# Calculate the angular momentum of this particle in rotational motion

• YanZhen
In summary, the conversation discusses the use of vectors i, j, and k to define angular momentum, with a reminder to use the vector valued cross product instead of regular multiplication. The anti-symmetry of the cross product is also mentioned, followed by a clarification on the cross product in two dimensions.

#### YanZhen

Homework Statement
The mass of the particle is m,the equation of motion is r=acosωti+bsinωtj.a and b are constant.Calculate the angular momentum of a particle at any time.
Relevant Equations
L=P*r=m*v*r
v=dr/dt
i,j,k arevector
I know L=P*r=m*v*r=m(acosωti+bsinωtj)*(-aωsinωti+bωcosωtj)=mabw((cos^2)ωt+(sin^2)ωt)k=mabωk.
but why m(acosωti+bsinωtj)*(-aωsinωti+bωcosωtj)=mabw((cos^2)ωt+(sin^2)ωt)k.I need some detail.

You should not be using regular multiplication to multiply vectors, you need to use the vector valued cross product to define the angular momentum:
$$\vec L = \vec p \times \vec r$$
where
$$\hat i \times \hat j = \hat k, \quad \hat j \times \hat k = \hat i, \quad \hat k \times \hat i = \hat j$$
and furthermore you have anti-symmetry so ##\vec u \times \vec v = - \vec v \times \vec u## for any vectors ##\vec u## and ##\vec v##.

IIRC $$\vec L\equiv \vec r \times\vec p$$

BvU said:
IIRC $$\vec L\equiv \vec r \times\vec p$$
Too busy arguing about the need to use the cross product to care about getting the right definition I suppose ...

berkeman and BvU
Orodruin said:
You should not be using regular multiplication to multiply vectors, you need to use the vector valued cross product to define the angular momentum:
$$\vec L = \vec p \times \vec r$$
where
$$\hat i \times \hat j = \hat k, \quad \hat j \times \hat k = \hat i, \quad \hat k \times \hat i = \hat j$$
and furthermore you have anti-symmetry so ##\vec u \times \vec v = - \vec v \times \vec u## for any vectors ##\vec u## and ##\vec v##.
wow,thanks for your wonderful answer,it solved my question.and it also requires a point of knowlege about vectors.(x1,y1)X(x2,y2)=x1y2-y1x2

YanZhen said:
(x1,y1)X(x2,y2)=x1y2-y1x2
Careful. This is not the cross product. It is a component (the third) of the cross product between two vectors that only have i and j components. The cross product as such does not exist in two dimensions (although there are generalisations of it that do - among them the wedge product, which would look something like what you wrote but is generally too advanced for basic introduction to vectors and their application). The general form of the cross product (in three dimensions) would read
$$(x_1,y_1,z_1) \times (x_2, y_2, z_2) = (y_1 z_2 - y_2 z_1, z_1 x_2 - z_2 x_1, x_1 y_2 - x_2 y_1)$$

YanZhen