Item 9 Part A A uniform disk with mass 394 kg and radius 0.210 m is pivoted at i

Question

Item 9 Part A A uniform disk with mass 394 kg and radius 0.210 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 31.5 N is applied tangent to the rim of the disk. What is the magnitude v of the tangential velocity of a point on the rim of the disk after the disk has tuned through 0.180 revolution? Express your answer with the appropriate units. Value Units Submit My Answers Give Up Part B what is the magnitude of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0 180 revolution? Express your answer with the appropriate units. aValue Units Submit My Answers Give Un

Explanation / Answer

Torque, T = F*r = 31.5N * 0.21m = 6.615 N·m

but also T = I*alpha = 0.5mr^2alpha,

So,

6.615 N·m = 0.5 * 39.4kg * (0.21 m)^2 * alpha

alpha = 6.615/(0.5*39.4*0.21^2)

   = 7.614 rad/s^2

w^2 = wo^2 + 2*alpha*s

w = sqrt(wo^2 + 2*alpha*s)

= sqrt(0 + 2*7.614*0.180*2*pi)

= 4.15 rad/s

A) v = wr = 4.15 rad/s * 0.21 m = 0.8715 m/s (1)

B) tangential at = alpha*r

= 7.614 rad/s^2 * 0.21m = 1.59 m/s^2

centripetal ac = w^2r

= (4.15 rad/s)^2 * 0.21 m = 3.62 m/s^2

mag a = sqrt(at^2 + ac^2)

= sqrt(1.59^2 + 3.62^2)

= 3.95 m/s^2 Answer

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