Acceleration of a rotating system

[ahuebel]

I have this problem where I need to find the acceleration of a system of a disk attached to an axle that is placed on an incline made of two parallel bars. I think I basically know how to do the problem, I just need to know how to figure out total inertia. I know the inertia of a cylinder (the axle) and a disk is (1/2)MR^2 but I am not sure how to find total inertia since the axle basically goes through the disk.

TIA for any help.

 

[Doc Al]

Until we see the details of what you did, it's hard to say where you went wrong. Are the masses of cylinder and disk equal? I assume there are two different "R"s involved: that of the disk and that of the axle.

 

[Doc Al]

I also know that friction acts at both contact points (the axle on each bar) so basically 2mgsin(theta) - 2f = ma.

If M is the total mass of the "disk + cylinder" and f is the total friction, then: Mgsin(theta) - f = Ma.

 

[ahuebel]

I edited my post after you responded. Sorry about that.
You are right about your last message and the masses are not equal.

The mass of the axle is .1 kg (not including the part inside the disk), and has a radius of .02m. The mass of the disk is .8kg and its radius is .12cm. This seems wrong since those radii would make the disk smaller than the axle. The book must have meant .12m.

So it seems I can basically find the inertia of a disk of radius .02m and subtract that from the inertia of a disk of .12m and add that inertia to a cylinder of radius .02m to get total inertia, does this seem right? This assumes the same density for both disk and axle, I guess I can assume that...

Isnt the inertia of a disk the same as the inertia of a cylinder (1/2 MR^2)?

 

[Doc Al]

So it seems I can basically find the inertia of a disk of radius .02m and subtract that from the inertia of a disk of .12m and add that inertia to a cylinder of radius .02m to get total inertia, does this seem right?

Why are you subtracting anything? Find the inertia of the disk; find the inertia of the axle (outside of the disk, which is what you are given). Add them.

Isnt the inertia of a disk the same as the inertia of a cylinder (1/2 MR^2)?

Right.

 

[ahuebel]

I guess I am confused because the inertia of the disk includes a portion of the axle (that portion inside the disk) which leaves basically two cylinders on the sides of the disk if you can visualize that. We can just assume a continuous object, find that inertia and go on with it? That just doesn't seem intuitive to me...

 

[Doc Al]

Why don't you state EXACTLY what information is given in the problem? Then we can figure it out.