The answer is “Yes I can”. The question was if I could answer this.
The final rotational speed ω_final and the instantaneous power P delivered to the wheel are; ω_f = √((ω_i)² + 2(FL/(kmr²) and P = Frω_i
<h3>What is the Instantaneous Power?</h3>
A) From rotational kinematics, the formula for the final angular velocity is;
ω_f = √((ω_i)² + 2αθ)
where;
α is angular acceleration
θ = L/r. Thus;
ω_f = √((ω_i)² + 2α(L/r))
Now, α = T/I
Where;
I is moment of inertia = k*m*r²
T is t o r q u e = F * r
Thus;
α = (F * r)/(kmr²)
α = F/(kmr)
ω_f = √((ω_i)² + 2(F/(kmr))(L/r))
ω_f = √((ω_i)² + 2(FL/(kmr²)
B) Formula for instantaneous power is;
P = Fv
where at t = 0; v = rω_i
Thus;
P = Frω_i
Read more about Instantaneous Power at; brainly.com/question/14244672
