Since the Equation for the Angular velocity of Precession of a Spinning Disk is:
$$\omega _{p}=\frac{m*g*r}{I*\omega _{spin}}$$
Where,
m -> Mass of the spinning Disk. [$kg$]
g -> Acceleration due to Gravity. [$m/(s^2)$]
r -> Distance of the Centroid of disk from the pivot. [$m$]
I -> Mass moment of inertia of the Disk. [$kg*m^2$]
$\omega _{spin}$ -> Angular velocity of the disk. [$rad/s$]
$$\therefore \, \, \omega _{p} (unit) = \frac{kg*\frac{m}{s*s}*m}{kg*m*m*\frac{rad}{s}}$$
Simplifying,
$$\omega _{p} (unit) = \frac{1}{rad*s}$$
Shouldn't it be:
$$\omega _{p} (unit) = \frac{rad}{s}$$
The Answer that I got is 'Since radian is the ratio of lengths, it has no dimension. Which means the expression $\frac{1}{rad*s}$ Simplifies to $\frac{1}{s}$ and again could be written as $\frac{rad}{s}$
So, instead of $\frac{rad}{s}$, I could write $\omega _{p} (unit) = \frac{rev.}{s}$, which are similar.
Now, I want to solve a problem with this.
A gyroscope has a 1-kg disk that spins at 100 rev/s. The center of mass of the disk is 10 cm from a pivot which is also the radius of the disk.
To find the precession angular velocity.
m = 1 [$kg$]
g = 9.8 [$m/(s^2)$]
r = 0.1 [$m$]
I (Disk) = $\frac{mr^2}{2}$ = $5*10^{-3}$ [$kg*m^2$]
$\omega _{spin}$ = 100 [$rev/s$]
On Solving, $$\omega _{spin} = 1.96 \, \frac{rev}{s}$$
since, 1 rev/s = 2$\pi$ rad/s
$$\omega _{spin} = 12.315 \, \frac{rad}{s}$$
Now, instead I want to give the input for $\omega _{spin}$ as $2 \pi * 100$ [$rad/s$]. (Solving the problem again with all the above given data.)
Which gives $$\omega _{spin} = 0.31194 \, \frac{rad}{s}$$
Which is kinda Contradictory.
Since rev/s and rad/s are similar units.
I used the same method above to derive the unit of $\omega _{p} (unit) \,\, as \,\, \frac{rev}{s}$
Now, please explain the catch in here..
