Calculating Moment of Inertia for an 80kg Plank on Two Workers' Shoulders

[jono90one]

I have a question but got two different answers by two different methods. The question:
"Two workers are holding an 80kg plank, one worker let's go. The weight is carried by 55% of the first worker. It is 2.5m long and no uniform. The angular acceleration is 5.5 rads/s^2, what is the moment of inertia of the plank about the axis perpendicular to the beam at the end held by the worker."

Method 1:
τ = F x r [1] (F=mg, r=2.5 x 0.45)
τ = Iα [2]
[1]=[2]
I = 161 kgm^2

Method 2:
I = ∑mx^2/l .δx between -0.45l and 0.55l
lim δx => 0
I = ∫mx^2/l .dx between -0.45l and 0.55l
I = 2060ml^2/8000
I = 128.75 kg/m^2
Using parallel axis theorem:
I = I1 +md^2
I = 128.75 + 80(0.45x2.5)^2
=230 kg/m^2

I do not know which method is the correct one, but unsure why the other would be wrong.

Can someone help me?

Thanks.

 

[Doc Al]

Your first method looks good. Can you explain what you were doing in your second method?

 

[jono90one]

Thanks for the responce, well from what i ahve been taught, i can work out the moment of inertia via integration. Though I am not 100% sure whether i can use the exact same method of the mass isn't uniform.

But the basis of the second method is work out the mass of a small piece = (m/l) δx

Then Moment of Inertia = ∑mr^2 = ∑((m/l)δx)x^2 with the appropriate limits of integration (that are in terms of l, hence l's cancle to give an l^2 term)

Then lim δx -> 0 that becomes dx and ∑ becomes ∫

Can this method not be used in this circumstance?

 

[Doc Al]

Can this method not be used in this circumstance?

No, since it assumes a uniform distribution of mass.

 

[jono90one]

Oh ok, just out of interest, is there a method with integration for this circumstance?

 

[Doc Al]

Oh ok, just out of interest, is there a method with integration for this circumstance?

No, not that I can see. You'd need to know how the mass was distributed.