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A 8.0-cm radius disk with a rotational inertia of 0.12 kg ·m2 is free to rotate on a horizontalaxis. A string is fastened to the surface of the disk and a 10-kgmass hangs from the other end.The mass is raised by using a crank to apply a 9.0-N·mtorque to the disk. The acceleration ofthe mass is:A. 0.50m/s2B. 1.7m/s2C. 6.2m/s2D. 12m/s2E. 20m/s2

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Final answer:

To find the acceleration of the mass, we can use Newton's second law for rotational motion and the given torque and moment of inertia. The angular acceleration is found to be 75 rad/s², and by using the relationship between angular acceleration and tangential acceleration, we determine that the acceleration of the mass is 6.0 m/s².

Step-by-step explanation:

To determine the acceleration of the mass, we need to apply Newton's second law for rotational motion, which states that the torque applied to an object is equal to the product of its moment of inertia and its angular acceleration.

The torque applied to the disk is given as 9.0 N·m. The moment of inertia of the disk is given as 0.12 kg·m². Therefore, we can use the formula τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration.

Substituting the given values, we have 9.0 N·m = 0.12 kg·m² × α. Solving for α, we find that the angular acceleration is 75 rad/s².

Since the angular acceleration is equal to a/R, where a is the tangential acceleration and R is the radius of the disk, we can rearrange the formula to solve for a. Thus, a = α × R = 75 rad/s² × 0.08 m = 6.0 m/s².

Therefore, the acceleration of the mass is 6.0 m/s².

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