A cord is wrapped around the rim of a solid uniform wheel 0.250 m in radius and of mass 9.20 kg. A steady horizontal pull of 40.0 N to the right is exerted on the cord, pulling it off tangentially from the wheel. The wheel is mounted on frictionless bearings on a horizontal axle through its center. (a) Compute the angular acceleration of the wh

Question

Avery Coffey

Physics

27 October, 11:36

A cord is wrapped around the rim of a solid uniform wheel 0.250 m in radius and of mass 9.20 kg. A steady horizontal pull of 40.0 N to the right is exerted on the cord, pulling it off tangentially from the wheel. The wheel is mounted on frictionless bearings on a horizontal axle through its center. (a) Compute the angular acceleration of the wh

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Ayana Barry · Answer 1

The angular acceleration of the wheel is 4.33m/s^2

Explanation:

F=40N, r=0.25m, m=9.20kg

V = √ (Fr : m) = √ (40*0.25 : 9.2) = √1.09 = 1.04m/s

Angular acceleration (a) = V^2/r = 1.04^2/0.25 = 1.0816/0.25 = 4.33m/s^2