Noob question about centripetal acceleration

Question

This is a very simple question, but one that has been bugging me since long.

Consider a rod of length $L$ and mass $m$ to be hinged at one end. It is vertical initially. It is displaced slightly from its position. I have to find the centripetal acceleration when the rod is displaced by 90°.

I can easily find the angular acceleration using torque due to the weight of the rod, and subsequently the angular velocity can be found as well by using equations of motion.

Now, the part where I have the problem is taking the distance "$l$" while calculating the centripetal acceleration in the equation $a_c=ω²l$. Is the distance "$l$" from the center of mass of the rod (i.e. "$L/2$"), or is it the total length of the rod (i.e. "$L$")?

Answer 1

Is the distance l from the center of mass of the rod (i.e. L/2), or is it the total length of the rod (i.e. L)?

It is $\frac L 2$.

We can just say that the centripetal force is acting on the COM of the rod, which is at $\frac L 2$.

We can also break the rod in $n$ little segments, $\Delta L$, each having mass $\Delta M$, and say that the total centripetal force, $F_c$ has to keep all of these segments on a circle, i.e.,

$F_c=\omega^2 (\Delta M\frac 1 2 \Delta L + \Delta M\frac 3 2 \Delta L + \Delta M\frac 5 2\Delta L...+\Delta M\frac {2n-1} 2\Delta L)=$

$\omega^2 \Delta M \Delta L \frac 1 2 \frac n 2 (1+2n-1)= \omega^2 \Delta M \Delta L \frac {n^2} 2 =\omega^2M\frac L 2$

From here, $a_c=\frac {F_c} M=\omega^2\frac L 2$.