Center of Gravity Calc: $\bar{y}$

[Dustinsfl]

The y center of gravity is found by
\[
\bar{y} = \frac{M_x}{M}.
\]
In my calculus book, they are always finding the center of gravity of a flat object or objects spaced apart (like planets).

Suppose we have a block with length and width \(1\) and height \(a\). On top of this block is another with height \(a\), width \(1\), and length \(1 - a\). This block is flush with the first block except it is shorter by \(a\) so one side has that gap. It has uniform mass.

Suppose the density is then \(\rho\). The mass of the first block is \(V\cdot \rho = a\rho\) and the of the second object is \(V\cdot\rho = a(1 - a)\rho\). The total mass is then \(M_1 + M_2 = a\rho(2 - a) = M\).

What is \(M_x\)?

 

[Sudharaka]

The y center of gravity is found by
\[
\bar{y} = \frac{M_x}{M}.
\]
In my calculus book, they are always finding the center of gravity of a flat object or objects spaced apart (like planets).

Suppose we have a block with length and width \(1\) and height \(a\). On top of this block is another with height \(a\), width \(1\), and length \(1 - a\). This block is flush with the first block except it is shorter by \(a\) so one side has that gap. It has uniform mass.

Suppose the density is then \(\rho\). The mass of the first block is \(V\cdot \rho = a\rho\) and the of the second object is \(V\cdot\rho = a(1 - a)\rho\). The total mass is then \(M_1 + M_2 = a\rho(2 - a) = M\).

What is \(M_x\)?

If the object is place with the $1-a$ side parallel to the x-axis the center of mass could be found by summing up the centre of mass of each individual object. So in this case, the x coordinate of the center of mass of the bottom object is 0.5 units. And for the second object it's $\frac{1-a}{2}$ units. Therefore,

\[\overline{y}=\frac{0.5+\frac{1-a}{2}}{M}=\frac{2-a}{2a\rho(2 - a)}=\frac{1}{2a\rho}\]

 

[HallsofIvy]

The center of mass of combined objects is the weighted average of the centers of masses of the two objects, weighted by their masses. That is if one object has mass $$m_1$$ and center of mass [tex](\overline{x}_1,\overline{y}_1, \overline{z}_1)[/itex] and the other mass $$m_2$$ and center of mass $$(\overline{x}_2, \overline{y}_2, \overline{z}_2)$$ then the combined object has center of mass
$$\left(\frac{m_1\overline{x}_1+ m2\overline{x}_2}{m_1+ m_2}, \frac{m_1\overline{y}_1+ m2\overline{y}_2}{m_1+ m_2}, \frac{m_1\overline{z}_1+ m2\overline{z}_2}{m_1+ m_2}\right)$$.

Here, assuming the smaller object is sitting on top of the larger with "left" (x-axis) sides flush then the lower has mass $$m_1= \rho a(1)(1)= \rho a$$ and center of mass $$(1/2, 1/2, a/2)$$ while the upper object has mass $$\rho a(1)(1- a)= \rho a(1- a)$$ and center of mass $$((1- a)/2, 1/2, (3/2)a)$$.

So the center of mass of the combined object is at
$$\left(\frac{\rho a(1/2)+ \rho a(1- a)(1-a)/2}{\rho a(2-a)}, \frac{\rho a(1/2)+ \rho a(1-a)(1/2)}{\rho a(2- a)},\frac{\rho a^2/2+ 3\rho a^2(1- a)}{\rho a(2- a)}\right)$$