Recall the formulas for linear speed and centripetal acceleration in uniform circular motion.

The linear speed ($v$) is related to the angular velocity ($\omega$) and radius ($r$) by:

$$v = r\omega$$

The centripetal acceleration ($a_c$) can be expressed in terms of linear speed and radius, or angular velocity and radius:

$$a_c = \frac{v^2}{r} = r\omega^2$$

Let the initial angular velocity be $\omega_1$.
Let the initial linear speed be $v_1 = V$.
Let the initial radius be $r_1$.
The initial centripetal acceleration is $a_{c1}$.
From the formulas:

$$V = r_1 \omega_1 \quad (1)$$ $$a_{c1} = r_1 \omega_1^2 \quad (2)$$

The angular velocity is doubled: $\omega_2 = 2\omega_1$.
The centripetal acceleration is unchanged: $a_{c2} = a_{c1}$.
Let the new linear speed be $v_2$.
Let the new radius be $r_2$.
From the formulas:

$$v_2 = r_2 \omega_2 = r_2 (2\omega_1) \quad (3)$$ $$a_{c2} = r_2 \omega_2^2 = r_2 (2\omega_1)^2 = 4r_2 \omega_1^2 \quad (4)$$

Since $a_{c1} = a_{c2}$:

$$r_1 \omega_1^2 = 4r_2 \omega_1^2$$

Divide both sides by $\omega_1^2$ (assuming $\omega_1 \neq 0$):

$$r_1 = 4r_2$$

This means the new radius is $r_2 = \frac{r_1}{4}$.

Substitute $r_2 = \frac{r_1}{4}$ into equation (3):

$$v_2 = \left(\frac{r_1}{4}\right) (2\omega_1)$$ $$v_2 = \frac{2 r_1 \omega_1}{4}$$ $$v_2 = \frac{1}{2} (r_1 \omega_1)$$

From equation (1), we know that $V = r_1 \omega_1$.
Substitute $V$ into the expression for $v_2$:

$$v_2 = \frac{1}{2} V$$

The new speed will be $\frac{V}{2}$.