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Where did the radian come from (Torque)?

  1. Jan 19, 2014 #1
    In the simple torque equation:

    [tex] \tau_z = \alpha_z \times I [/tex]

    [tex] \frac{\tau_z}{I} = \alpha_z [/tex]


    [tex] \frac{10 N.m}{10 kg.m^2} = \alpha_z [/tex]

    [tex] \frac{10 kg.m^2}{10 kg.m^2.s^2} = \alpha_z [/tex]

    [tex] 1 \frac{rad}{s^2} = \alpha_z [/tex]
  2. jcsd
  3. Jan 19, 2014 #2
    Do you use Google?
  4. Jan 19, 2014 #3
    Yes. I understand that a "radian" is an angle measure. But radian is nonexistent in the formula.
  5. Jan 19, 2014 #4


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    What are the units of angular velocity? Of angular acceleration?
  6. Jan 19, 2014 #5
    Angular velocity: rad/s

    Angular acceleration: rad/s^2
  7. Jan 19, 2014 #6
    maybe i should use an analogy

    In the formula for acceleration:

    [tex] a = \frac{\Delta V}{time} [/tex]

    [tex] a = \frac{\frac{m}{s} - \frac{m}{s}}{s} [/tex]

    The formula explicitly shows that [tex] a = \frac{m}{s^2}[/tex]

    In the formula for torque, torque (τ) and the moment of momentum (I) do not have the units of "radians", hence my question: Where did the radians come from?
  8. Jan 19, 2014 #7
    Radians are really dimensionless. The units of angular velocity are simply ## \text{s}^{-1} ##. We usually add "radians" so that we can distinguish that from other possible dimensionless units, such as "revolutions" or "degrees".
  9. Jan 19, 2014 #8
    Yes, I see. Two questions: if the answer for acceleration (given the torque equation) was written as ## x \frac{rad}{s} ##, why doesn't the value change from ## 1 \text{s}^{-1} ## to ## \frac{1 rad}{2\pi sec} ##. Would ## x \frac{deg}{s} ## be any different?

    How do you use "#..#"?
    Last edited: Jan 19, 2014
  10. Jan 19, 2014 #9


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    hi raddian! welcome to pf! :smile:
    (you mean /s2 :wink:)

    yes! you need to convert from radians to degrees, using π = 180° :wink:
    you type two #s and two #s after :smile:
  11. Jan 19, 2014 #10
    Just like 2 degress is very different from 2 radians, 2 degrees per second is very different from 2 radians per second. You convert between the two exactly as you would convert between degrees and radians for simple angles. So ## 1 \ \text{rad} \ \text{s}^{-2} ## which you got previously gets converted to ## 57 \ ^ \circ \text{s}^{-2} ## by multiplying the former with ## 180^\circ \over \pi ##.

    You may be asking yourself "how do I know which dimensionless unit I should use"? It is very simple: in science, anything related to angles is done in radians by default. Then the result may be converted to degrees or revolutions or something else.

    Regarding the use of #..#, this is just a shorthand for the itex tag, just like $ .. $ is a shorthand for the tex tag. Note you need two #'s and two $'s on each side.
  12. Jan 19, 2014 #11
    But because ## \alpha = \frac{10 N.m}{10 kg.m^{2}} ##

    shouldn't the answer technically be ## 1 \frac{}{s^{2}} ##? And isn't ## 1 \frac{}{s^{2}} ## very different from ## 1 \frac{rad}{s^{2}} ##?
  13. Jan 19, 2014 #12
    The answer as ## 1 \ \text{s}^{-2} ## is perfectly correct. We know that it means ##1 \ {\text{rad} \over \text{s}^2} ## because we know that our default measure for angles is radians.
  14. Jan 19, 2014 #13
  15. Jan 19, 2014 #14


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    The angular acceleration can be [itex]α=\frac{rad}{s^{2}}[/itex], because the angular velocity is [itex]ω=\frac{rad}{s}[/itex] (how many rads per second the matter moves on a circle).
    Now the rad on a circle is connected to meters, because knowing the radius of the circle, you can determine how many meters the particle has moved, just by knowing how radians have changed. So for an obvious one, the full circle would be 2π (rads) and the length in meters is the well know 2πR
    The nice thing is that radians are dimensionless (the dimensions were absorbed into the radius)
  16. Jan 19, 2014 #15


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    The radian is a natural unit of arc - scientists and mathematicians on another planet / continent / age would all eventually find themselves using radians, quite independently (not the same name, of course). The use of 360 degrees or 100 divisions or whatever, is quite arbitrary.
  17. Jan 19, 2014 #16
    d[itex]\frac{sin(x)}{dx}[/itex]=cos(x) if x is expressed in radians. That is why many formulae use the radians unit.
  18. Jan 20, 2014 #17
    Now I (think) understand your question.

    The answer is that when we derive the formulae used in circular movement we suppose the angles are measured in radians.

    From wiki:
    "Let R be the radius of the circle, θ is the central angle in radians"

    The arc length is s = [itex]\frac{\theta}{R}[/itex]

    Theta is dimentionless, you could use degrees or turns, but the formula is derived supposing it is measured in radians.

    You can measure angles in degrees (even in dB) but then you should use another formulae.
  19. Jan 20, 2014 #18
    The current that flows through a capacitor is
    i =C[itex]\times[/itex][itex]\frac{dV}{dt}[/itex]

    If the voltage is v(t) = V[itex]_{0}[/itex][itex]\times[/itex]sin(w.t)

    the current will be i(t) = C.w.cos(wt)

    and the impedance of the capacitor will be w.C if w is expressed in radians
  20. Jan 21, 2014 #19
    The impedance of a capacitor is 1/(w.C). My mistake.
  21. Jan 21, 2014 #20
    From wiki:

    The result of this formula has to be expressed in dB. Why? It is dimentionless. Where did the dB come from? From the way we used to derive it.

    Honestly, I do not feel comfortable with my own answers.
  22. Jan 21, 2014 #21
    help me please

    please solve the integral in the attached file

    Attached Files:

    • B2.docx
      File size:
      27.7 KB
  23. Jan 21, 2014 #22


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    "dB" means wrt I0

    it's a way of saying what the log of I/I0 is :smile:

    (Basma Elyan, please start a new thread)
  24. Jan 21, 2014 #23
    I think the question posed by the OP is "Where did the dB/radians come from?"
    Not "What is a dB/radian?"
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