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 ofa point 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 ofa point 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

Radius r= 36 cm = 0.36 m

Initial angular speed w = 18 rev / s

Angular speed after turn 120 revolutions w ‘ = 6 rev/ s

Angular displacement = 120 rev

From the relation w’^ 2 – w^ 2 =2

Angular acceleration  = [ w’^ 2- w^ 2] / 2

                                  = -1.2 rev / s^ 2

                                  = -1.2* 2(pi) rad / s^ 2

                                  = -7.539 rad / s^2

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