Tangential velocity of rotating rod

AI Thread Summary
The discussion revolves around the tangential velocity of a rotating rod and the derivation of its final velocity equation. Initial calculations for tension and normal force are confirmed as correct, leading to a derived expression for the final angular velocity. A discrepancy arises when comparing the calculated final velocity with a textbook value, prompting a request for clarification. The error is identified as using the incorrect moment of inertia, as the rod rotates about its end rather than its center of mass. The discussion concludes with an acknowledgment of the mistake and appreciation for the clarification provided.
lorenz0
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Homework Statement
A homogeneous rod of mass ##m## and length ##L## can rotate without friction around the pin ##O##. The rod is balanced in the situation in the figure thanks to the presence of an ideal horizontal rope fixed to the vertical wall, which causes the angle between the wall and rod to be ##\theta##. Calculate: 1) Tension ##T## of the rope and the reaction ##N## of the pin; 2) the kinetic energy of the rod when it reaches the vertical position after the rope has been cut; 3) the tangential speed of the free end of the rod when it reaches the vertical position
Relevant Equations
##\tau=r\times F, F=ma, E=U_{grav}+E_K, E_{K_rot}=\frac{1}{2}I\omega^2##
1) ##LT\sin(\frac{\pi}{2}-\theta)-\frac{L}{2}mg\sin\theta=0\Rightarrow T=\frac{mg}{2}\tan\theta##.

##N_{x}-T=0, N_{y}-mg=0\Rightarrow N=\sqrt{N_x ^2+N_y ^2}=mg\sqrt{(\frac{\tan\theta}{2})^2 +1}##

2) ##E_{k_{fin}}=mg\frac{L}{2}[1+\cos\theta]##

3) ##mg\frac{L}{2}[\cos\theta+1]=\frac{1}{2}I_O \omega_f^2=\frac{1}{2}(\frac{1}{12}mL^2) \omega_f^2\Rightarrow \omega_f=\sqrt{\frac{12g[\cos\theta +1]}{L}}\Rightarrow v_f=L\omega_f=\sqrt{12gL[1+\cos\theta]}##

Now the first two results are correct, but for ##v_f## the book gives ##v_f=\sqrt{3gL[1+\cos\theta]}## and I don't see why so I would appreciate an explanation. Thanks

---

##mg\frac{L}{2}[\cos\theta+1]=\frac{1}{2}I_O \omega_f^2=\frac{1}{2}(I_{CM}+m(\frac{L}{2})^2 \omega_f^2=\frac{1}{2}((\frac{1}{12}+\frac{1}{4})mL^2 ) \omega_f^2=\frac{1}{6}mL^2\omega_f^2\Rightarrow \omega_f=\sqrt{\frac{3g[\cos\theta +1]}{L}}\Rightarrow v_f=L\omega_f=\sqrt{3gL[1+\cos\theta]}##
 

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Physically how could the result not depend upon theta?
How did you eliminate it?
 
hutchphd said:
Physically how could the result not depend upon theta?
How did you eliminate it?
I don't get what you are referring to: every single one of the results depends on ##\theta##.
 
lorenz0 said:
I don't get what you are referring to: every single one of the results depends on θ.
Oops sorry it got cut off on my display!
 
lorenz0 said:
##\dots## and I don't see why so I would appreciate an explanation. Thanks
You used the wrong moment of inertia. The rod is rotating about its end, not about its CM.
 
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kuruman said:
You used the wrong moment of inertia. The rod is rotating about its end, not about its CM.
Ah, of course! Such a stupid mistake. Thank you very much!
 
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