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

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SUMMARY

The angular velocity of a process control motor is defined by the equation (21−(1/2t^2)) rad/s. The motor reverses direction at t = 6.5 seconds, determined by setting the angular velocity equation to zero. To find the angle through which the motor turns before reversing, one must integrate the angular velocity function, resulting in the angular displacement equation θ(t) = 21t - (1/6)t^3 + C. Evaluating this integral from t = 0 to t = 6.5 seconds provides the total angular displacement.

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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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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.
 
ok, I'll integrate the function
 
Last edited:
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?
 
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?
 

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