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=kmr^2, 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.
(a) Suppose that after a certain time t_L, the string has been pulled through a distance L, What is the final rotational speed ω_final of the wheel?
(b) What is the instantaneous power P delivered to the wheel via the force F at time t=0?

A) The final rotational speed ω_final of the wheel is; ω_f = √((ω_i)² + 2(FL/(kmr²)

B) The instantaneous power P delivered to the wheel via the force F at time t=0 is; P = Frω_i

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 = kmr²

T is torque = 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

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a) , b)

Explanation:

a) The final rotational speed is:

But:

Lastly, the angular acceleration is replace in the expression for the final angular velocity:

b) The power is rate of change of work in terms of time:

Instantaneous power is obtained by differentiating the previous expression with respect to time:

The instantaneous power delivered to the wheel at t = 0 is:

see earlier answer