Calculating Rod's Speed After Free Fall

[jdstokes]

Homework Statement

From the GRE 0177 practice exam:

A thin uniform rod of mass M and length L is positioned vertically above an anchored frictionless pivot point, as shown above, and then allowed to fall to the ground. With what speed does the free end of the rod strike the ground?

The Attempt at a Solution

$\frac{MgL}{2} = \frac{1}{2}I \omega^2 \implies$
$MgL = \frac{1}{12} M L^2 \omega^2 \implies$
$\omega = \sqrt{\frac{12g}{L}} \implies$
$v_t = L\omega = \sqrt{12gL}$

Answer
$v_t=\sqrt{3gL}$

Any help would be appreciated.

 

[nvn]

0.5*m*g*L + 0 + 0 = 0 + 0.5*m*v_c^2 + 0.5*I_o*omega^2, where v_c = tangential velocity of rod centroid, and I_o = rod centroidal mass moment of inertia.

 

[jdstokes]

Hi nvn,

Your equation is incorrect. Why have you put a point mass term on the RHS?

 

[jdstokes]

Oh I understand now, the moment of inertia requires a modification because the object is not rotating about the center of mass. Thanks.

 

[Dick]

The moment of inertia of a rod is only (1/12)*M*L^2 if it's rotating around the center. If it's rotating around an end it's (1/3)*M*L^2. Ach, I see you have already figured it out. Good job!

 

[nvn]

jvicens: The mass moment of inertia of the rod is centroidal as stated in post 2. The rod mass is distributed along its length, not concentrated at its center. Try looking up I_o again; your formula appears incorrect.

The answer to your second question is, yes, we should, and we are. See post 2. Your current answer appears to be incorrect.

 

[nvn]

jvicens: You can instead work the problem that way, which is a good method, but it gives the same answer (which is stated in the last line of post 1), not the answer you listed.