Question

4. A wagon of mass 14 tonnes is hauled up an incline of 1 in 20 by a rope which is parallel to the incline and is being wound round a drum of 1 m diameter. The drum, in turn, is driven through a 40 to 1 reduction gear by an electric motor. The frictional resistance to the movement of the wagon is 1.2 kN, and the efficiency of the gear drive is 85 per cent. The bearing friction at the drum and motor shafts may be neglected. The rotating parts of the drum have a mass of 1.25 tonnes with a radius of gyration of 450 mm and the rotating parts on the armature shaft have a mass of 110 kg with a radius of gyration of 125 mm.
At a certain instant the wagon is moving up the slope with a velocity of 1.8 m/s and an acceleration of 0.1 m/s2. Find the torque on the motor shaft and the power being developed. [T= 154 Nm, P=22,24 kW]

Answer 1

TORQUE ON MOTOR SHAFT,

There are four torques generated ,

1. Torque on drum shaft to accelerate load
2. Torque on drum to accelerate drum-shaft
3. Torque on armature to accelerate drum and load
4. Torque on armature to accelerate armature shaft

Now tension in rope = total forces P1

component of weight + inertia force + frictional resistance (Horizontal forces)

(Where frictional resistance=1.2kN and acceleration is 0.1m/s^2)

SO the torque on the drum to accelerate load is given by ,

Similarly because of presence of inertia, the TORQUE required on drum shaft to accelerate drum shaft,

where is angular acceleration

( Where I1 is mass moment of inertia for drum )

Now torque on armature shaft of motor to accelerate load as well as drum,

As the drum is driven by electric motor with gear ratio (G.R)=40:1

  

And Torque to accelerate armature shaft,

  (Where I2 is mass moment of inertia for armature shaft)

   (Because armature rotates 40 times than that of drum)

So, torque on motor shaft is goven by

2) POWER DEVELOPED BY MOTOR

POWER DEVELOPED WILL BE,