A 4.80kg disk is free to rotate about an axis through itscenter (+z); the plane

Question

A 4.80kg disk is free to rotate about an axis through itscenter (+z); the plane of the disk is horizontal and the radius ofthe disk is 36.0cm. It initially rotates with an angularvelocity of (18.0 rev/s, -z). Due to torques acting on thedisk, the angular velocity is (6.00 rev/s, -z) after the disk hasturned through 120 revolutions. Determine (a) the angularacceleration during this time; (b) the elapsed time; and(c) the linear velocity of apoint on the edge of the disk after the disk turned through 60revolutions. I am having trouble understanding how todetermine C. If you respond please include adetailed explanation; not just an answer. A 4.80kg disk is free to rotate about an axis through itscenter (+z); the plane of the disk is horizontal and the radius ofthe disk is 36.0cm. It initially rotates with an angularvelocity of (18.0 rev/s, -z). Due to torques acting on thedisk, the angular velocity is (6.00 rev/s, -z) after the disk hasturned through 120 revolutions. Determine (a) the angularacceleration during this time; (b) the elapsed time; and(c) the linear velocity of apoint on the edge of the disk after the disk turned through 60revolutions. I am having trouble understanding how todetermine C. If you respond please include adetailed explanation; not just an answer.

Explanation / Answer

M = 4.8 o = 18 rev/s = 6 rev/s 0 = 0 = 120 rev 1 rev/s = 2 rad/s You will have to convert through out the problem = ? t = ? Vtan = ? a.) = 754 rad 2 = 02 +2()
=-7.54 rad/s2

b.) = 0 +t t = 10 secs

c.) Vtan =r 60 rev = 377 rad 2 =02 +2() = 84.3rev/s x r = 40.7m/s

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