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

Question

Quizplus · Accepted Answer

The force causing the mass to accelerate downwards is given by:
F = mg = 3.63 kg x 9.81 m/s^2 = 35.55 N
The torque causing the wheel to rotate is given by:
τ = Fr = 35.55 N x 0.610 m = 21.66 N·m
The angular acceleration of the wheel is given by:
α = τ/I = 21.66 N·m / 2.71 kg·m^2 = 7.996 rad/s^2
The tangential acceleration of the mass is given by:
a = rα = 0.610 m x 7.996 rad/s^2 = 4.64 m/s^2
Therefore, the acceleration of the falling weight is 4.64 m/s^2, which is closest to 3.26 m/s^2.

A Cord Attached to a 3