Finding the Center of Mass of a System of Rods

[Litcyb]

Homework Statement

Three uniform thin rods, each of length L = 22 cm, form an inverted U. The vertical cords each have a mass of 14 grams. The horizontal rod has a mass of 42 grams. What are the x coordinate and y coordinate of the system's center of mass?

Homework Equations

xcom= (x₁m₁+x₂m₂+x₃m₃)/M
ycom= (y₁m₁+y₂m₂+y₃m₃)/M

The Attempt at a Solution

I managed to solve this question, but I think I'm a bit confused on the way I solved it.

I set up a coordinate system in which the left top corner of the table is (0,0).
The way I managed to do this is by first finding the center of mass of each part of the table. For instance, the point of the center of mass of the left leg is (0,L/2); for the top of the table, it is (L/2,0); and for the right leg, it would be (L,L/2). And then I would apply it to the equation above.

Now, my question is, do we always have to find the center of mass of every side first in order to calculate the center of mass of the whole system?

Thank you in advance.

 

[Sunil Simha]

1.While solving such center of mass problems, it always helps to consider symmetry (if any) of the situation. For example, in the mentioned problem we can draw an imaginary plane bisecting the table along the y axis. One side of the table is a mirror image of the other (regarding its mass) and thus its center of mass must lie on that plane. Thus the x coordinate can be obtained without any calculation.

2.It always helps to know the center of mass of common mass distributions like a uniform sphere, a uniform rod etc. Thus the system can be split into such simple parts and then the center of mass can be found using these known centers.

 

[vela]

Homework Statement

Three uniform thin rods, each of length L = 22 cm, form an inverted U. The vertical cords each have a mass of 14 grams. The horizontal rod has a mass of 42 grams. What are the x coordinate and y coordinate of the system's center of mass?

Homework Equations

xcom= (x₁m₁+x₂m₂+x₃m₃)/M
ycom= (y₁m₁+y₂m₂+y₃m₃)/M

The Attempt at a Solution

I managed to solve this question, but I think I'm a bit confused on the way I solved it.

I set up a coordinate system in which the left top corner of the table is (0,0).
The way I managed to do this is by first finding the center of mass of each part of the table. For instance, the point of the center of mass of the left leg is (0,L/2); for the top of the table, it is (L/2,0); and for the right leg, it would be (L,L/2). And then I would apply it to the equation above.

Now, my question is, do we always have to find the center of mass of every side first in order to calculate the center of mass of the whole system?

Thank you in advance.

No, but it's the most straightforward way. Did you have another way of solving this kind of problem in mind?

 

[Litcyb]

No. at first i thought we just had to take the position of each side and multiply it by the mass. but according to the book, we first had to find center of mass of the sides. Whats the appropriate way of solving these kinds of problems? because i would really like to have a good overall idea on how to approach these problems.

 

[vela]

What do you mean by "position of each side"? Isn't that the same as the location of the center of mass of each side? It sounds like you're agreeing with what the book is saying.

 

[Chestermiller]

No. at first i thought we just had to take the position of each side and multiply it by the mass. but according to the book, we first had to find center of mass of the sides. Whats the appropriate way of solving these kinds of problems? because i would really like to have a good overall idea on how to approach these problems.

There is a more general way of expressing the location of the center of mass in terms of volume integrals involving the density. Have you had enough math to understand how to do volume integrals?

 

[Litcyb]

Yes, I have. so you're telling me to take the triple integral? but we don't have the density.

 

[Doc Al]

Yes, I have. so you're telling me to take the triple integral? but we don't have the density.

Since the rods are presumed "thin", you'd use the linear density (λ) instead of the volume density (ρ).

But that will just lead to finding the center of mass of each rod, then using that to find the overall center of mass.

 

[Chestermiller]

Yes, I have. so you're telling me to take the triple integral? but we don't have the density.

If the density is uniform, then you don't need to know the density.

Calculating the center of mass is sort of like calculating your grade point average. You take a weighted average of the grades, weighted in terms of the number of credits.

If $\vec r$ represents a position vector from an arbitrary origin to a given parcel of mass, and $\vec r_c$ represents the position vector to the center of mass, then

$$\vec r_c=\frac{\int_V{ \rho \vec r dV}}{\int_V{ \rho dV}}$$

where ρ is the density. Notice that, if ρ is constant, it cancels out.