A bug of mass 0.020 kg is at rest on the edge of a solid cylindrical disk (M = 0.10 kg, R = 0.10 m) rotating in a horizontal plane around the vertical axis through its center. The disk is rotating at 10.0 rad/s. The bug crawls to the center of the disk. (a) What is the new angular velocity of the disk?
(b) What is the change in the kinetic energy of the system?
(c) If the bug crawls back to the outer edge of the disk, what is the angular velocity of the disk then?
(d) What is the new kinetic energy of the system? (e) What is the cause of the increase and decrease of kinetic energy?

Id = 1/2 Md * R^2  = 1/2 * .1 * .1^2 = .0005 kg m^2   inertia of disk

Ib = Mb * R^2 = .02 * .1^2 = .0002 kg m^2   inertia of bug at edge

(Id + Ib) w1 = Id w2      conservation of angular momentum

w2 = .0007 / .0005 * 10 = 14 /sec     angular speed with bug at center

KE1 = 1/2 I1 * w1^2 = 1/2 * .0007 * 10^2 = .035 kg m^2 / s^2

KE2 = 1/2 * I2 w2*2 = (.0005 / 2 ) ^ 14^2 = .049 kg m^2 / s^2

The bug has to exert radial force on the disk to maintain its                                                    

centripetal acceleration. As the bug crawls to the center of the disk it

does work against this centripetal force which appears as an increase

of rotational energy of the disk. As the the bug crawls back to the edge

of the disk, the disk does work on the bug and loses KE.

a, c, d, e are all applying about success strategies.

as we probably am aware that our atmosphere or air have various levels, each level has different gasses, and each gas has another porousness. at the point when the light originates from lower porousness to higher penetrability, it alters its course. with the goal that implies that when the light originates from the moon to the earth, it alters it's course a lot more occasions. it essentially implies that the state of the light piece of the moon is changed on the earth.

not copied too! hope this !

joke:

h o h o hope you have a merry christmas!