# Time to rotate system 20 degrees

1. Feb 5, 2015

### D.B0004

1. The problem statement, all variables and given/known data
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.

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

3. 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.

2. Feb 5, 2015

### Simon Bridge

Where is problem 2.6 and what is the system it proposes?

3. Feb 5, 2015

### D.B0004

Im sorry. I completely forgot to add those! See file attached.

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4. Feb 5, 2015

### D.B0004

Wow and now I posted the WRONG PICTURE. Thought I only had one screen shot.

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5. Feb 5, 2015

### D.B0004

This problem probably shouldn't be in the introductory section...but hey maybe someone will get it.

6. Feb 5, 2015

### D.B0004

Here are equations listed below the figure.

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7. Feb 6, 2015

### haruspex

In the time the motor rotates once, how often does the load rotate?
It's constant during each of the two phases, acceleration and deceleration.
Constants are the easiest of all to integrate.