Acceleration of the midpoint of a light rod

[Bling Fizikst]

Homework Statement

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Relevant Equations

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I assumed the length of the rod to be $6l$ for simpler calculations . Here , $$a_{cm}=\frac{F}{m+2m}=\frac{F}{3m}$$
Here , CM is located at a distance of $2l$ from the $2m$ mass . Writing the moment equation : $$\alpha \times \left[ m(4l)^2+2m(2l)^2\right]=Fl\implies \alpha=\frac{F}{24ml}\implies \alpha l=\frac{F}{24m}$$ $$a_{\text{mid}} = a_{CM}+\alpha l=\frac{F}{3m}+\frac{F}{24m}=\frac{3F}{8m}$$ which doesn't match the answer key .

 

[erobz]

If the rod is spinning about its center of mass, replace the masses with the tension developed in each bar (they say inextensible cords - but I don't think that is consistent ) in the FBD of the rod, and you should be using the thin rods moment of inertial about the center of mass then apply $\sum \tau = I_{cm} \alpha$

Scratch that, massless rod. Looks fine to me. Did you notice they put a mass of $3m$ in the diagram?

Another thing, maybe they do actually mean cords, i.e. they don't accept compression? It wouldn't hurt if you gave the answer they expect.

 

[Bling Fizikst]

My bad , the answer matches if I took the mass as given in the diagram , i.e, $3m$ instead of $2m$ . The answer turns out to be : $\frac{F}{3m}$ .

 

[erobz]

My bad , the answer matches if I took the mass as given in the diagram , i.e, $3m$ instead of #2m$. The answer turns out to be :$\frac{F}{3m}## .

That's not your bad, it's their bad!