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!

 

[jvicens]

jdstokes, nvn, Dick,

I was searching the analysis for this particular falling rod problem as I'm currently trying to figure out the speed of the free end of the rod just before it hits the ground.

If we are assuming all the mass of the rod as located at its center of mass when we calculate the potential energy (Ep=M.g.L/2), why are we not considering the same thing (mass located at its center of mass) when we calculate Moment of Inertia? In this case I=M(L/2)^2?

The next question is about the kinetic energy before the rod hits the ground.
Shouldn't we consider two components of the kinetic energy: one due to angular velocity and another due to tangential velocity?

Taking into considerations these two assumptions I got that Vend=SQRT(2gL)

Jvicens

 

[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.

 

[jvicens]

nvn,

I understand what you said for my second question but I was initially still confused about the moment of inertia that should be used in these calculations
I thought that the expression for potential energy 0.5xMxgxL was derived assuming the entire mass of the rod located at its center of mass and this was the wrong assumption. With calculus is easy to see that the above mentioned expression is the result of integrating dEp along the length of the rod from zero to L. As we have not considered the total mass concentrated in the center of mass of the rod then we should not consider either the moment of inertia as coming from a concentrated mass located in the center of mass. I should have taken Io as the one from a rod rotating by one of its ends (Io= 1/3xMxL^2).
The new calculation yields Vend = SQRT(3xgxL/4). Is this in agreement with your calculations?
Jvicens

 

[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.