Exercise 10.10 Part A What is the magnitude U of the tangential velocity of a po

Question

Exercise 10.10 Part A What is the magnitude U of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.120 revolution? A uniform disk with mass 44.2 kg and radius 0.250 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 28.5 N is applied tangent to the rim of the disk. Express your answer with the appropriate units. Value Units Submit My Answers Give Up 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 0.120 revolution? Express your answer with the appropriate units. Value Units Submit My Answers Give Up

Explanation / Answer

torque = F*r = 28.5N * 0.25 m = 7.125 N.m

but also torque = I*a = (1/2)mr^2*a, so

7.125 N.m = (1/2) * 44.2 kg * (0.25m)^2 * a

a = 5.158 rad/s^2

now using equation

w^2 = w0^2 + 2a*theta = 0 + 2 * 5.158 rad/s^2 * 0.120rev * 2*pi rads/rev = 7.78 rad^2/s^2

w = 2.79 rad/s

v = wr = 2.79 rad/s * 0.25m = 1.945 m/s [Answer of Part A]

tangential acceleration at = ar = 5.158 rad/s^2 * 0.25m = 1.289 m/s^2

centripetal acceleration ac = w^2r = (2.79 rad/s)^2 * 0.25 m = 1.946 m/s^2

mag a = sqrt(at^2 + ac^2) = 2.334 m/s^2 [Answer of Part B]

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