That's almost there. Reduce the value of the 4.7K to 2.2K. If that doesn't work, the 741 is not able to supply enough current to drive the transistor. You may have to add an amplifier stage.

I reduced the value of 4.7K to 2.2K. Same thing. If the motor is not connected to the gear, it spins when the output of 741 is hight, and it doesn't if if the output is low.. If it's connected to the gear, it doesn't do anything. The voltage at the base is 0.62V and 0.47V. By the way, I ended up destroying a motor. Luckily, I have some more.

I reduced to 1K. And when not connected to the gear, if the output of 741 is low, it spins, if it's high, it spins more forcefully. If I connect it to the gear, it doesn't do anything.

EDIT: It actually works with the 1K||1K voltage divider. The reason why I thought it didn't is because the battery drains too fast. I was using a brand new 9V battery, and after a few trials it went down to ~7.20V. How can I power this circuit so the battery doesn't last just a few minutes?

 

Use a battery with a higher capacity and lower internal resistance. You do not say what kind of 9 volt battery you are using or the motor current under load so I am assuming it is the very small PP3 type battery. I think a battery pack made up of 6 AA size alkaline cells would probably work for a reasonable time. This is just a guess without knowing the motor current to do a calculation.

Les.

 

The battery drains too fast. I was using a brand new 9V battery, and after a few trials it went down to ~7.20V. How can I power this circuit so the battery doesn't last just a few minutes?

The little 9V battery is designed to power a smoke detector that uses an extremely low current that has no electric motor.
The size of the battery determines how long it will run the motor.

 

Now you have it working, you should experiment with it to make it work better, especially with the LDR. I am assuming that you are using a cadmium sulfide sensor. They have a very large change in resistance from dark to light. They are effected by ambient light which is probably what is causing yours to produce such a small output change. It helps to make it more directional by putting it in a short tube. That reduces the detection of ambient light and will give a lower "off" voltage at the output of the op-amp. Increasing the value of the series resistor from 1K to 10K will also help to increase the voltage change from off to on.
Have fun with your project and keep on learning.

 

" That reduces the detection of ambient light and will give a lower "off" voltage at the output of the op-amp. "

How does that work with the Op-Amp configured as a comparator?

 

Now you have it working, you should experiment with it to make it work better, especially with the LDR. I am assuming that you are using a cadmium sulfide sensor. They have a very large change in resistance from dark to light. They are effected by ambient light which is probably what is causing yours to produce such a small output change. It helps to make it more directional by putting it in a short tube. That reduces the detection of ambient light and will give a lower "off" voltage at the output of the op-amp. Increasing the value of the series resistor from 1K to 10K will also help to increase the voltage change from off to on.
Have fun with your project and keep on learning.

Yes, I'm using a regular LDR. I ended up using a 12V battery with a 7809 voltage regulator. I left the Rb unsoldered so I can remove it and replace it with another value. Thanks for your help!