Question

only final asnwer

If the system is released from rest, determine the speed of the 30-kg cylinders A and B after B has moved upward a distance of 0.50 m. The differential pulley has a mass of 20 kg with a radius of gyration about its center of mass of ko = 100 mm. 200 m 100 mm B

Answer 1

Given

Wo =0 rad/s

D = 0.5 m

Ma = Mb = 30Kg

K= 0.1 m

Ra = 0.2m

Rb = 0.1 m

Mp = 20 Kg

Moment of inertia about center point =Ic

Ic = Mp*K^2+Ma* R^2+Mb* Rb^2

Ic= 20*(0.1)^2+30*(0.2)^2+30*0.1^2

Ic = 1.7 Kg-m^2

Net torque = I *alpha = Ma*g*Ra- Mb *g*Rb

Alpha =angular acceleration = 30*9.81*0.1÷(1.7)

Alpha =17.312 rad/s^2

Then ,

W^2 = Wo^2+2*alpha *theta

W^2 = 0+ 2* 17.312* (0.5÷0.1)

W = angular velocity= 13.1574 rad/s^2