Example 3.2. A winding drum xaises a cage of mass 500 kg through a height of 100 metres. The mass of the winding drum is 250 kg and has an effective radius of 0.5 m and radius of gyration is 0.35 m . The mass of the rope is 3kg/m The cage has, at first.an acceleration of 1.5m/s^2 until a velocity of 10m/s is reached, after which the velocity is constant until the cage nears the top and the final retardation is 6m/s^2 Find 1. The time taken for the cage to reach the top, 2 . The torque which must be applied to the drum at starting; and 3. The power at the end of acceleration period. Solution. Given : Vert _(C)=500kg;s=100m;m_(D)=250kg;r=0.5m;k=0.35m m=3kg/m

1. To find the time taken for the cage to reach the top, we can use the kinematic equation: where:- is the distance (100 m)- is the initial velocity (0 m/s)- is the acceleration (1.5 m/s²)- is the timePlugging in the values, we get: So, the time taken for the cage to reach the top is approximately 11.55 seconds.2. To find the torque which must be applied to the drum at starting, we can use the formula for torque: where:- is the torque- is the moment of inertia- is the angular accelerationThe moment of inertia for a solid disk is given by: where:- is the mass of the drum (250 kg)- is the radius of the drum (0.5 m)Plugging in the values, we get: Now, we can find the angular acceleration using the formula: where:- is the acceleration (1.5 m/s²)- is the radius of the drum (0.5 m)Plugging in the values, we get: Now, we can find the torque: So, the torque which must be applied to the drum at starting is 187.5 N m.3. To find the power at the end of the acceleration period, we can use the formula for power: where:- is the power- is the force- is the velocityThe force can be calculated using Newton's second law: where:- is the mass of the cage (500 kg)- is the acceleration (1.5 m/s²)Plugging in the values, we get: Now, we can find the power: So, the power at the end of the acceleration period is 7500 W.