# Moment of inertia of 4 particle system

Four particles are connected by rigid rods of negligible mass. The origin is at the center of the rectangle. The system rotates in the $xy$ plane about the z axis with an angular speed of $6$ rad/s. Calculate the moment of inertia of the system about the $z$ axis. The system looks as follows: where the $d(m_1, m_2) = 4$, $d(m_1, m_3) = 6$ and * represents the origin.The solution that I have seems to calculate the moment of inertia about that $z$-axis, that is:

The distance from each mass to the origin: $r^2 = (3m)^2+ (2m)^2 = 13m^2$ and $\sum\limits_{i=1}^4 m_ir^2 = 3kg*13m^2 + 2kg*13m^2 + 2kg*13m^2 + 4kg*13m^2= 143kg \ m^2$.

This does not seem correct, since if we calculate the center of mass we find that $\bar{x} = \frac{1}{11kg} 3kg*(-2m) + 2kg *(2m) + 2kg*(-2m) + 4kg*(2m) = \frac{2}{11}m$ $\bar{y} = \frac{1}{11kg} 3kg*(3m) + 2kg *(3m) + 2kg*(-3m) + 4kg*(-3m) = -\frac{3}{11}m$

Since the center of mass is not located at the origin but at $\Big(\frac{2}{11}, -\frac{3}{11}\Big)$, shouldn't we use the parallel axis theorem - where $I = I_{cm} + Md^2$ - to compute the moment of inertia?

Hence my solution would be:

distances:

\begin{aligned} r_1^2 &= \Big(\frac{35}{11}\Big)^2 + \Big(\frac{36}{11}\Big)^2 = \frac{2521}{121}\quad \text{(upper left particle)}\\ r_2^2 &= \Big(\frac{31}{11}\Big)^2 + \Big(\frac{36}{11}\Big)^2 = \frac{2257}{121}\quad \text{(upper right particle)}\\ r_3^2 &= \Big(\frac{35}{11}\Big)^2 + \Big(\frac{30}{11}\Big)^2 = \frac{2125}{121}\quad \text{(lower left particle)}\\ r_4^2 &= \Big(\frac{31}{11}\Big)^2 + \Big(\frac{30}{11}\Big)^2 = \frac{1861}{121}\quad \text{(lower right particle)}\\ r_5^2 &= \Big(\frac{2}{11}\Big)^2 + \Big(\frac{3}{11}\Big)^2 = \frac{13}{121}\quad \text{(origin to center of mass)} \end{aligned}

and $Md^2 = \sum\limits_{i=1}^4 m_ir_i^2$

I'm not sure about this one, but for the moment of inertia at the center of mass $I_{cm}$ I'm thinking that it could be modeled as a single particle of mass $11kg$ rotating about the $z$-axis which would give $I_{cm} = \frac{1}{2}mr_5^2 = \frac{1}{2}(11kg)*\frac{13}{121} = \frac{143}{242}kg \ m^2$. Putting it all together we would then arrive at:

$$I = \Big(\frac{143}{242} + \sum\limits_{i=1}^4 m_ir_i^2\Big) \ kg \ m^2$$

• The vectors $\mathbf{r}_{i}$ in the Figure are position vectors from the $xy\:$ origin, not to be confused with the distances $r_{i}$ from the center of mass used by the question owner in his trial. – Frobenius May 4 '16 at 23:12
• I believe the parallel axis theorem overcomplicates the problem. This question is similar to solving the moment of inertia for an unevenly weighted dumbbell (2 point masses on a massless rod), which you surely wouldn't use the parallel axis theorem to solve. – Snyder005 May 4 '16 at 23:44

## 3 Answers

shouldn't we use the parallel axis theorem ... to compute the moment of inertia?

You could...if you already had the moment of inertia of the object about its center of mass. Since you don't, it's far easier to simply sum the moments of inertia about the $z$ axis. You have made some errors in you calculation of distances. Let $$\mathbf{r}^{\prime}_{k}=\mathbf{r}_{k}-\mathbf{r}_{_{CM}}$$

Then

\begin{aligned} r^{\prime\;2}_1 &= \Big(\frac{24}{11}\Big)^2 + \Big(\frac{36}{11}\Big)^2 = \frac{1872}{121}\quad \text{(upper left particle)}\\ r^{\prime\;2}_2 &= \Big(\frac{20}{11}\Big)^2 + \Big(\frac{36}{11}\Big)^2 = \frac{1696}{121}\quad \text{(upper right particle)}\\ r^{\prime\;2}_3 &= \Big(\frac{24}{11}\Big)^2 + \Big(\frac{30}{11}\Big)^2 = \frac{1476}{121}\quad \text{(lower left particle)}\\ r^{\prime\;2}_4 &= \Big(\frac{20}{11}\Big)^2 + \Big(\frac{30}{11}\Big)^2 = \frac{1300}{121}\quad \text{(lower right particle)}\\ r^{\prime\;2}_{_{CM}} & = \Big(\frac{2}{11}\Big)^2 + \Big(\frac{3}{11}\Big)^2 = \frac{13}{121}\quad \text{(origin to center of mass)} \end{aligned}

Then $m_1 r^{\prime\;2}_1 +m_2 r^{\prime\;2}_2 +m_3 r^{\prime\;2}_3 +m_4r^{\prime\;2}_4 + (m_1+m_2+m_3+m_4)r^{\prime\;2}_{_{CM}} = 143$ as before showing how much easier it is to evaluate the moment of inertia about the $z$-axis without using the parallel axis theorem.

• Can we follow the same procedure when the point masses lie only in one quadrant such that the axes (X & Y ) don't intersect within the area bounded by them ? If not , what should we do ? – Nehal Samee Jun 20 '18 at 15:37
• Why would it not work? – Farcher Jun 20 '18 at 17:26
• I was confused reading Wikipedia article that the XY planes must lie on the object ... – Nehal Samee Jun 21 '18 at 5:53

Your first calculation is correct. Nothing wrong with it. The mystery for me is why you go to so much trouble trying to disprove it! You give no reason for your impression that "this does not seem correct".