Resultant torque of a solid object

[Lone Wolf]

Homework Statement

A solid object rotates around an axis perpendicular to the plane of the figure and that goes through point A, due to the action of two forces applied in opposite sides of the object and with opposite directions; the resultant torque is proportional to the time squared. At the instant t = 0, Θ = 0 and ω = 0, and 5 seconds later, ω = 100 rad/s. The kinetic energy of the solid increases 1064 J between t = 3 and t = 2 s.
Find:
a) The moment of inertia of the solid relative to the axis of rotation.
b) The torque at the instant where the angular acceleration is 38.4 rad/s².

Relevant Equations

τ = Iα
K = 1/2 * I * ω²

a) ΔK = 1064 = 1/2 * I * (ω²(3) - ω²(2))
I = 2*1064/(ω²(3) - ω²(2))
If I assume α = n*t²:
α = dω/dt --> ω = ∫ n* t² dt = n*t³/3 + ω0 = n*t³/3
ω(5) = 100 = n*5³/3 --> n = 2.4
α = 2.4 t² and ω = 2.4 * t³/3
ω(3) = 21.6 rad/s and ω(2) = 6.4 rad/s. Replacing the values:
I = 2*1064/(21.6²-6.4²) = 5.0 kg m²
b) Using the previous value:
τ = I α;
τ = 5.0 * 38.4 = 192 N m
Does that seem okay?

 

[Doc Al]

I didn't check your arithmetic, but your approach looks good to me.

 

[haruspex]

I didn't check your arithmetic

I have. All good.