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A 6.60 kg stone is tied to a thin, light wire wrapped around the outer edge of the uniform cylindrical pulley of the same mass as the stone shown in the figure below. The outer radius of the pulley is 1.10 m. The inner radius of the pulley is R=0.35m meters. The system is released from rest, and there is no friction at the axle of the pulley. (The moment of inertia for the pulley is calculated to be 6.806 kg-m2 ). What is the linear acceleration of the hanging stone?

User Mondieki
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Final answer:

To calculate the linear acceleration of the 6.60 kg stone, we use the equation a = (m*g*R²)/I, substituting in the given values for the mass of the stone, the outer radius of the pulley, and the pulley's moment of inertia.

Step-by-step explanation:

To determine the linear acceleration of the hanging stone, we must apply Newton's second law for rotation. The only force causing the pulley to rotate is the weight of the stone, which creates a torque on the pulley. We can set up the following equation: τ = Iα, where τ is torque, I is the moment of inertia of the pulley, and α is the angular acceleration. Since the wire is wrapped around the outer edge of the pulley, the torque produced by the stone's weight is τ = m*g*R, where m is the mass of the stone, g is acceleration due to gravity (9.81 m/s²), and R is the outer radius of the pulley. However, linear acceleration of the stone a is related to the angular acceleration of the pulley by the equation α = a/R. Substituting these relationships into our original equation, we get:

m*g*R = I*(a/R) → a = (m*g*R²)/I.

Using m = 6.60 kg, R = 1.10 m, and I = 6.806 kg-m², we can calculate the linear acceleration a. Therefore, the linear acceleration of the hanging stone is 6.60 kg * 9.81 m/s² * 1.10 m² / 6.806 kg-m² which gives us the final value of a.

User MBarton
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