# Angular motion of a particle

1. Nov 18, 2007

### karnten07

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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, im 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 im 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

2. Nov 18, 2007

### nrqed

to be more clear, this is

$$\vec{ r} \times \vec{ p} = (r_y p_z - r_z p_y) \hat{i} - (r_x p_z - r_z p_x) \hat{j} + (r_x p_y - r_y p_x) \hat{k}$$
In the equation above, r_x, p_y, etc are the components of your vectors $\vec{r}, \vec{p}$.

3. Nov 18, 2007

### karnten07

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?

Last edited: Nov 18, 2007
4. Nov 18, 2007

### nrqed

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

5. Nov 18, 2007

### karnten07

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

6. Nov 18, 2007

### karnten07

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

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

for a im getting it = 2ax + 6bt y

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

7. Nov 18, 2007

### nrqed

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

8. Nov 18, 2007

### karnten07

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