# I have a question about a problem of conservation of momentu

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1. Nov 21, 2014

### Queren Suriano

Momentum in rigid bodies; In this problem when I draw all the forces acting on the disc A, would the reactions: Ax, Ay; W, Friction, normal force; I do not understand is how to check that Ay = W, that's what I understand from the solution. Anyone have any tips on how to treat reactions in such problems

2. Nov 21, 2014

### Simon Bridge

I don't see any Ay or W on your diagrams, so I don't know what these labels refer to.

When you are doing these problems, you work out the relations by using your knowledge of physics.
If Ay = W was not true, what would happen?

3. Nov 21, 2014

### Queren Suriano

The picture of the diagrams is the solution, in my diagram I draw in direcction "Y": normal force, Ay, and W; So if apply sum of forces in "Y" direction: Ay- W +N=0; where Ay it's not W.What is it wrong that I'm considering? I dont understand my mistake

4. Nov 21, 2014

### Simon Bridge

You are not being clear.
None of your diagrams have an Ay or a W or anything labelled "normal force".
Therefore, the references to your diagrams are no help in understanding you.

Lets see if I can demonstrate what sort of description I am looking for:

If W is weight, then $\vec W=-mg\hat\jmath$ - acting at the center of mass of the wheel.
(You have an arrow labelled "mgt" in the right place for the weight ... what is "t" here?)

To write that equation I am implicitly defining $\hat\jmath$ as the unit vector pointing upwards.
Using the same definition:

If I use N for the force normal to the surface, pressing on the bottom of the wheel, then $\vec N = N\hat\jmath$.
Since the wheel has no vertical acceleration, $\vec N + \vec W = 0 \implies \vec N = mg\hat\jmath$
... so where does $A_y$ come in?

In context it is a force - so which force is it?
The only "A" in any of your diagrams is on the top one where it labelled the wheel as "wheel A".
Other common meanings for A are "area" and "amplitude" - which don't seem to apply.
So your notation is not as intuitive as you seem to think.