Time to rotate system 20 degrees

  • Thread starter D.B0004
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  • #1
D.B0004
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Homework Statement


In the system of Problem 2-6, we need move the load by 20 degrees and bring the system to rest. Assuming a triangular speed profile of the load with equal acceleration and deceleration rates (starting and ending at zero speed). Assume a coupling efficiency of 100%. The magnitude of the electromagnetic torque (positive or negative) available from the motor is 500 Nm.

nL/nM= 3
JL = 8.3 kg*m2
JM = 1.4 kg*m2

What is the time (in seconds) needed to rotate the load by an angle of 20o? Give the correct answer to 3 or more decimal places.

Homework Equations


T(em) = [Jm + (Wl/Wm)^2 * Jl]*dw/dt

The Attempt at a Solution


I see a dt in there so I think this is my hint. I know I can set T(em) to 500. Jm is given, and I believe W (omega) of load is 3 times smaller than the motor, w/w would be (1/3)^2. But dw/dt is constant due to the triangle torque constraint isn't it? Cant integrate that right? Could someone explain how I am thinking of this wrong with a push in the right direction? Thanks.
 

Answers and Replies

  • #2
Simon Bridge
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In the system of Problem 2-6,...
Where is problem 2.6 and what is the system it proposes?
 
  • #3
D.B0004
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Where is problem 2.6 and what is the system it proposes?
Im sorry. I completely forgot to add those! See file attached.
 

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  • #4
D.B0004
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Im sorry. I completely forgot to add those! See file attached.
Wow and now I posted the WRONG PICTURE. Thought I only had one screen shot.
 

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  • #5
D.B0004
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This problem probably shouldn't be in the introductory section...but hey maybe someone will get it.
 
  • #6
D.B0004
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Here are equations listed below the figure.
 

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  • #7
haruspex
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I believe W (omega) of load is 3 times smaller than the motor
In the time the motor rotates once, how often does the load rotate?
dw/dt is constant
It's constant during each of the two phases, acceleration and deceleration.
Cant integrate that right?
Constants are the easiest of all to integrate.
 

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