Moment of inertia of a cylinder with a drilled hole

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I can't wrap my head around this problem?
I think that when you drill a whole like that, the cylinder will roll back to attain its equilibrium so need to hold it with a certain torque to keep it in place, but i do not know how to find the torque

What i have in mind is:
since T=dL/dt, but L=Iw(w is angular velocity) so T=Ia(a is angular acceleration). So I'm trying to find a balanced torque but i do not know how to find the new moment of inertia? Do we use the parallel axis theorem? or it doesn't work because there's hole in it?? HELP!
 
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This is the same thing as if you took the original cylinder and applied an upward force of 15N at the location where you drilled the hole (i.e., at the location of the center of the hole).

Chet
 
Can you explain how you got that? that'd be swelled thanks
 
xphysics said:
Can you explain how you got that? that'd be swelled thanks

Removing material is the same as adding negative material (or gluing on negative weight material).

Chet
 
xphysics said:
but.. the right answer isn't 15N it's 6N?
You need to put the 15 N force in where I said, and then do a balance of moments to figure out what additional force you need to apply tangentially at a to get no net moment. For example, the moment of the 15 N force is 15(2a)/5.

Chet
 
OH! I GET IT! THANKS DUDE! still kinda not intuitive about the idea of that 15N upward force, what if the shape it different or cutting 2 shapes of different of different size?? is there a mathematical way to determine that 15N?
 
xphysics said:
OH! I GET IT! THANKS DUDE! still kinda not intuitive about the idea of that 15N upward force, what if the shape it different or cutting 2 shapes of different of different size?? is there a mathematical way to determine that 15N?
I don't think you need anyone's help answering either of these questions. Just take some time to mull them over (say, while you're walking the dog, brushing your teeth, or taking a shower). If you're still not sure after another day, come back and I'll help you.

Chet