Mechanical model of the proton (angular momentum etc.)

Question:
The proton's mass and radius are 1.67 $\times$ 10$^{-27}$ kg and 1.00 $\times$ 10$^{-15}$ m. This system is governed by quantum and classical regimes. Namely, the proton is assumed to have a rotational uniform solid spherical body and a half spin of quantum angular momentum. Find the equatorial velocity of the proton.

Answer:
In terms of classical sense, the moment of inertia of proton is
\[ I = \frac{2}{5}mr^2 \]
Therefore, the angular momentum becomes
\[ L = I\omega = \frac{2}{5}mr^2\omega \]
For the quantum perspective, the spin angular momentum is given as
\[ S = \frac{1}{2}\hbar \]
Therefore,
\[ \frac{2}{5}mr^2\omega  = \frac{1}{2}\hbar \]
Solve for the angular velocity.
\[ \omega = \frac{1}{2}\hbar \frac{5}{2mr^2} \]
The equatorial velocity is given by
\[ v = r\omega =  \frac{5\hbar}{4mr} \\
      = \frac{5}{4}\frac{1.055\times 10^{-34}\mathrm{Js}}{1.673\times 10^{-27}\mathrm{kg}\times 1.00\times 10^{-15}\mathrm{m}} \]
Then we have
\[ v = 7.88 \times 10^7 \mathrm{m/s} \]

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