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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...

Verified Answer :

a)♣ the string being pulled, 
the angular speed is: w1=w0 +w’t, hence t=(w1-w0)/w’; 
the angular path is: b=w0t+0.5w’t^2, where angular acceleration 
w’=T/J, torq T=Fr, and br=L; 
♦ thus b=w0(w1-w0)/w’ +0.5w’((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)/(Fr/J); 
2(Fr/J)L/r =w1^2 –w0^2, hence 
w1^2=2FL/J +w0^2; 
b)♣ the power is P=F(w0r);

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