Angular Momentum Homework: Magnitude of Object's L at (0,9,0)

In summary, the problem involves finding the magnitude of the angular momentum of a 1.2 kg object traveling with constant velocity (4i - 2j) m/s at the point (0 m, 9 m, 0 m). The equation p = m*v and L = r*p are used, but the answer obtained is twice the actual value. The correct approach is to use the vector cross product, with the angle between the r and p vectors being determined by their respective directions. The radius is in the Y direction and the momentum is in the X direction. The i,j,k notation may also be useful in solving the problem.
  • #1
yankees26an
36
0

Homework Statement



A 1.2 kg object travels with constant velocity (4i - 2j) m/s. When the object is at the point (0
m, 9 m, 0 m), what is the magnitude of its angular momentum about the origin?


Homework Equations



p = m*v
L = r*p

The Attempt at a Solution



v = sqrt(4^2 + 2^2)
p = m*v = 5.36

L = r*p = 9*5.36 = 48.2

Interestingly the answer I get this way is 2x the right answer. I wonder if this is a coincidence of not. Not sure how to do it the 'right' way.
 
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  • #2
yankees26an said:
L = r*p

These are not scalars, so this

L = r*p = 9*5.36 = 48.2

is wrong.
 
  • #3
L is NOT simply given by r multiplied by p. The definition of L is

[tex] \textbf{L} = \textbf{r} \times \textbf{p} [/tex]​

Where boldface quantities are vectors and the multiplication symbol signifies the vector cross product. Since L is the cross product of the r and p vectors, components of the momentum parallel to r do not contribute to the angular momentum. Are you sure that the right answer is 24.1, because I get a different answer both ways (i.e. by doing the full vector cross product or by using the fact that components of the velocity || to the position vector can be ignored)?
 
  • #4
Borek said:
These are not scalars

Ok. On a diagram, 9 just turns out to be the height/radius. Not sure if that means anything.

x and z are 0 so they don't affect anything, so I just have the value of y left. If it's not scalar then do you have any suggestions how to continue the problem. :-p

cepheid said:
The answer was 24.0, which was 0.1 off but I thought it was acceptable to call it 2x because of the magnititude of the uncertainty

So a cross product would be r*Fsin(pheta). But how do I get to pheta?
 
Last edited:
  • #5
First of all, to answer your question, the angle in the sine factor is the angle between the two vectors in the cross product.

Second of all, that is probably not the approach you want to take to calculate the cross product. It only tells you the magnitude, not the direction. You need to look up the definition of a cross product in Cartesian coordinates. Ie for input vectors of given x,y, and z components, what are the x y and z components of the result? This is closely related to the right hand rule.

Thirdly (minor point), there is no such Greek letter as 'pheta'. Perhaps you were thnking of theta?
 
  • #6
cepheid said:
Thirdly (minor point), there is no such Greek letter as 'pheta'. Perhaps you were thnking of theta?

Ok you got me =P

cepheid said:
First of all, to answer your question, the angle in the sine factor is the angle between the two vectors in the cross product.

How would I go about finding the angle between momentum and r?
 
  • #7
How would I go about finding the angle between momentum and r?[/QUOTE]

You would find it based upon their respective directions. But like I said, that is not the best approach. You should really read the second of my three points (the one you did not quote)
 
  • #8
cepheid said:
How would I go about finding the angle between momentum and r?

You would find it based upon their respective directions. But like I said, that is not the best approach. You should really read the second of my three points (the one you did not quote)[/QUOTE]

Ok so the radius is in the Y direction? The momentum in the x direction?

I also remember some stuff about i,j,k, but seriously how do I apply it here :bugeye:
 

Related to Angular Momentum Homework: Magnitude of Object's L at (0,9,0)

Q: What is angular momentum?

Angular momentum is a physical quantity that measures the rotational motion of an object around a fixed axis. It is represented by the symbol L and is equal to the product of an object's moment of inertia and its angular velocity.

Q: How is angular momentum calculated?

Angular momentum is calculated using the formula L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity. The moment of inertia is a measure of an object's resistance to changes in its rotational motion, while the angular velocity is the rate at which the object is rotating.

Q: What is the magnitude of an object's angular momentum?

The magnitude of an object's angular momentum is equal to the product of its mass, linear velocity, and the distance from the axis of rotation. It can also be calculated using the formula |L| = mvr, where m is the mass, v is the linear velocity, and r is the distance from the axis of rotation.

Q: How is the magnitude of an object's angular momentum affected by its position?

The magnitude of an object's angular momentum is directly proportional to its distance from the axis of rotation. This means that as the distance from the axis increases, the magnitude of the angular momentum also increases. Additionally, the direction of the angular momentum is always perpendicular to the plane of rotation.

Q: How can I find the magnitude of an object's angular momentum at a specific point?

To find the magnitude of an object's angular momentum at a specific point, you will need to know the moment of inertia, angular velocity, and the distance from the axis of rotation at that point. Then, you can use the formula |L| = Iωr to calculate the magnitude. Alternatively, you can also use the formula |L| = mvr, where m is the mass, v is the linear velocity, and r is the distance from the axis of rotation at that point.

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