Rotating motor dynamics: reactions on supports.

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Homework Statement



Hi everyone!
I'm struggling to understand how to solve this problem, I searched on the web, but I didn't find anything similar.

I have a motor delivering a power P=5kW at n=1500rpm. This motor is mounted on 4 supports, each having a rigidity k=15kN/m. I know that distance between the supports is 250mm.

I need to determine the angle of which the motor frame rotates, and the sense of rotation.

I've added a drawing, hoping it helps to understand the situation

Homework Equations


The Attempt at a Solution

I've computed the angular speed in m/s and then I've used the formula for power to determine the torque produced by the motor. At this point I got stuck: I've considered the free body diagram of the disk of the rotor and I've put these forces on it:
- m*g acting on the baricenter of the disk (center)
- H and V as horizontal and vertical reactions of the bearing on the rotor
- M, the motor torque
- I*ω' as the inertial reaction of the rotor, then opposite to M

My idea was that, to produce an inclination of the frame I must have a non uniformly distributed vertical action on the supports, then I thought there should be an intertial action "m*a" on the baricenter of the rotor, but I can't really find out how to compute it (if that's the way to solve the exercise). Another problem is ofcourse that I don't have any information about the mass and the radius of the rotor, so I can't compute neither I nor ω'.

I think I have to use angular momentum equations to compute the intertial reactions, but I really can't find out how to!
If anyone could help me I would really appreciate that! Than you!
 

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Your drawing looks like an electric motor. The thing you are missing is "what causes the torque on the motor shaft". The answer is the electromagnetic forces between the rotating and static parts of the motor, not the forces acting at the rotor bearings.

If you use the power and RPM to find the torque T on the rotor, there is an equal and opposite torque -T on the motor casing.

Start by drawing a free body diagram of the motor casing, with the torque -T and the forces from the supports.

The are no "ma" forces involved, because the motor casing is not moving. You can ignore the weight of the motor, because if the supports are symmetrical about the center of mass of the motor, its weight will compress them all equally but won't cause any rotation of the casing.
 
The motor is at rest so the torque at the shaft must cancel out the torque produced by the springs.
 
Omg, feeling so dumb :D
Thank you very much!