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

## 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'

#### Attachments

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rcgldr
Homework Helper
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.

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

SammyS
Staff Emeritus
Homework Helper
Gold Member

## 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 ?

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

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mfb
Mentor
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?

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
Mentor

Opposite to the momentum?

haruspex
Homework Helper
Gold Member
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.

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

SammyS
Staff Emeritus
Homework Helper
Gold Member
Do you know the Right Hand Rule for cross-product ?

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) [Broken]

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

Last edited by a moderator:
mfb
Mentor
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!

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

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mfb
Mentor
It would be going out of the page then!?!?!?
Right.

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

So who is right here? I only have one try

SammyS
Staff Emeritus
Homework Helper
Gold Member
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
Staff Emeritus
Homework Helper
Gold Member
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.

The answer was out of the page, thanks everyone.

haruspex
Homework Helper
Gold Member
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
Homework Helper
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).

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