Try using an NPN transistor with the pot in series with a suitable base resistor and the cap across BE. E to GND, C in series with relay to +V. Add a diode from base to the other end of the pot to discharge the cap rapidly when the input goes low.
The actual component values can be calculated by the OP to suit the application.
p.s. Should have added a diode across the relay coil. Use your imagination and pretend it is there
If I understand correctly, in this diagram the relay will take place of the LM555 in that when I apply voltage it will be transferred through the mosfet immediately. But because of the slow build-up due to the capacitor/resistor, relay won't turn on until a certain critical voltage is reached.
Well, it is a BJT, not a MOSFET. However, you could also do this with a MOSFET.
What is happening is voltage of the input would normally turn on the transistor to activate the relay. However, the capacitor is "stealing" the current that makes it through the resistors to charge. Once the capacitor gets full, the current it requires goes down and is used to activate the transistor. The time to reach this point is determined by the values of R and C.The transistor is just acting as the switch when we reach a threshold. For a MOSFET, it wouldn't be the current flowing through the transistor, but the gate voltage of the transistor that does the actual switching, but essentially the same thing occurring.
The relay isn't acting as the LM555 in other schematics, the transistor is. It is acting similar to what exists in the LM555 at the threshold pin, just much simpler.
You MAY required protection on the input side with a small resistor in front of the diode, if the discharging capacitor through the diode is too high of a current surge.