Question

Course Contents» Last: HW #11 (11/21 Th 11:59 Notes Feedback Timer Evaluate Print M, a solid cylinder (M 2.27 kg, R=0.115 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.690 kg mass, i.e., F = 6.769 N. Calculate the angular acceleration of the cylinder. R Tries 0/12 Submit Answer MM

Answer 1

Moment of Inestia → Icm = me = 2.27 kg x .115m)? = 0.015010375 torque, I = RF kgm² also I - Icma (@angular acceleration) → a= RF - Icm > a = 2F . MR a = 2x 6.769N 151.86 rad's 2.27 kg X 0.115m Now, m= 0.690 kg 2 g = 9,81m/s² force, F = mg = 0.690 kg x 9,8 m/s² = F = 6.769 N again [a = 51.86 rad/s]

acceleration of mass m, narra a = 0.115m x 51.86 rad/s a = 5.9639 m/s² from rest distance travelled by mass in in stime t . d = vit + 1 at? I d = – at? (: V; 20). » dz-d, = 2 ( + - +²) da-di = 5.9639 m/s² (10.650s)²-(0.450s)") =0.656 m Nowo, again d = vit + 1 act? => 0.454m = 0 + cxax (0.510s)? . a = 3.4910 m/s² (acceleration) angular acceleration of cylinder - az a * RF = a => Ecm = Farkle Ce64 6.769801152 3.4910 (of mass in I cm = 0.025643089 Kg m² !