-1
$\begingroup$

I am trying to drive that $[\alpha]=\mathrm{rad}/\mathrm{s^2}$ from $$\tau=I\cdot\alpha.$$ My try: $N\cdot m=\mathrm{kg}\cdot m^2 \cdot [\alpha]$, and as Newton is $\mathrm{kg}\cdot m/\mathrm{sec}^2$ then I got $$[\alpha]=\frac{m\cdot \mathrm{kg}\cdot \frac{m}{\mathrm{sec}^2}}{\mathrm{kg}\cdot m^2}=\frac{1}{\mathrm{sec}^2}?$$

Thanks

Improve this question
$\endgroup$
9
  • 1
    $\begingroup$ That is correct (although you have a typo in the first mention of the Newton unit. The metre isn't squared in the numerator). The radian is a unit-less unit so what you got here is an equivalent result. $\endgroup$
    – Steeven
    Commented Jun 7, 2021 at 16:50
  • $\begingroup$ Angular acceleration is expressed in $rad/s^2$. Radians are dimensionless, and are used to tell you how angles are measured. $\endgroup$
    – David White
    Commented Jun 7, 2021 at 16:53
  • $\begingroup$ @Steeven: May you explain why a radian is a unitless unit? $\endgroup$
    – Illuminate
    Commented Jun 7, 2021 at 16:54
  • $\begingroup$ rad/s² is same as 1/s² $\endgroup$
    – Akshaj Bansal
    Commented Jun 7, 2021 at 16:54
  • $\begingroup$ why does rad/$s^2$ be the same as $1/s^2$? $\endgroup$
    – Illuminate
    Commented Jun 7, 2021 at 16:56

1 Answer 1

Reset to default
1
$\begingroup$

You are correct; the "rad" term does not appear in your answer because the number of radians an angle is is defined as the ratio of the arc of the sector it sweeps through to the radius of that sector. This ratio between two lengths ends up being a dimensionless number with no units, as it's a quantity in metres divided by a quantity in metres, despite the fact that its meaning may most effectively be communicated with the companion term "radians".

You can think of your answer as equivalent to (m/m)/s²; in informal terms, you might think of this as "the acceleration (m/s²) experienced by a point in an object on its circular path around the centre of mass, per unit length of the distance of that point away from the center of mass (1/m)", which has the same meaning as "the angular acceleration of the points in the object".

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.