmetries present. Then, treat each each rod as a point particle of mass ##\alpha L## located at the centre of where the rod was, with ##\alpha## just being a scaling constant (if you'd prefer, it's fine here to just give each rod unit mass).
In each case, the centre of mass satisfies$$\mathbf{x}_{\text{CM}} = \frac{1}{M} \sum_i m_i \mathbf{x}_i$$where ##M## is the total mass, and the summation is over all of the rods. Now it's your turn! Please show an attempt.
@etotheipi is right in his post #2, however his method is algebraic/analytic and it requires setting up equations.
If you prefer there is a pure geometrical method, with pencil and ruler you can use to find the CoM in cases like this.
To outline the geometrical method:
We know that the CoM of each rod separately is at its center. So mark the center of each rod with a pencil. Then choose (any) two centers and draw the line segment between them with a ruler. Then mark the center of this line segment, that's where the combined CoM of the previously two chosen rods is. Then choose this center and the center of the third rod and draw the line between them. At the center of this line is where the combined CoM of the three rods is.
Hope the above was helpful to you.
#4
etotheipi
Yes it's equivalent to determining the centroid of the points at the centre of each rod. For the last two that's just the centre of a line segment, for the first two you can use that the centroid is a third of the way up the median of the triangle
yes well my method at post #2 is not entirely correct, it gives correct results only for (3) and (4), for (1) and (2) follow post #4 please.
#6
etotheipi
Delta2 said:
yes well my method at post #2 is not entirely correct, it gives correct results only for (3) and (4), for (1) and (2) follow post #4 please.
It's because when you find the centre between the first two points, you're effectively replacing those two point masses with a single mass of ##2m##, not ##m##, at that new point.
So to find the CM you'd need to split the second line segment (between the new ##2m## mass and the remaining ##m## mass) into thirds instead.
If you start from the basis that the location is only intended in two dimensions and if you already know the location of the center of mass of each segment of length L assuming constant density, you can reduce the calculation for each figure to two equations
$$ \rho_i = \dfrac {M_i} {L_i}= \dfrac {M} {L}= \dfrac {\sum\limits_i^nM_i} {\sum\limits_i^nL_i}$$
$$ \displaystyle \mathbf x_{CMT} = \dfrac {\sum\limits_i^n (\mathbf x_{CM_i} \cdot M_i) }{\sum\limits_i^n M_i} $$
(Let's all remember that this is the student's homework problem. We should let them respond to see if they can use the hints they've been given so far...)