# Angular motion of a particle

## 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

## 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

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nrqed
Homework Helper
Gold Member

## 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

## 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)
to be more clear, this is

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

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 $\vec{r}, \vec{p}$.
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:
nrqed
Homework Helper
Gold Member
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.

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?

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

nrqed
Homework Helper
Gold Member
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?
The z component of a is zero . (if you want, the derivative of $c \hat{k}$ is zero. )

The z component of a is zero . (if you want, the derivative of $c \hatk}$ is zero. )
Thankyou, i had convinced myself of that also after a while. Just had to check.