# How can I find the rotational inertia of two joined bars with respect of an axis?

I found this problem in a book whose author is unknown as it is merely a collection of riddles in mechanics, and it has left me very confused on how to approach the rotational inertia when two objects are joined.

The problem is as follows:

The figure from below shows two homogeneous bars $$OA$$ and $$BC$$ has $$48\,kg$$ of mass each are joined at $$A$$ making a single body. Find the rotational inertia (moment of inertia) in $$kg\cdot m^2$$ with respect of an axis perpendicular to the point $$O$$.

The alternatives given in my book are as follows:

$$\begin{array}{ll} 1.&17\,kg\cdot m^2\\ 2.&5\,kg\cdot m^2\\ 3.&12\,kg\cdot m^2\\ 4.&24\,kg\cdot m^2\\ 5.&10\,kg\cdot m^2\\ \end{array}$$

I'm not very sure how to assess this problem in particular. What I think it comes into play is the Steiner theorem which relates that the rotational inertia around an axis from a distance as follows:

$$I_{\textrm{new center}}=I_{\textrm{point}}+Md^2$$

But this problem in particular is making me confused.

What I've attempted to do is to treat the system as two independent bars and then using the principle of superposition of moment of inertia.

For the rod about one end the rotational inertia would be:

$$I=\frac{1}{3}ML^2$$

But the problem arises for the bar which is vertical. Should it be okay to think this bar works as a particle?.

$$I=MR^2$$

Then this would become into: (for purposes of brevity I'm ommiting units)

$$I=\frac{1}{3}ML^2 + MR^2$$

$$I=\frac{1}{3}(48)\left(0.5\right)^2 + 48\left(0.5\right)^2$$

$$I=49\,kg\cdot m^2$$

But this doesn't check with any of the alternatives, what could I be doing wrong here?.

• What does "an axis perpendicular to the point O" mean exactly? It sounds nonsensical to me. – NickD Dec 23 '19 at 3:49
• @NickD I was also confused about the same. In one of the answers it has pointed that Steiner theorem can be used also for an axis of rotation perpendicular to the center of mass of the object. – Chris Steinbeck Bell Dec 23 '19 at 5:59
• "Perpendicular to the CM", "perpendicular to a point": these are meaningless statements. A line is perpendicular to another line or to a surface, not to a point. – NickD Dec 23 '19 at 16:06

Remember that the rotational inertia of an object depends heavily on how the mass is distributed along the rotational axis, so it wouldn't make much sense to treat the BC bar as a particle. You are almost close to the answer though. You can use the theorem you mentioned to calculate the rotational inertia of the BC bar. The theorem propertly reads: $$I_{\text{new}}=I_{\text{center of mass}}+Md^2$$ where $$I_{\text{center of mass}}$$ is the rotational inertia the bar would have if it rotated around its center of mass, and $$d$$ is the distance from the center of mass of the BC bar to the new rotatioanl axis. Thus, the rotational inertia of the BC bar would be $$I_{\text{new}}=\frac{1}{12}ML^2+Md^2$$ Since in this case, $$L=d=0.5\text{ m}$$ $$I_{\text{new}}=\frac{13}{12}ML^2$$ Thus, $$I=\frac{1}{3}ML^2+\frac{13}{12}ML^2=\frac{17}{12}ML^2=\frac{17}{12}(48\text{ kg})(0.5\text{ m})^2=17\text{ kg}\cdot\text{m}^2$$
Edit: Since Steiner's theorem works only if the new axis is parallel to the axis located at the center of mass, I considered that the new axis was located at point O and was perpendicular to the plane of the sketch (this is a bit hard to tell and is pretty much an assumption, since the problem doesn't specify this very well). The inertia $$\frac{1}{12}ML^2$$ that I used is for a bar whose rotational axis is perpendicular to its length, making this axis parallel to the new axis I mentioned.
• Interesting!, I believe you should include your last comment as part of your answer because it does clarify the intended use of the Steiner's theorem as it was not clear how was understood the axis of rotation. What I was seeing was an axis going along the line connecting $O$ and the marker of the ruler located over the title in my sketch. If you look the axis in that way is perpendicular, but if you put it going outside the plane (in other words crossing the computer screen) it will make sense with the stated theorem. Am I right with this?. – Chris Steinbeck Bell Dec 23 '19 at 8:53