Angular momentum of a particle

[peachy112girl]

Homework Statement

The position of a particle of mass m = 2.0kg is given at time t by the equation, r = 3t i - t² j + 2 k, where r is measured in meters from the origin, and t is in seconds. What is the angular momentum of the particle with respect to the origin at time t = 2 s?

Homework Equations

L = Iω = (mr²)(ω)

The Attempt at a Solution

I kind of understand the general idea of the problem, I think it's just the presence of i, j, k that is confusing me.

From the radius equation, I took the derivative to get the angular velocity equation.
ω = 3 i - 2t j

From there, I calculated r(2) and ω(2).
r(2) = 6i - 4j + 2k
ω(2) = 3i - 4j

Using the equation L = Iω = mr²ω
L = (2)(6i - 4j + 2k)²(3i - 4j)

Then, I just multiplied through to get the answer. But, the numbers look way too big and I'm not too sure I did it correctly.

Any help is appreciated!

 

[ehild]

The angular momentum is the cross product (vector product) of the momentum of the particle and its position vector. You can determine the velocity of the particle by differentiating r(t) with respect time; multiply it by mass and then calculate the cross product with the position vector.

ehild

 

[kerimek]

Hello,

Your approach is correct, but there are a few things you can improve in your solution. First, let's clarify what the i, j, k represent in this problem. They are unit vectors in the x, y, and z directions, respectively. So, when you take the derivative of the position equation, you should get the angular velocity vector, not just a scalar value.

ω = 3i - 2tj

Next, you correctly calculated r(2) and ω(2). However, when using the equation L = Iω, it is important to note that the moment of inertia, I, is a scalar value, not a vector. So, you should only use the magnitude of the angular velocity vector, not the entire vector.

L = (2)(6i - 4j + 2k)2√(3^2 + (-2)^2) = (2)(6i - 4j + 2k)2(√13)

Finally, you can simplify this further by noting that the cross product of i and j is equal to k, and the cross product of i and k is equal to -j. So, you can rewrite the angular momentum as:

L = (2)(12k)(√13) = 24√13k

This is the final answer for the angular momentum of the particle with respect to the origin at time t = 2s.

I hope this helps clarify the problem for you. Keep up the good work!