As a plane gets slower, why does a certain bank angle make you turn faster?
Rate of turn is dependent on the following two items:


The horizontal component of lift (centripetal force)
The tangential velocity of the aircraft (true airspeed)


The rate or turn is directly proportional to the horizontal component of lift and inversely proportional to the tangential velocity of the aircraft.

For a given angle of bank, the vertical and horizontal components of lift will be the same, regardless of airspeed in level flight.

Consequently the airplane will experience the same centripetal acceleration, regardless of airspeed.

Since the tangential velocity is slower, any kind of centripetal force will produce a greater rate of turn for a slower flying aircraft as opposed to a faster moving aircraft and this can be shown by the centripetal acceleration equation

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

so both slow flying airplane with a true airspeed $v_s = 100$ knots and fast flying airplane with a true airspeed $v_f = 200$ knots experience the same centripetal acceleration. 

$$\dfrac{v_s^2}{r_s} = \dfrac{v_f^2}{r_f} = 4\ \dfrac{v_s^2}{r_f}$$

or, $$\dfrac{1}{r_s} = \dfrac{4}{r_f}$$

Consequently  $r_s &lt; r_f$; in this case $r_f = 4\ r_s$

Since the angular velocity is equal to the tangential speed divided by the radius.

$$\omega = v/r$$

the angular speed of the slower aircraft will be greater than the faster aircraft.

$$\omega_s = v_s/r_s$$

and 

$$\omega_f = \dfrac{v_f}{r_f} = \dfrac{2 \  v_s}{4 \ r_s} = \frac{1}{2}w_s$$

So our twice as slow airplane turns twice as fast as the faster one does under these conditions.