A string is wrapped tightly around a fixed pulley that has a moment of inertia of 0.0352 kg m² and a radius of 12.5 cm. The string is pulled away from the pulley with a constant force of 5.00 N. As the string unwinds the pulley begins to rotate. What is the speed of the string after it has unwound 1.25 m? A) 2.09 m/s B) 2.36 m/s C) 1.18m/s D) 3.18 m/s E) 4.95 m/s

Question

Quizplus · Accepted Answer

First, we need to calculate the torque applied to the pulley by the force pulling the string. The torque is given by:

τ = rF = (0.125 m)(5.00 N) = 0.625 Nm

Next, we need to find the angular acceleration of the pulley using the moment of inertia:

τ = Iα

α = τ/I = 0.625 Nm / 0.0352 kg m2 = 17.76 rad/s2

The final angular velocity of the pulley can be found using the kinematic equation:

ωf^2 = ωi^2 + 2αΔθ

Since the pulley rotates in the same direction as the unwinding string, the angle of rotation is given by the distance the string has unwound divided by the pulley radius:

Δθ = Δx / r = 1.25 m / 0.125 m = 10 rad

Plugging in the values, we get:

ωf^2 = 0 + 2(17.76 rad/s2)(10 rad)
ωf^2 = 355.2
ωf = 18.85 rad/s

Finally, we convert the angular velocity to linear velocity by multiplying by the pulley radius:

v = ωf r = (18.85 rad/s)(0.125 m) = 2.36 m/s

Therefore, the answer is B.

A String Is Wrapped Tightly Around a Fixed Pulley That