Where did the radian come from (Torque)?

[raddian]

In the simple torque equation:

$$\tau_z = \alpha_z \times I$$

$$\frac{\tau_z}{I} = \alpha_z$$

Finally:

$$\frac{10 N.m}{10 kg.m^2} = \alpha_z$$

$$\frac{10 kg.m^2}{10 kg.m^2.s^2} = \alpha_z$$

$$1 \frac{rad}{s^2} = \alpha_z$$

 

[Malverin]

In the simple torque equation:

$$\tau_z = \alpha_z \times I$$

$$\frac{\tau_z}{I} = \alpha_z$$

Finally:

$$\frac{10 N.m}{10 kg.m^2} = \alpha_z$$

$$\frac{10 kg.m^2}{10 kg.m^2.s^2} = \alpha_z$$

$$1 \frac{rad}{s^2} = \alpha_z$$

Do you use Google?

 

[raddian]

Yes. I understand that a "radian" is an angle measure. But radian is nonexistent in the formula.

 

[SteamKing]

What are the units of angular velocity? Of angular acceleration?

 

[raddian]

Angular velocity: rad/s

Angular acceleration: rad/s^2

 

[raddian]

maybe i should use an analogy

In the formula for acceleration:

$$a = \frac{\Delta V}{time}$$

$$a = \frac{\frac{m}{s} - \frac{m}{s}}{s}$$

The formula explicitly shows that $$a = \frac{m}{s^2}$$

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?

 

[voko]

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".

 

[raddian]

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".

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 "#..#"?

 

[tiny-tim]

hi raddian! welcome to pf!

if the answer for acceleration (given the torque equation) was written as x rad/s, would x °/s be any different?

(you mean /s² )

yes! you need to convert from radians to degrees, using π = 180°

How do you use "#..#"?

you type two #s and two #s after

 

[voko]

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.

 

[raddian]

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 $1 \frac{rad}{s^{2}}$?

 

[voko]

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.

 

[raddian]

Voko your answer led me on the right path to http://webcache.googleusercontent.com/search?q=cache:w8p0JeQq9gMJ:mathforum.org/library/drmath/view/64034.html+&cd=3&hl=en&ct=clnk. Thank you all for your answers.

 

[ChrisVer]

The angular acceleration can be $α=\frac{rad}{s^{2}}$, because the angular velocity is $ω=\frac{rad}{s}$ (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)

 

[sophiecentaur]

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.

 

[alva]

d$\frac{sin(x)}{dx}$=cos(x) if x is expressed in radians. That is why many formulae use the radians unit.

 

[alva]

maybe i should use an analogy

In the formula for acceleration:

$$a = \frac{\Delta V}{time}$$

$$a = \frac{\frac{m}{s} - \frac{m}{s}}{s}$$

The formula explicitly shows that $$a = \frac{m}{s^2}$$

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?

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 = $\frac{\theta}{R}$

Theta is dimensionless, 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.

 

[alva]

d$\frac{sin(x)}{dx}$=cos(x) if x is expressed in radians. That is why many formulae use the radians unit.

The current that flows through a capacitor is
i =C$\times$$\frac{dV}{dt}$

If the voltage is v(t) = V$_{0}$$\times$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

 

[alva]

and the impedance of the capacitor will be w.C if w is expressed in radians

The impedance of a capacitor is 1/(w.C). My mistake.

 

[alva]

From wiki:

Sound intensity level or acoustic intensity level is a logarithmic measure of the sound intensity (measured in W/m2), in comparison to a reference level.

The measure of a ratio of two sound intensities is

L= 10 . log$_{10}$$\frac{I1}{I0}$ dB

where I0 is the reference level.

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

Honestly, I do not feel comfortable with my own answers.

 

[Basma Elyan]

help me please

please solve the integral in the attached file

 

[tiny-tim]

From wiki:

where I0 is the reference level.

The result of this formula has to be expressed in dB. Why? It is dimensionless. Where did the dB come from?

"dB" means wrt I₀

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

(Basma Elyan, please start a new thread)

 

[alva]

"dB" means wrt I₀

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

I think the question posed by the OP is "Where did the dB/radians come from?"
Not "What is a dB/radian?"