A disk of negligible mass and radius ....?

Question

A disk of negligible mass and radius 30 cm is constrained to rotate on a frictionless axis about its center. The disk remains in a vertical plane with its axis horizontal. On the disk are mounted masses of 21 grams at a distance of 22.2 cm from the center, 10 grams data distance of 12.9 cm from the center and 41 grams at a distance of 11.4 cm from the center. A force is applied at the rim of the disk by a mass of 4.071 grams attached to by a light string around the rim.
•As the mass descends 53 cm, with the disk originally at rest, by how much does the potential energy of the system change?
•What therefore will be the angular velocity attained by the disk and the velocity attained by the descending mass?

Answer 1

♠ work done by mass m=4.071g is E=0.004071*g*0.53, where g=9.8m/s^2; thus E=0.021145 Joules; and this is “how much does the potential energy of the system change”;
♣ now according to energy conserv law this change has flown into kin energy of the system E=0.5*(J1+J2+J3)*w^2 +0.5*m*v^2, where
J1=21*22.2^2·10-7, J2=10*12.9^2·10-7, J3=41*11.4^2·10-7 are inertia moments of mounted masses, w is angular speed of disk, while v=w*0.30 is linear speed of mass m;
♦ thus 0.021145=0.5*(21*22.2^2 +10*12.9^2 +41*11.4^2 +0.004071*0.3^2) *10-7 *w^2, hence w=4.938 rad/s; and v=w*0.30 =1.481m/s;
it’s in my personality to muddle this or that; so pull your hands out of pockets, check me and click me if I’m wrong so we could work it out 2gether.

Answer 2

as fast as fast can be