The angular velocity of a process control motor is (21−(1/2t^2)) rad/s

In summary, the angular velocity of a process control motor is given by the function (21−(1/2t^2)) rad/s, where t is in seconds. When the equation is set equal to 0, the time at which the motor reverses direction is found to be 6.5 seconds. To find the total angle turned by the motor between t=0s and t=6.5s, the function can be integrated to get (21t - (1/6t^3) + c), with the constant of integration becoming irrelevant when evaluating a definite integral. The resulting answer is -0.8125 rad, representing the total angular displacement of the motor.
  • #1
zapzapper
3
0

Homework Statement


The angular velocity of a process control motor is (21−(1/2t^2)) rad/s, where t is in seconds.

1.) At what time does the motor reverse direction?

2.) Through what angle does the motor turn between t =0 s and the instant at which it reverses direction?

Homework Equations


angular velocity of a process control motor is (21−(1/2t^2)) rad/s

The Attempt at a Solution


For part 1 I just set the equation equal to 0 and got a time of 6.5 sec, which is correct

For part 2, I have no idea what to do. Nothing is working, so if you could help, that would be great! I have tried taking the derivative to find acceleration, which is -t, and then plugging that into θf = θi + ωi(t) + 1/2a(t^2). I set θi = 0 because I figured there would initially be no angle. I set ωi = 21 because at t=0, ω = 21. I set t= 6.5, and a=-6.5...so my equation was θf = 0 + 21(6.5s) + 1/2(-6.5)(6.5s^2), but the answer came to like -0.8125 rad? is that right? it seems odd to me...
 
Last edited:
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  • #2
zapzapper said:

Homework Statement


The angular velocity of a process control motor is (21−(1/2t^2)) rad/s, where t is in seconds.

1.) At what time does the motor reverse direction?

2.) Through what angle does the motor turn between t =0 s and the instant at which it reverses direction?


Homework Equations


angular velocity of a process control motor is (21−(1/2t^2)) rad/s


The Attempt at a Solution


For part 1 I just set the equation equal to 0 and got a time of 6.5 sec, which is correct

For part 2, I have no idea what to do. Nothing is working, so if you could help, that would be great! I have tried taking the derivative to find acceleration, which is -t, and then plugging that into θf = θi + ωi(t) + 1/2a(t^2). I set θi = 0 because I figured there would initially be no angle. I set ωi = 21 because at t=0, ω = 21. I set t= 6.5, and a=-6.5...so my equation was θf = 0 + 21(6.5s) + 1/2(-6.5)(6.5s^2), but the answer came to like -0.8125 rad? is that right? it seems odd to me...

Hi zapzapper, Welcome to Physics Forums.

It appears that you're familiar with calculus since you mentioned taking a derivative.

Suppose this were a linear motion problem and you were given a velocity function v(t). How would you go about finding d(t) using calculus? You can apply the same method to rotational motion where velocity is represented by the variable ω and distance by θ.
 
  • #3
ok, I'll integrate the function
 
Last edited:
  • #4
gneill said:
Hi zapzapper, Welcome to Physics Forums.

It appears that you're familiar with calculus since you mentioned taking a derivative.

Suppose this were a linear motion problem and you were given a velocity function v(t). How would you go about finding d(t) using calculus? You can apply the same method to rotational motion where velocity is represented by the variable ω and distance by θ.

oh ok, so I just integrate the function then?, If I integrate it then I get (21t - (1/6t^3) + c), what do i do with the constant c though?
 
  • #5
You are trying to find the total angular displacement between two known times. What happens to the constant of integration when you evaluate a definite integral?
 

Related to The angular velocity of a process control motor is (21−(1/2t^2)) rad/s

1. What is the significance of the angular velocity in a process control motor?

The angular velocity of a process control motor refers to the rate at which the motor rotates around its axis. It is an important factor in determining the speed and efficiency of the motor in controlling a process.

2. How is the angular velocity of a process control motor calculated?

The angular velocity can be calculated by dividing the change in angle (in radians) by the change in time. In this case, the formula for the angular velocity is (21 - (1/2t^2)) rad/s.

3. What does the term "rad/s" mean in relation to angular velocity?

The term "rad/s" stands for radians per second and is the unit of measurement for angular velocity. It represents the change in angle (in radians) per unit of time.

4. How does the angular velocity of a process control motor affect the overall process?

The angular velocity directly impacts the speed and accuracy of the process control motor. A higher angular velocity means the motor can rotate faster, allowing for quicker and more precise adjustments in the process.

5. Can the angular velocity of a process control motor be adjusted?

Yes, the angular velocity of a process control motor can be adjusted by changing the parameters of the motor, such as the voltage or current. This can be done to optimize the motor's performance for a particular process.

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