Engineering Equation for a DC motor driving an arm

AI Thread Summary
The discussion centers on the dynamics of a DC motor driving an arm, with emphasis on the relationship between the motor angle (θ_m) and the load angle (θ_L). Clarification is sought on why these angles are not equal, particularly in the absence of a gearbox that would alter their relationship. Key parameters include the torsional spring constant (K) and the viscous-friction coefficient (B), which are suggested to be positioned between the motor and the arm. The original poster is struggling to develop the necessary dynamic equations for their attempted solution involving differential equations. Understanding these relationships is crucial for accurately modeling the system's behavior.
YinSan
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Homework Statement
Develop dynamic equations of the system with ia(t).
Relevant Equations
Tm=KiIa
https://d2vlcm61l7u1fs.cloudfront.net/media/9b2/9b24aa39-e135-4529-9eaf-29a3cd057502/phpubWKNH.png
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Welcome to the PF. :smile:

It would help if you would put into words what the problem is and what your approach is.

And could you please clarify why ##\theta_m## and ##\theta_L## are not equal? I don't see any sort of gearbox that will gear down the motor to make the arm rotate differently from the shaft.

1594651927636.png
 
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K is the torsional spring constant
B = viscous-friction coefficient of shaft
between the motor and arm
TL is the disturbance on the torque arm.
K& B I think it should be located between the arm and motor. Subscript m refers to the motor. Subscript L refers to the load.
The picture I posted is my attempted solution for differential equations. I just don't know how to develop dynamic equations.
 
Thread 'Have I solved this structural engineering equation correctly?'

Thread 'Have I solved this structural engineering equation correctly?'

Hi all, I have a structural engineering book from 1979. I am trying to follow it as best as I can. I have come to a formula that calculates the rotations in radians at the rigid joint that requires an iterative procedure. This equation comes in the form of: $$ x_i = \frac {Q_ih_i + Q_{i+1}h_{i+1}}{4K} + \frac {C}{K}x_{i-1} + \frac {C}{K}x_{i+1} $$ Where: ## Q ## is the horizontal storey shear ## h ## is the storey height ## K = (6G_i + C_i + C_{i+1}) ## ## G = \frac {I_g}{h} ## ## C...
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