1. Given vectors

$\vec v_1 = \sqrt{3}p\,\hat i + \hat j = (\sqrt{3}p,1),$ $\vec v_2 = 2\hat i + (p+1)\hat j = (2,p+1).$

Since $\vec v_2$ is obtained by rotating $\vec v_1$ about the origin, their magnitudes are equal.

2. Use equality of magnitudes

$|\vec v_1|^2 = |\vec v_2|^2$

$3p^2 + 1 = 2^2 + (p+1)^2$

$3p^2 + 1 = 4 + p^2 + 2p + 1$

$3p^2 + 1 = p^2 + 2p + 5$

$2p^2 - 2p - 4 = 0$

$p^2 - p - 2 = 0$

$(p-2)(p+1)=0$

Given $p>0$, so

$p=2.$

3. Substitute $p=2$

Then

$\vec v_1 = (2\sqrt{3},1), \qquad \vec v_2=(2,3).$

4. Find angle of rotation using tangent formula

If $\theta$ is the angle from $\vec v_1$ to $\vec v_2$, then

$\tan\theta = \frac{x_1y_2-y_1x_2}{x_1x_2+y_1y_2}.$

Here,

$x_1=2\sqrt3,\ y_1=1,\ x_2=2,\ y_2=3.$

So,

$x_1y_2-y_1x_2 = (2\sqrt3)(3)-(1)(2)=6\sqrt3-2,$

$x_1x_2+y_1y_2 = (2\sqrt3)(2)+(1)(3)=4\sqrt3+3.$

Hence,

$\tan\theta = \frac{6\sqrt3-2}{4\sqrt3+3}.$

This is given as

$\tan\theta = \frac{\alpha\sqrt3-2}{4\sqrt3+3}.$

Comparing numerators,

$\alpha\sqrt3 -2 = 6\sqrt3 -2$

so

$\alpha=6.$

5. Final answer

$\boxed{6}$