What is the direction of the red ball's angular momentum

[Westin]

Homework Statement

What is the direction of the red ball's angular momentum about the point A?
(see figure)

Same as the momentum.
Out of the page.
Zero magnitude.
Opposite to the momentum.
Same as rA.
Into the page. [/B]

Homework Equations

Angular Momentum[/B]

The Attempt at a Solution

I believe the answer is same as the momentum because the angular momentum is 'the moment of momentum'

 

[rcgldr]

Is the ball rolling up the plane or down the plane? If this is during an instant where the ball is stopped, then it has zero momentum.

 

[Westin]

Is the ball rolling up the plane or down the plane? If this is during an instant where the ball is stopped, then it has zero momentum.

Unfortunately that info is not given, but if we assumed it was down wouldn't the answer just be same as momentum?

 

[mfb]

How do you determine angular momentum of a point-mass?

 

[SammyS]

Homework Statement

What is the direction of the red ball's angular momentum about the point A?
(see figure)

Same as the momentum.
Out of the page.
Zero magnitude.
Opposite to the momentum.
Same as rA.
Into the page. [/B]

Homework Equations

Angular Momentum[/B]

The Attempt at a Solution

I believe the answer is same as the momentum because the angular momentum is 'the moment of momentum'

The figure helps tremendously.

It looks to me like the ball is at the position of the red dot and is moving horizontally.

How do one calculate angular momentum ?

 

[Westin]

L = r x p
L = same as momentum then ?

 

[mfb]

r x p and p are not the same, and do not point in the same direction.
What do you know about the cross-product?

 

[Westin]

r x p and p are not the same, and do not point in the same direction.
What do you know about the cross-product?

Cross product is r⃗A × F⃗

This helps you find torque

 

[mfb]

And what is your result?

 

[Westin]

And what is your result?

Opposite to the momentum?

 

[haruspex]

Opposite to the momentum?

If you have two vectors and take their cross product, $\vec c = \vec a \times \vec b$, what is the angle between $\vec a$ and $\vec c$? It is always the same answer.

 

[Westin]

And what is your result?

If you have two vectors and take their cross product, $\vec c = \vec a \times \vec b$, what is the angle between $\vec a$ and $\vec c$? It is always the same answer.

it is 180 degrees, I'm still a little lost of what the answer could be though..

 

[mfb]

It is not 180 degrees. Check the definition and basic properties of the cross-product.

 

[SammyS]

Do you know the Right Hand Rule for cross-product ?

 

[Westin]

Do you know the Right Hand Rule for cross-product ?

If you have two vectors and take their cross product, $\vec c = \vec a \times \vec b$, what is the angle between $\vec a$ and $\vec c$? It is always the same answer.

Vector C will be perpendicular to both A & B. The angle will be between 0 and 180 degrees

• 
  A x B = A B sin([PLAIN]https://www.physics.uoguelph.ca/tutorials/torque/theta.gif)

not familiar with right hand rule but I'm trying to learn it right now

 

[mfb]

Vector C will be perpendicular to both A & B.

Right.

The angle will be between 0 and 180 degrees

You can know the exact angle between a and c. This is not the angle between a and b!

 

[Westin]

Vector C will be perpendicular to both A & B. The angle will be between 0 and 180 degrees

not familiar with right hand rule but I'm trying to learn it right now

Right.
You can know the exact angle between a and c. This is not the angle between a and b!

It would be going out of the page then??!?

because it will be pointing upwards

 

[mfb]

It would be going out of the page then??!?

Right.

 

[rcgldr]

If the ball is rolling down hill (left to right), then the angular momentum vector points into the page (right hand rule).

 

[Westin]

So who is right here? I only have one try

 

[SammyS]

If the ball is rolling down hill (left to right), then the angular momentum vector points into the page (right hand rule).

@ rcgldr,
Look at the figure.

There is no downhill. The momentum of the ball is to the right, so the ball is moving to the right (horizontally). The ball is at a position below and to the right of point A.

There is no indication that the ball itself is rotating at all.

 

[SammyS]

So who is right here? I only have one try

That all depends upon who is interpreting the problem as it was intended.

Looking at the figure it appears that you may have previously tried the "Into the page " answer.

 

[Westin]

The answer was out of the page, thanks everyone.

 

[haruspex]

If the ball is rolling down hill (left to right), then the angular momentum vector points into the page (right hand rule).

I see nothing in the statement about the ball's rotating, nor any indication of a vertical direction. It is shown as moving directly left to right, not at an angle to the page edge, so I don't think it can be intended as 'rolling downhill'.

 

[rcgldr]

I see nothing in the statement about the ball's rotating, nor any indication of a vertical direction. It is shown as moving directly left to right, not at an angle to the page edge, so I don't think it can be intended as 'rolling downhill'.

So is this a 3d image mapped into a 2d image and the vector labeled as "rA" is either pointed out from the page or into the page? The momentum vector is to the right, so maybe we assume the ball is rolling clockwise (rolling to the right), in which case the angular momentum vector should be into the page (right hand rule).

 

[SammyS]

So is this a 3d image mapped into a 2d image and the vector labeled as "rA" is either pointed out from the page or into the page? The momentum vector is to the right, so maybe we assume the ball is rolling clockwise (rolling to the right), in which case the angular momentum vector should be into the page (right hand rule).

The ball is not rotating. The ball is not rolling. It's as if the ball is a point mass.

All motion is in the plane of the page, and the ball is at the location of the red dot. The big gray circular thing must just be some other object.

 

[rcgldr]

All motion is in the plane of the page, and the ball is at the location of the red dot.

So effectively, A is the center of rotation, rA is the current position of the center of mass of the ball, and p/mass = velocity, so the rotation is counter-clockwise, and the direction of the angular momentum vector is out of the page.

 

[mfb]

So effectively, A is the center of rotation, rA is the current position of the center of mass of the ball, and p/mass = velocity, so the rotation is counter-clockwise, and the direction of the angular momentum vector is out of the page.

Right.