Question

Three uniform spheres are placed in the xy plane as follows: The center of mass of an 8.10-kg sphere is located at (2.00, 2.00) m, the center of mass of a 5.00-kg sphere is located at (0, 4.70) m, and the center of mass of a 2.00-kg sphere is located at (3.20, 0) m. Where should a 6.00-kg sphere be placed if the center of mass of the four-sphere system is to be located at the origin? m m Need Help? Read it

Answer 1

Approximately
`(-3.77,\, -6.62)\; {\rm m}`.

Step-by-step explanation:

The
`x`-coordinate of the center of mass of this system is the average
`x\!`-coordinate of the spheres, weighted according to the mass of each sphere.

For example, if a sphere of mass
`m_(A)` is at
`(x_(A),\, y_(A))`, a sphere of mass
`m_(B)` is at
`(x_(B),\, y_(B))`, a sphere of mass
`m_(C)` is at
`(x_(C),\, y_(C))`, and a sphere of mass
`m_(D)` is at
`(x_(D),\, y_(D))`, the
`x`-coordinate of the center of mass would be at:

`\begin{aligned} (\sum\limits_(j=1)^(n) \left[m_(j)\, x_(j)\right])/(\sum\limits_(j=1)^(n) \left[m_(j)\right]) = (m_(A)\, x_(A) + m_(B)\, x_(B) + m_(C) \, x_(C) + m_(D)\, x_(D))/(m_(A) + m_(B) + m_(C) + m_(D))\end{aligned}`.

Similarly, the
`y`-coordinate of the center of mass would be the weighted average of the
`y\!`-coordinates of the spheres:

`\begin{aligned} (\sum\limits_(j=1)^(n) \left[m_(j)\, y_(j)\right])/(\sum\limits_(j=1)^(n) \left[m_(j)\right]) = (m_(A)\, y_(A) + m_(B)\, y_(B) + m_(C) \, y_(C) + m_(D)\, y_(D))/(m_(A) + m_(B) + m_(C) + m_(D))\end{aligned}`.

In this question, the mass of each sphere has been given. The position of sphere
`A`,
`B`, and
`C` are also given.

The goal is to find the coordinates
`(x_(D),\, y_(D))` of sphere
`D` such that the center of mass is at the origin
`(0,\, 0)`. To do so, set the
`x`- and
`y`-coordinates of the center of mass to
`0` and obtain two equations. Solve these two equations to obtain
`x_(D)` and
`y_(D)`.

Given the mass and position of the other spheres, the expression for the
`x`-coordinate of the center of mass would be:

`\begin{aligned} &(m_(A)\, x_(A) + m_(B)\, x_(B) + m_(C) \, x_(C) + m_(D)\, x_(D))/(m_(A) + m_(B) + m_(C) + m_(D)) \\ =\; & (8.10\, (2.00) + 5.00\, (0) + 2.00\, (3.20) + (6.00)\, x_(D))/(8.10 + 5.00 + 2.00 + 6.00)\end{aligned}`.

Set this expression to
`0` (
`x`-coordinate of the origin) and solve for the
`x_(D)`:

`\displaystyle (8.10\, (2.00) + 5.00\, (0) + 2.00\, (3.20) + (6.00)\, x_(D))/(8.10 + 5.00 + 2.00 + 6.00) = 0`.

Note that the denominator can be eliminated:

`8.10\, (2.00) + 5.00\, (0) + 2.00\, (3.20) + (6.00)\, x_(D) = 0`.

`\begin{aligned} x_(D) &= -(8.10\, (2.00) + 5.00\, (0) + 2.00\, (3.20))/(6.00)\\ &\approx -3.77\end{aligned}`.

In other words, the
`x`-coordinate of sphere
`D` should be approximately
`(-3.77)` meters.

Similarly, set the expression for the
`y`-coordinate of the center of mass to
`0` and solve for
`y_(D)`:

`\begin{aligned} &(m_(A)\, y_(A) + m_(B)\, y_(B) + m_(C) \, y_(C) + m_(D)\, y_(D))/(m_(A) + m_(B) + m_(C) + m_(D)) \\ =\; & (8.10\, (2.00) + 5.00\, (4.70) + 2.00\, (0) + (6.00)\, y_(D))/(8.10 + 5.00 + 2.00 + 6.00)\end{aligned}`.

`\begin{aligned} (8.10\, (2.00) + 5.00\, (4.70) + 2.00\, (0) + (6.00)\, y_(D))/(8.10 + 5.00 + 2.00 + 6.00) = 0\end{aligned}`.

`\begin{aligned} 8.10\, (2.00) + 5.00\, (4.70) + 2.00\, (0) + (6.00)\, y_(D) = 0\end{aligned}`.

`\begin{aligned} y_(D) &= -(8.10\, (2.00) + 5.00\, (4.70) + 2.00\, (0))/(6.00)\\ &\approx -6.62\end{aligned}`.

In other words, the
`y`-coordinate of sphere
`D` should be approximately
`(-6.62)` meters.

Therefore, the position of this
`6.00\; {\rm kg}` sphere should be approximately
`(-3.77,\, -6.62)` meters.