How Do You Calculate Angular Momentum for a Particle in Non-Linear Motion?

[karnten07]

Homework Statement

A particle of mass m follows a path given by

r = (Xo + at^2)x + bt^3 y + ct z

where o, a, b and c are constants, t is the time. x, y , z should have hats on them showing they are position vectors (unit vectors).

1. Find the angular momentum L of the particle about the origin.
2. Find the force F required to produce this motion and verify explicitly that dL/dt = r x F

Homework Equations

The Attempt at a Solution

L = r x p

p = mv

to get v differentiate r w.r.t time to get v = 2at.x + (3bt^2)y + cz

Subsitute into the equation:

L = ((Xo + at^2)x + bt^3y + ctz) x m(2at.x + (3bt^2)y + cz)

Now to perform the cross product, I am a bit unsure here. For the cross product i have this formula:

r x p = x(ry.pz - rz.py) - y(rx.pz - rz.px) + z(rx.py - ry.px)

where the x,y,z are unit vectors and the other x,y and z's are in subscript showing the particular component of each vector. So I am a bit confused because do i just subsitute the constants from my equations of r and p or do i include their respective unit vecotrs associated with each component? But if i did that i would get for example x x y = z so would make quite a difference.

Any thought guys, thanks

 

[nrqed]

Homework Statement

A particle of mass m follows a path given by

r = (Xo + at^2)x + bt^3 y + ct z

where o, a, b and c are constants, t is the time. x, y , z should have hats on them showing they are position vectors (unit vectors).

1. Find the angular momentum L of the particle about the origin.
2. Find the force F required to produce this motion and verify explicitly that dL/dt = r x F

Homework Equations

The Attempt at a Solution

L = r x p

p = mv

to get v differentiate r w.r.t time to get v = 2at.x + (3bt^2)y + cz

Subsitute into the equation:

L = ((Xo + at^2)x + bt^3y + ctz) x m(2at.x + (3bt^2)y + cz)

Now to perform the cross product, I am a bit unsure here. For the cross product i have this formula:

r x p = x(ry.pz - rz.py) - y(rx.pz - rz.px) + z(rx.py - ry.px)

to be more clear, this is

[tex]
\vec{ r} \times \vec{ p} = (r_y p_z - r_z p_y) \hat{i} - (r_x p_z - r_z p_x) \hat{j} + (r_x p_y - r_y p_x) \hat{k} [/tex]

where the x,y,z are unit vectors and the other x,y and z's are in subscript showing the particular component of each vector. So I am a bit confused because do i just subsitute the constants from my equations of r and p or do i include their respective unit vecotrs associated with each component? But if i did that i would get for example x x y = z so would make quite a difference.

Any thought guys, thanks

In the equation above, r_x, p_y, etc are the components of your vectors [itex] \vec{r}, \vec{p} [/itex].

 

[karnten07]

to be more clear, this is

[tex]
\vec{ r} \times \vec{ p} = (r_y p_z - r_z p_y) \hat{i} - (r_x p_z - r_z p_x) \hat{j} + (r_x p_y - r_y p_x) \hat{k} [/tex]

In the equation above, r_x, p_y, etc are the components of your vectors [itex] \vec{r}, \vec{p} [/itex].

Ok thankyou for explaining, i thought i was overthinking it. So i think that i get my value of L from r x p as:

x(mc(Xo + at^2 - 3bt^3)) - y(mc(Xo - at^2)) + z(mb(Xo + t^2(a+3-2at^2))

How do i find the Force required to produce this motion?

Edit: do i just take the double derivative of r to get the acceleration and use F = ma. Then by taking the derivative of L i can show that my F multiplied by r should equal this derivative?

 

[nrqed]

Ok thankyou for explaining, i thought i was overthinking it. So i think that i get my value of L from r x p as:

x(mc(Xo + at^2 - 3bt^3)) - y(mc(Xo - at^2)) + z(mb(Xo + t^2(a+3-2at^2))

How do i find the Force required to produce this motion?

the force vector is simply the second derivative with respect to time of you rposition vector.

 

[karnten07]

the force vector is simply the second derivative with respect to time of you rposition vector.

I need to multiply it by m after differentiating it twice to get F right?

 

[karnten07]

the force vector is simply the second derivative with respect to time of you rposition vector.

Okay, I am taking the secomd derivative of r, but a bit sure on my math. I have

v = 2atx + 3bt^2 y + cz

for a I am getting it = 2ax + 6bt y

but i wasn't sure if the z component completely disappeared or if the cz become just z - sorry I am a bit rusty on this stuff. Anyone?

 

[nrqed]

Okay, I am taking the secomd derivative of r, but a bit sure on my math. I have

v = 2atx + 3bt^2 y + cz

for a I am getting it = 2ax + 6bt y

but i wasn't sure if the z component completely disappeared or if the cz become just z - sorry I am a bit rusty on this stuff. Anyone?

The z component of a is zero . (if you want, the derivative of [itex] c \hat{k} [/itex] is zero. )

 

[karnten07]

The z component of a is zero . (if you want, the derivative of [itex] c \hatk} [/itex] is zero. )

Thankyou, i had convinced myself of that also after a while. Just had to check.