Voltage = 400v
I = C*dv/dt. dv/dt is rate of change of voltage in volts per second - which means voltage would drop 1/.000047 or about 2130 volts per second. It will be back to zero volts in about one-tenth of a second.
Capacitor can only discharge slowly by limiting current or adding capacitance.
Charged capacitors can deliver high current at high voltage and is hazardous.
EDIT: The energy stored in the cap can be computed as follows:
E=1/2CV^2 = 0.5 ×47⋅10−6F × (400V)2=3.8J
You could add a resistor in series with the motor, so to drop the voltage to a sane level and limit the current (thus increasing the discharge time), but then only a fraction of that energy gets delivered to the motor, rather than just dissipated in the resistor!
For comparison, a AA battery (let's say 1.5 V, 2.4Ah) stores as much as 13000 joules!
Adding resistor with various resistance for the 47uF @ 400 Volt capacitor to light up LED
10 Ohm - Still fast release
100 Ohm - Still fast release
270 Ohm - Still fast release
1KOhm - Still fast release
20KOhm - Release a little slower