Question

a rotor weighing 2 kn is supported on bearings a and b which are 1 m apart. the centre of mass of the rotor is at a distance 0.4 m from bearing a. it is observed that there is an unbalanced couple of magnitude 300 n-m which *leaves the shaft balanced statically. the dynamic reactions at the supports will be

Answer 1

The dynamic reactions at the supports can be calculated using the principle of moments. Since the unbalanced couple leaves the shaft balanced statically, we can assume that the dynamic reactions at the supports will also be balanced.

Let's denote the dynamic reaction at support A as RA and at support B as RB. Since the rotor is in equilibrium, the sum of the moments about any point should be zero.

Considering the moments about support A:

Clockwise moment due to the unbalanced couple: 300 N-m

Counter-clockwise moment due to the weight of the rotor: 2 kN * 0.4 m = 800 N-m

Counter-clockwise moment due to the reaction at support B: RB * 1 m

Setting up the equation:

300 N-m - 800 N-m - RB * 1 m = 0

Simplifying the equation:

-500 N-m - RB = 0

RB = -500 N-m

Since the dynamic reactions at the supports are balanced, RA will be equal in magnitude but opposite in direction to RB:

RA = 500 N-m

Therefore, the dynamic reactions at the supports will be -500 N-m at support A and 500 N-m at support B.