Exercise 10.10 Constants Part A A uniform disk with mass 38.3 kg and radius 0.29

Question

Exercise 10.10 Constants Part A A uniform disk with mass 38.3 kg and radius 0.290 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 33.5 N is applied tangent to the rim of the disk Whel is the magnitude u of the l?ngert a ve ticly of ? port on the rim of the diskafler the disk has turned through 0.130 revolution? Express your answer with the appropriate units. Submit Previous Answers Request Answer Incorrect; Ty Again; 3 attempts remaining Part B What is the magnitude a of the resuitant acceleration of a point on the rim of the disk after the disk has turned through 0.130 revolution? Express your answer with the appropriate units. 1 =1 Value Units Submit Request Answer

Explanation / Answer

Part A

torque = Fxr

=33.5N * 0.29m = 9.715 N·m
but also

torque = moment of inertia*angular acceleration

= ½mr²(angular acceleration),

so

9.715N = ½ * 38.3kg * (0.3m)² *(angular acceleration)

Angular acceleration= 5.636rad/s²
Now applying third equation of rotatirot motion

(Angular speed)2= 0 + 2 * 5.636rad/s² * 0.130rev * 2*3.14rads/rev

Angular speed = 3.033rad/s
v = (angular speed)r = 3.033rad/s * 0.29m = 0.879 m/s
Part B
tangential at = (angular acceleration* r )= 5.636rad/s² * 0.29m = 1.634 m/s²
centripetal ac = (angular speed)²r = (3.033rad/s)² * 0.29m = 2.667 m/s²

a = ?(at² + ac²)

= 3.128 m/s²

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