Question

A bicycle wheel is mounted on a fixed, frictionless axle, with a light string wound around its rim. The wheel has moment of inertia i=kmr2, where m is its mass, r is its radius, and k is a dimensionless constant between zero and one. The wheel is rotating counterclockwise with angular speed ω0, when at time t=0 someone starts pulling the string with a force of magnitude f. Assume that the string does not slip on the wheel

Answer 1

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

What is the Instantaneous Power?

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