Machine Dynamics

Classical Mechanics Level pending

Consider a hypothetical rotating machine, whose angular position θ \theta varies as follows:

τ i n τ o u t α ( θ ˙ θ ˙ r e f ) = I θ ¨ \large{\tau_{in} - \tau_{out} - \alpha \, (\dot{\theta} - \dot{\theta}_{ref}) = I \, \ddot{\theta}}

In the above equation, τ i n \tau_{in} and τ o u t \tau_{out} are the input and output torques, α \alpha is a damping parameter, θ ˙ r e f \dot{\theta}_{ref} is the machine's reference speed, and I I is the machine's moment of inertia.

Initially, the input and output torques are at their nominal values, and the machine rotates at the reference speed.

Then, the output torque increases to 150 % 150 \% of its nominal value, and the machine approaches a new steady-state speed.

How many rad/s \text{rad/s} below the reference speed is the new steady-state speed?

Nominal Parameter Values:
τ i n = τ o u t = 10 N m α = 3 N m s I = 4 N m s 2 θ ˙ r e f = 120 π rad/s \tau_{in} = \tau_{out} = 10 \, \text{N m} \\ \alpha = 3 \, \text{N m s} \\ I = 4 \, \text{N m s}^2 \\ \dot{\theta}_{ref} = 120 \pi \, \text{rad/s}


The answer is 1.6667.

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1 solution

Steven Chase
Jan 30, 2018

The machine asymptotically approaches a new steady-state, in which θ ¨ = 0 \ddot{\theta} = 0 . In the limit:

τ i n τ o u t α ( θ ˙ θ ˙ r e f ) = I θ ¨ 10 15 3 ( θ ˙ θ ˙ r e f ) = 0 θ ˙ θ ˙ r e f = 5 3 \tau_{in} - \tau_{out} - \alpha \, (\dot{\theta} - \dot{\theta}_{ref}) = I \, \ddot{\theta} \\ 10 - 15 - 3 \, (\dot{\theta} - \dot{\theta}_{ref}) = 0 \\ \dot{\theta} - \dot{\theta}_{ref} = -\frac{5}{3}

Thus, the new steady-state speed is approximately 1.667 rad/s \text{rad/s} slower than the reference speed.

Steven! This is an excellent dynamics problem if we study its step response to external torque changes!

Max Yuen - 2 years, 1 month ago

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Yes, I agree. Would you like to post that one?

Steven Chase - 2 years, 1 month ago

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