Question

Pulling a wheel with a string?
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 = k m r^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 velocity omega_0, 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, the string has been pulled through a distance L. What is the final rotational speed omega_final of the wheel? Express answer in terms of L, F, I, and omega_0. I found that the final kinetic energy of the wheel is K_f = .5I omega_0^2 + FL, but I can't find an equation that relates this to final rotational speed.
b. What is the instantaneous power P delivered to the wheel via the force F at time t = 0?

Answer 1

a)♣ the string being pulled, 
the angular speed is: w1=w0 +w’*t, hence t=(w1-w0)/w’; 
the angular path is: b=w0*t+0.5*w’*t^2, where angular acceleration 
w’=T/J, torq T=F*r, and b*r=L; 
♦ thus b=w0*(w1-w0)/w’ +0.5*w’*((w1-w0)/w’)^2 = 
= (w1-w0)*(w0 +0.5w1 -0.5w0)/w’ =0.5*(w1^2 –w0^2)/w’; 
♠ and b=L/r =0.5*(w1^2 –w0^2)/(F*r/J); 
2(F*r/J)*L/r =w1^2 –w0^2, hence 
w1^2=2F*L/J +w0^2; 
b)♣ the power is P=F*(w0*r);