How Is Angular Momentum Calculated in Particle Motion?

In summary, the problem involves finding the magnitude of a particle's angular momentum given its mass, velocity, and position vector. The solution involves scaling the velocity vector by the mass to get a new momentum vector, and then taking the cross product of the position vector with the new momentum vector. The answer should not be negative, as magnitudes cannot be negative.
  • #1
thehiggsboson
4
0

Homework Statement



A 1.47kg particle moves in the xy plane with a velocity of v = (4.59i - 3.28j)m/s. Determine the magnitude of the particle's angular momentum when its position vector is r = (1.35i + 2.57j)m.


Homework Equations



p = mv

L = r x p (the x is supposed to be a cross product and not a variable)

L = r x mv

The Attempt at a Solution



First I scaled the velocity vector: v = (4.59i - 3.28j)m/s by the mass, 1.47 kg, to get a new momentum vector (6.75i - 4.82j)kg*m/s.

Then I took the cross product of the r vector with the new momentum vector:
(1.35i + 2.57j)m x (6.75i - 4.82j)kg*m/s (I let a=1.35, b=2.57, c=0, d=6.75, e=-4.82, and f=0, the got the k vector cross product by doing k=ae-bd)

The answer I got was -23.9 kg*m^2/s, which wasn't right.

What did I do wrong? Am I even anywhere near the correct solution/answer?
 
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  • #2
thehiggsboson said:

Homework Statement



A 1.47kg particle moves in the xy plane with a velocity of v = (4.59i - 3.28j)m/s. Determine the magnitude of the particle's angular momentum when its position vector is r = (1.35i + 2.57j)m.


Homework Equations



p = mv

L = r x p (the x is supposed to be a cross product and not a variable)

L = r x mv

The Attempt at a Solution



First I scaled the velocity vector: v = (4.59i - 3.28j)m/s by the mass, 1.47 kg, to get a new momentum vector (6.75i - 4.82j)kg*m/s.

Then I took the cross product of the r vector with the new momentum vector:
(1.35i + 2.57j)m x (6.75i - 4.82j)kg*m/s (I let a=1.35, b=2.57, c=0, d=6.75, e=-4.82, and f=0, the got the k vector cross product by doing k=ae-bd)

The answer I got was -23.9 kg*m^2/s, which wasn't right.

What did I do wrong? Am I even anywhere near the correct solution/answer?

Did you specify the direction of the angular momentum? (It's a pseudo-vector, so it has a magnitude and direction).
 
  • #3
I didn't, but I just went back and tried it and it picked up the k as being part of the unit. So, it's not that, I don't think. The question also just asks for magnitude.
 
  • #4
thehiggsboson said:
I didn't, but I just went back and tried it and it picked up the k as being part of the unit. So, it's not that, I don't think. The question also just asks for magnitude.
Okay, just the magnitude. What answer did you type in for that?
 
  • #5
-23.9 kg*m^2/s
 
  • #6
Are magnitudes ever negative?
 
  • #7
Nope. Wow, I feel a little stupid now. Thanks!
 
  • #8
thehiggsboson said:
Nope. Wow, I feel a little stupid now. Thanks!

Heh. We've all been there, done that, got the T-shirt. Cheers.
 

1. What is angular momentum cross product?

Angular momentum cross product is a mathematical operation that calculates the vector product of two vectors, in this case angular momentum and position vectors. It is represented by the symbol "x" and is used to calculate the direction and magnitude of a resulting vector.

2. How is angular momentum cross product calculated?

To calculate the angular momentum cross product, you first need to find the cross product of the two vectors by taking the determinant of a 3x3 matrix. Then, you can use the right-hand rule to determine the direction of the resulting vector. Finally, the magnitude can be calculated by taking the product of the magnitudes of the two vectors and the sine of the angle between them.

3. What is the relationship between angular momentum and torque?

Angular momentum and torque are related by the formula L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity. This means that the angular momentum of an object is directly proportional to its moment of inertia and angular velocity.

4. How does angular momentum cross product affect rotational motion?

Angular momentum cross product is a key concept in rotational motion as it helps to determine the direction and magnitude of the resulting vector. It is also used in the equations of motion for rotational dynamics, such as the conservation of angular momentum and the relationship between torque and angular acceleration.

5. What are some real-life applications of angular momentum cross product?

Angular momentum cross product has many practical applications, including in physics, engineering, and astronomy. It is used to calculate the motion of rotating objects, such as gyroscopes and planets, and is also used in the design of machines and vehicles that involve rotational motion.

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