A cord attached to a 3.63-kg mass is wrapped around a wheel of radius 0.610 m and released. The moment of inertia of the wheel is 2.71 kg · m². If the wheel rotates on frictionless bearings, the acceleration of the falling weight is A) 3.26 m/s². B) 1.04 m/s². C) 2.44 m/s². D) 1.95 m/s². E) 4.27 m/s².

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Quizplus · Accepted Answer

The acceleration can be found using the equation a = (m*g) / (m + I/r^2), where m is the mass, g is the acceleration due to gravity (9.81 m/s²), I is the moment of inertia, and r is the radius. Plugging in the values, a = (3.63*9.81) / (3.63 + 2.71/0.610²) = 3.26 m/s².

A Cord Attached to a 3