Solving Rotational Energy Problem | 1.6m Stick, .4-.070kg/m^2

[Jack86]

I am having problems with a rotational energy problem.

It reads...
a think stick of length 1.6m is dense at one edn than at teh other: it's mass desity is p=.4kg/m-.070kg/m^2x, where x is the distance from the heavier end of teh stick. The stick rotates about an axis perpindicular to the heavier end with a period of 1.1s. Determine teh Rotational kinetic energy of teh stick.

I used K=1/2Iw^2. I found w by using the period. However, when I try to use the integral I=(integral)pR^2 dV, I cannot find R becuase it is not stated. How do I find R or is there another way around the problem?

 

[Fermat]

R is the length of the stick.

And your integral is wrong. It should be I=(integral)pr² dr, not I=(integral)pR^2 dV,

 

[Päällikkö]

I disagree: it should be dV (I can't think why it shouldn't, enlighten me ).

(The distance from the rotational axis) R² = x² + y². As the stick's thin: y,z $\to$ 0.

 

[Jack86]

R is the radius of the stick, I am sure of this. And the integral has to be with respect to volume, it's a 3d object

 

[Fermat]

I disagree: it should be dV (I can't think why it shouldn't, enlighten me ).

(The distance from the rotational axis) R² = x² + y². As the stick's thin: y,z $\to$ 0.

We have a stick rotating about one end. A small element δm at a distance r giving a small inertia δI = r²δm. Puting δm = ρ*δr and substituting gave me my final result. Hmm, how is it derived with dV, I'm not familiar with that ?

 

[Päällikkö]

δm = ρ*δr

I feel this should be:
δm = ρ*δV

Then again as the stick's thin, δm = ρ*δV basically reduces to δm = y*z*ρ*δr, where y and z are the thicknesses into different directions (I don't think you can jus cancel them out).

The attachement will take a while before it is approved (thus I cannot view it), maybe host the file on http://www.imageshack.us" ? (EDIT: You already did )

If you use your method to derive the I for a uniform thin rod, you will come up with (1/3)ρR³, which is wrong. Either of the methods I've mentioned should give the correct (1/3)MR²

EDIT: Lots and lots of edits , small typos of all kinds.

 

[Fermat]

I feel this should be:
δm = ρ*δV

ρ is the linear density, in kg/m

 

[Jack86]

The integral is still from 0 to 1.6 though right?

 

[Fermat]

Yeah, I think we would both agree on that