Question

A disk with a rotational inertia of 6.75 kg·m² rotates like a merry-go-round while undergoing a torque given by τ = (5.75 + 4.30t) N · m. At time t = 1.00 s, its angular momentum is 9.09 kg·m²/s. What is its angular momentum at t = 3.00 s?

Answer 1

In rotational dynamics, the relationship between α (acceleration) and ω (velocity) is that dω(t)/dt = α
and torque is related to α as T=I*α where I is the moment of inertia of the object about the axis of rotation.
In this case, Torque is time dependent so using an integral:

5.75 + 4.30t=I*α(t)
I*ω(t)=5.75*t+0.5*4.30*t^2 +I*ω0
That is the time dependent relationship for this case

using t=1 to find I*ω0
I*ω(1)=5.75+0.5*4.30+I*ω0
9.09=7.9+I*ω0
I*ω0= 1.19
therefore
I*ω(3)=5.75*3+.5*4.30*9 + 1.19
or
37.8 kg m2 s-1