# Calculating Rotational Inertia Using Parallel Axis Theorem

I am working on the following physics problem and have run into some trouble

The figure above shows particles $1$ and $2$, each of mass $m$, attached to the ends of a rigid massless rod of length $L_1 + L_2$, with $L_1 = 20cm$ and $L_2 = 80cm$. The rod is held horizontally on the fulcrum and then released. What are the magnitudes of the initial accelerations of (a) particle $1$ and (b) particle $2 \space ?$

My Approach:

So I first considered the net torque when the system is at rest so that I could get to the angular acceleration using the equation $\tau_{net} = I \alpha$. Given that particle $2$ would induce clockwise motion, it is given a negative sign while particle $1$ is given a positive sign because it induces counter-clockwise motion so $\tau_{net} = F_{t2}r_2 – F_{t1}r_1 = I \alpha$. (Where $F_{ti}$ represents the tangent force acting on particle $i$)

Solving for alpha we have that $\alpha = \frac{F_{t2}r_2 – F_{t1}r_1}{I} = \frac{mgL_2 - mgL_1}{I}$. The next step is then to find the rotational inertia. Now after consulting with my solutions manual I see that this can be found by simply treating the fulcrum as the axis of rotation, but I didn't see this approach when solving the problem. Instead I used the parallel axis theorem $I = I_{com} + Mh^2$. Now even though this approach is a waste I'm trying to figure out why I didn't arrive at the same answer anyway, so I've included my work computing the rotational inertia in this way.

Computing Rotational Inertia Using Parallel Axis Theorem

First I computed the center of mass of the rod as follows: $x_{com} = \frac{m_w * 0 + m_w* 0.80 m}{m_w + m_w}= 0.4m$ (Note: I use $m_w$ to denote mass while I use $m$ to denote distance). Next I computed $I_{com}$ as follows: $I_{com} = \Sigma \space m_{wi} \cdot \space r_i^2 = 2\space m_w \cdot (0.4 m)^2$.

And by the parallel axis theorem $I = I_{com} + Mh^2 = 2\space m_w \cdot (0.4 m)^2 + 2\space m_w (0.2m)^2$. After inputting this value into the original equation for angular acceleration I arrive at an invalid value. What have I done wrong? Also, how do we know definitively when to use the parallel axis theorem?

Any help understanding my problem would be appreciated greatly

Note: When either rotational inertia value is inputted into the angular acceleration formula the masses will cancel so using the correct approach and canceling the mass $\frac{I}{m_w} = (0.2m)^2 + (0.8m)^2 = 0.68 m^2$ while using my original approach I have $\frac{I}{m_w} = 2((0.4 m)^2 + (0.2m)^2) = 0.4m^2$

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I hope I don't need to use this for my question because this is way beyond my scope of knowledge! –  physicsaddict Sep 22 '14 at 6:34

Your problem is in the calculation of the COM. You first need to define the origin from where, you will get the COM distance by using the formula. Let the left end(particle 1) be the origin. Now, defining all distances from this origin:- $$x_{com}=\frac{1}{M_{Total}}\Sigma{m_i r_i}=m_w(L_1+L_2)/2m_w=0.5m$$ WHICH MEANS THE com is exactly at the centre of the rod, $0.3M$ right of the fulcrum. (unlike the $o.4m$ you got). Now, for $I_{com}=2m_w(0.5)^2$ which gives $$I_{fulcrum}=I_{com}+M_{total}(0.3)^2=2m_w(o.5^2+0.3^2)$$ which should give you the same answer as in the solutions manual.
Yeah, because it was so obvious - Not A Math Guy –  physicsaddict Sep 22 '14 at 6:35