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storchak [24]
4 years ago
12

The controlled variable in a closed-loop system is the direction of a robot arm. Initially, it is at 50o; then it is commanded t

o go to 25o. What is the error when the arm position is at 15o?a. 15 deg
b. 25 deg
c. 10 deg
d. 35 deg
Engineering
1 answer:
stellarik [79]4 years ago
5 0

Answer:

c. 10 deg

Explanation:

Error is the difference between the commanded reference and the actual output, E(s) = R(s) - Y(s)

the measured output is 15°

and the desired output 25°

25 - 15 = 10°

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Answer:

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Explanation:

Step1

Given:

Inner diameter is 2.00 in.

Gap between cups is 0.2 in.

Length of the cylinder is 2.5 in.

Rotation of cylinder is 10 rev/min.

Torque is 0.00011 in-lbf.

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Step2

Calculation:

Tangential force is calculated as follows:

T= Fr

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F = 0.00011 lb.

Step3

Tangential velocity is calculated as follows:

V=\omega r

V=(\frac{2\pi N}{60})r

V=(\frac{2\pi \times10}{60})\times1

V=1.0472 in/s.

Step4

Apply Newton’s law of viscosity for dynamic viscosity as follows:

F=\mu A\frac{V}{y}

F=\mu (\pi dl)\frac{V}{y}

0.00011=\mu (\pi\times2\times2.5)\frac{1.0472}{0.2}

\mu =1.3374\times 10^{-6}lb-s/in².

Step5  

Kinematic viscosity is calculated as follows:

\upsilon=\frac{\mu}{\rho}

\upsilon=\frac{1.3374\times 10^{-6}}{0.00095444}

\upsilon=1.4012\times 10^{-3} in2/s.

Thus, the dynamic viscosity and kinematic viscosity are 1.3374\times 10^{-6} lb-s/in2 and 1.4012\times 10^{-3} in2/s.

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Answer:

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