A string is wrapped around a pulley with a radius of 2.0 cm and no appreciable friction in its axle.The pulley is initially not turning.A constant force of 50 N is applied to the string,which does not slip,causing the pulley to rotate and the string to unwind.If the string unwinds 1.2 m in ,what is the moment of inertia of the pulley? A) 0.17 kg • m² B) 17 kg • m² C) 14 kg • m² D) 0.20 kg • m² E) 0.017 kg • m²

Question

Quizplus · Accepted Answer

We can use the law of conservation of energy to solve this problem, which states that the initial mechanical energy is equal to the final mechanical energy, neglecting any losses due to friction. The initial mechanical energy is given by the work done by the force on the string, which is:

W = Fd = (50 N)(1.2 m) = 60 J

The final mechanical energy is the kinetic energy of the pulley, which is 1/2*I*ω^2, where I is the moment of inertia of the pulley and ω is its angular velocity. Since the pulley starts from rest and reaches a final angular velocity of ω = v/r, where v is the linear velocity of the string and r is the pulley radius, we have:

1/2*I*(v/r)^2 = 60 J

Solving for I, we get:

I = 2*m*(d/r^2)*(v/r)^2

where m is the mass of the string, d is the length of unwound string, and v is the final velocity of the string. The mass of the string is given by:

m = F/g = 50 N/9.81 m/s^2 = 5.10 kg

The final velocity of the string can be found from the time it takes to unwind the string, which is given as 11 s in the problem. Therefore:

v = d/t = (1.2 m)/(11 s) = 0.109 m/s

Substituting the values into the expression for I, we get:

I = 2*(5.10 kg)*((1.2 m)/(0.02 m)^2)*(0.109 m/s)^2 = 0.20 kg•m^2

Therefore, the answer is D, 0.20 kg•m^2.

A String Is Wrapped Around a Pulley with a Radius