Angular Momentum of a 3kg Particle at t=5s

[Nickluvn]

Homework Statement

A 3kg particle has its position as a function of time given by
(−3m + 1 ms t)xˆ + (4m + 2 ms t)yˆ − 3 ms tzˆ . What is the particle’s angular momentum about the origin at t=5s?

Homework Equations

m1v1+m2v2=m1v1f + m2v2f

The Attempt at a Solution

I plugged in the value for t = 5s

gave me (-3m + 5 m^s)x^ + ( 4m + 10 m^s)y^-15 m^s z^

what i am unsure about is the usage of the cross product.

 

[collinsmark]

Homework Statement

A 3kg particle has its position as a function of time given by
(−3m + 1 ms t)xˆ + (4m + 2 ms t)yˆ − 3 ms tzˆ . What is the particle’s angular momentum about the origin at t=5s?

Homework Equations

m1v1+m2v2=m1v1f + m2v2f

The Attempt at a Solution

I plugged in the value for t = 5s

So far so good. On way to approach this problem is to first find the particle's position r at t = 5 sec.

gave me (-3m + 5 m^s)x^ + ( 4m + 10 m^s)y^-15 m^s z^

You lost me on your units.

Are you sure the position isn't actually given as:

\vec r = (-3 \ [\mathrm{m}] + (1 \ [\mathrm{m/s}])t ) \hat x \ \<br /> + \ \ (4 \ [\mathrm{m}] + (2 \ [\mathrm{m/s}])t ) \hat y \ \<br /> + \ \ (-3 \ [\mathrm{m/s}]})t \hat z \ \ ?

That would make all the units come out be meters. I really don't know what units of m^s are (something doesn't look right there).

Once you have the position at time t = 5 sec, the next thing is to find the particle's velocity at time t = 5 sec. (It's easy in this case since the particle is traveling at a constant velocity.)

what i am unsure about is the usage of the cross product.

Once you have the particle's position (about the origin) and velocity at time t = 5 sec, find the angular momentum about the origin.

\vec L = \vec r \times m \vec v = \left| <br /> \begin{array}{ccc} <br /> \hat x &amp; \hat y &amp; \hat z \\<br /> r_x &amp; r_y &amp; r_z \\<br /> mv_x &amp; mv_y &amp; mv_z<br /> \end{array} \right|

 

[Nickluvn]

My bad it should be m/s. Thanks again man this helped me alot, I am just confused on the placement of R in the matrix you provided.

Velocity because it is constant would mean that we would take the derivative of the function given and then plug in a value for t?

 

[collinsmark]

My bad it should be m/s. Thanks again man this helped me alot, I am just confused on the placement of R in the matrix you provided.

The r components go in the middle. It's the definition of the cross product.

<br /> \vec A \times \vec B = \left| <br /> \begin{array}{ccc} <br /> \hat x &amp; \hat y &amp; \hat z \\<br /> A_x &amp; A_y &amp; A_z \\<br /> B_x &amp; B_y &amp; B_z<br /> \end{array} \right| <br />

That's true for any vectors, A and B. (Excuse my mismatching notation. I'm using either boldface or the upper arrow to indicate vectors.)

And by the way, it's not just a matrix. It's the determinant of a matrix. Do an Internet search on "determinant of a matrix" if you're not sure where to go from here.

Velocity because it is constant would mean that we would take the derivative of the function given and then plug in a value for t?

Yes, that will work fine!

All I'm saying it it turns out to be pretty simple in this case (The variable t goes away in each component in this particular case, leaving a constant velocity independent of t). But yes, taking the derivative of r with respect to t will give you the velocity vector v.