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"Two workers are holding an 80kg plank, one worker lets 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.