The elevator system given in the figure moves up and down with a velocity of v=1 m/sec. As the elevator starts to move, it reaches to its steady state angular velocity in 1 seconds. It is known that all the friction effects within the system are neglected. The other information related with the system are given as below: Mass moment of inertia about motor shaft M-M Mass moment of inertia about the elevator shaft K-K Mass of the elevator cabin Counter mass Radius of the pulley Gear ratio Gravitational acceleration : : : : : : Da JM=1.25 kgm JK=1 kgm M=400 kg m=100 kg R=0.25 m i=13 g=10 m/s a. Calculate the angular velocity of motor shaft and elevator shaft. b. Find the equivalent mass moment of the system that is lumped at motor shaft. c. Find the total moment acting on the motor shaft (in Newton-meters) in steady state. 2 TA K