I'm not sure if I can drive a 24V (0.5A) dc motor with the following circuits?
I fear that there will be burned too much power in the transistor when the supply is 24V (I will become super hot).
Here is a link to the datasheet for the IRLZ44N: http://www.irf.com/product-info/datasheets/data/irlz44n.pdf
Thanks for the help!
The short answer is that it will work electrically. Vgsth for 0.5A when Vds=24V is quite low (<2.5V).
The main concern here is thermal. So, there is a thing called thermal resistance. It tells you how much the temperature will increase inside the device under various conditions. If you just stick the FET on some perfboard, the main parameter here is \$R_{\theta JA}\$, and what it says is that your junction temp will increase 62degC for every watt the device dissipates.
The other piece to the puzzle is the Rds on figure. This tells you what kind of resistor the FET behaves like under certain conditions. Here, your Vgs plays a big role. Let's assume it is small... like 3.3V. You can see that Rds on increases when Vgs decreases. That means your Rds on number will be higher than what is listed on the datasheet. Let's ballpark 0.05\$\Omega\$ 
Putting it all together that means you have: \$ T_{ambient} + I^2*R_{DSon} * R_{\theta JA} = T_{junction} \rightarrow 25^\circ C + 0.5A^2 * 0.05\Omega * 62^\circ C = 25.775^\circ C \$
At 0.5A, with Vgs>3V @ 25C, this load is a walk in the park for this FET.
From this datasheet Rds-on at 5V= 0.028Ohm
Lets be conservative and say 10x that at 5V drive.
Power = i * U = R * iˆ2
Power = 0.28R * 0.5A * 0.5A = 0.07W
Junction to Ambient thermal resistance is 62 ºC per W.
So considering 10* Rds-on your temp rise is:
62C/W * 0.07W = 4ºC
So you should not be worried about overheating.
I'd add a gate resistor and use a faster diode as winny said.
