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A string is wrapped around a uniform solid cylinder of radius r = 2 meter. The cylinder can rotate freely about its axis. The loose end of the string is attached to a block. The block has a mass of 10 kg, the cylinder has a mass of 5 kg. Note that the positive y direction is downward and counterclockwise torques are positive. Find the magnitude of the angular acceleration α as the block descends. Begin by sketching a free body diagram for the disk and for the block

User Pdenlinger
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Final answer:

To find the magnitude of the angular acceleration α as the block descends, we can start by sketching a free body diagram for the disk and for the block. We can then apply Newton's second law for rotational motion to the disk, which states that the torque on an object is equal to the moment of inertia times the angular acceleration. The magnitude of the angular acceleration α as the block descends can be found using the equation: Tension * Radius = moment of inertia * α

Step-by-step explanation:

To find the magnitude of the angular acceleration α as the block descends, we can start by sketching a free body diagram for the disk and for the block. For the disk, the forces acting on it are the force of gravity and the tension in the string. For the block, the only force acting on it is the tension in the string.

We can then apply Newton's second law for rotational motion to the disk, which states that the torque on an object is equal to the moment of inertia times the angular acceleration. The torque on the disk is given by the tension in the string times the radius of the disk. The moment of inertia of the disk can be calculated using the formula for the moment of inertia of a solid cylinder. Setting up an equation with the torques, we can solve for the angular acceleration.

The magnitude of the angular acceleration α as the block descends can be found using the equation:

  • Tension * Radius = moment of inertia * α
User Blake Taylor
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