I have a question about a problem of conservation of momentu

In summary: The area under the curve from the center of the wheel to the top of the disc is the force you are looking for, which is labelled as "A_y" on your diagram.
  • #1
Queren Suriano
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0
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

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  • #2
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
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 don't understand my mistake
 
  • #4
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.
 

FAQ: I have a question about a problem of conservation of momentu

What is conservation of momentum?

The law of conservation of momentum states that in a closed system, the total momentum remains constant. This means that the total amount of momentum before a collision or interaction is equal to the total amount of momentum after the collision or interaction.

Why is conservation of momentum important?

Conservation of momentum is important because it is a fundamental principle in physics that helps us understand the behavior of objects in motion. It allows us to predict the outcome of collisions and interactions between objects, and is crucial in fields such as mechanics, thermodynamics, and astrophysics.

How is momentum conserved?

Momentum is conserved through the transfer of momentum between objects in a closed system. In a collision, the total momentum of the objects before and after the collision must be equal. This can happen through transfer of momentum from one object to another, or through changes in the velocities of the objects.

What are some real-life examples of conservation of momentum?

Some real-life examples of conservation of momentum include: a billiard ball breaking a rack of stationary balls, a car collision, a bouncing ball, and the recoil of a gun after firing a bullet. In each of these scenarios, the total momentum before and after the event remains constant.

How does conservation of momentum relate to Newton's Third Law of Motion?

Newton's Third Law states that for every action, there is an equal and opposite reaction. This relates to conservation of momentum because when objects interact, they exert equal and opposite forces on each other, resulting in a transfer of momentum. This is why the total momentum remains constant in a closed system.

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