A uniform thin bar formed into a L-shaped object of mass $m=2.5kg$ with a longer side of length $l=0.8m$ and a shorter side of length $l/2$. Initially the object is positioned with one end at the origin and the longer side along the $x$ axis. The centre of mass of the object has coordinates $r_{cm}=\frac{2l}{3} \hat i -\frac{l}{12} \hat j$ . I should also add that the object is held in position by a massless wire that makes an angle $\phi=50 ^{\circ} $ with the longer side of the object.

The object is attached to the wall by a pivot (at the origin). Compute the moment of inertia of the object about an axis through the pivot perpendicular to the plane of the object.

I know that moment of inertia is equal to $I=r^2 m $. I broke down the moment of inertia into two components, one calculating $I_1$ over the longer side of L (length of $l$), and the other calculating $I_2$ over the shorter side of L ( length of $ \frac{l}{2} $ ). However, the correct answer provided clearly states that $I_1=\frac{2}{9}ml^2$ and $I_2= \frac{13}{12}\frac{1}{3} ml^2$ (using the parallel axis theorem). I don't know how to achieve these results, my reasoning behind calculating $I_1$ is:

Since $I=mr^2$,

$I_1=\frac{2}{3}m l^2$ (since the longer side is twice as long as the shorter side)

which clearly gives me an incorrect result.

If someone could explain the logic behind calculating the total moment of inertia of this type of object that would be great!

  • $\begingroup$ The equation is ${\rm d} I_z = (x^2+y^2)\,{\rm d}m$ which you need to integrate over the cross section domain. Your $I=m r^2$ is incorrect, unless $r$ is exactly the radius of gyration of the section. $\endgroup$ Aug 20, 2015 at 14:36
  • $\begingroup$ I tried to draw the situation given but can't reconcile items in the description. "One end at the origin", "longer side along the X axis" and "coordinates of center of mass" are mutually exclusive. Can you add a diagram? $\endgroup$
    – Floris
    Aug 20, 2015 at 15:32
  • $\begingroup$ I found a diagram (very roughly) similar in the following link: d2vlcm61l7u1fs.cloudfront.net/… Ignoring the lengths provided on the L-bar in the picture, in my question the length of the vertical bar is half that of the horizontal bar, and there is also a string attached from the far right side of the horizontal bar to the top of the wall the whole L-bar is attached to, making an angle of $50$ degrees with the horizontal bar. @Floris $\endgroup$
    – mnmakrets
    Aug 20, 2015 at 15:53

3 Answers 3


Without doing integration you need to have a plan:

1) Draw good diagram with a well-defined coordinate system.

2) Consider the object to be composed of two slender rods of uniform density. Based on their lengths, you can assign masses to each rod. Also find the coordinates of the CM of each rod.

3) Calculate the MOI of each rod about its own CM. (Easily found in any physics text)

4) Use the parallel axis theorem and calculate the MOI of each rod about the pivot point.

5) Add these final MOIs together.

  • 1
    $\begingroup$ "Draw a good diagram". Cannot be said enough. $\endgroup$
    – Floris
    Aug 20, 2015 at 15:21
  • $\begingroup$ The centre of mass of the horizontal bar is $\frac{l}{2} $, hence the MOI about its centre of mass would be $I_{cm}\frac{1}{6} l^2 m $ where $m$ is the total mass of the system. Using the parallel axis theorem and calculating the MOI of the horizontal rod about the pivot point gives: $I_{parallel \space axis}= I_{cm}+Md^2 = \frac{1}{6} l^2 m +\frac{2}{3} m (\frac{l}{2})^2 = \frac{2}{6} l^2 m $. However this is the wrong answer for the MOI of the horizontal bar about the pivot, I am not sure what I am doing incorrectly. $\endgroup$
    – mnmakrets
    Aug 20, 2015 at 16:28
  • $\begingroup$ @mnmakrets - the MOI about the center of mass of the horizontal element is $\frac{1}{12}\frac{2m}{3}\ell^2$; to get MOI about the pivot point you have to add $\frac{2m}{3}\left(\frac{\ell}{2}\right)^2$ to that. See my answer for details. $\endgroup$
    – Floris
    Aug 20, 2015 at 16:46

Consider the general case:


Consider a cross section, with density $\rho$ and thickness $w$. The domain is split into two rectangular areas:

  1. The horizontal leg with:

    • Mass $m_1 =\rho w t_1 (\ell_1-t_2)$
    • Centroid $(x_1,y_1) = \left( \frac{\ell_1+t_2}{2}, \frac{t_1}{2} \right)$
    • MMOI about centroid $I_1 = \frac{m_1}{12} \left(t_1^2+(\ell_1-t_2)^2\right)$
  2. The vertical leg with:

    • Mass $m_2 =\rho w t_2 \ell_2$
    • Centroid $(x_2,y_2) = \left( \frac{t_2}{2}, \frac{\ell_2}{2} \right)$
    • MMOI about centroid $I_2 = \frac{m_2}{12} \left(t_2^2+\ell_2^2\right)$
  3. Combined properties

    • Mass $m = m_1+m_2 = \rho w \left( \ell_1 t_1 + t_2(\ell_2 - t_1) \right)$
    • Centroid $ (x_c,y_c) = \left( \frac{m_1 x_1+m_2 x_2}{m_1+m_2}, \frac{m_1 y_1+m_2 y_2}{m_1+m_2} \right)$
    • MMOI about centroid $I_c =\sum \limits_{i=1}^2 \left[I_i + m_i \left( (x_c-x_i)^2 + (y_c-y_i)^2 \right) \right]$

A diagram of the situation:

enter image description here

The moment of inertia of the entire system can be written as the sum of the moment of inertia of each element about its center of mass, plus a component due to the fact that they are rotating about a point that is not its center of mass.

This gives us for $I_1$:

$$\begin{align}I_1 &= \frac{1}{12} \frac{2m}{3} L^2 + \frac{2m}{3} \left(\frac{L}{2}\right)^2\\ &= \frac29 m L^2\end{align}$$

as given in your solution.

A similar approach can be used for $I_2$

  • $\begingroup$ How is the $\frac{1}{12}$ value obtained? $\endgroup$
    – mnmakrets
    Aug 20, 2015 at 17:07
  • $\begingroup$ That is a standard equation for moment of inertia of a rod - you can obtain it by integrating. $\endgroup$
    – Floris
    Aug 20, 2015 at 19:37

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