An annular disk of mass M , inner radius a and outer radius b is...

As we know, $\overset{\to}{J} = ∆ \overset{\to}{P}$

$⇒ {J}_{0} = MV . . . \left(i\right)$

About center of the disk, we can also write

${J}_{0} \times h = {I}_{CM}\omega ⇒ {J}_{0} \times h = \left[\dfrac{M}{2}\left({a}^{2} + {b}^{2}\right)\right]\omega . . . \left(ii\right)$

If $h = {h}_{m} ⇒ \omega = \dfrac{V}{b}\left(\text{pure rolling}\right)$

From equation(ii), we can write

$⇒ {J}_{0} \times {h}_{m} = \left(\dfrac{M}{2}\right)\left({a}^{2} + {b}^{2}\right)\dfrac{V}{b} . . . \left(iii\right)$

By dividing equation $( i )$ with the equation $( iii )$, we get

$⇒ {h}_{m} = \dfrac{{a}^{2} + {b}^{2}}{2b}$

Now, considering all cases:

If $a \to 0 ⇒ {h}_{m} = \dfrac{b}{2}$

If $a \to b ⇒ {h}_{m} = b$

If $h = {h}_{m} ⇒ \omega$ is independent of $a$(As we get $\omega = \dfrac{V}{b}$)

If $\mu = 0\text{and}h = 0 ⇒$Wheel will slide without rolling.