Part A A uniform disk with mass 43.9 kg and radius 0.280 m is pivoted at its cen

Question

Part A A uniform disk with mass 43.9 kg and radius 0.280 m is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force 27.5 N is applied tangent to the rim of the disk. What is the magnitude o of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.400 revolution? Express your answer with the appropriate units. . Value Units Incorrect; Try Again; 4 attempts remaining Part B What is the magnitude a of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0.400 revolution? Express your answer with the appropriate units Value Units Submit Incorrect; Try Again; 5 attempts remaining

Explanation / Answer

a)moment of inertia = 1/2 M R2 = 1/2 * 43.9 * 0.2802 = 1.72 kg.m2

alpha = F R / I = 27.5 * 0.280 / 1.72 = 4.47 rad/s2

w = sqrt [2 * alpha * delta theta] = sqrt [2 * 4.47 * 0.400 * 2pi] = 4.74 rad/s

v = R w = 0.280 * 4.74

= 1.33 m/s

b) ar = v2/R = 1.332 / 0.280 = 6.3 m/s2

at = R alpha = 0.280 * 4.47 = 1.25 m/s2

magnitude = sqrt [6.32 + 1.252]

= 6.42 m/s2

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