Question

A 1.30-kg particle moves in the xy plane with a velocity of v with arrow = (4.50 î − 3.30 ĵ) m/s. Determine the angular momentum of the particle about the origin when its position vector is r with arrow = (1.50 î + 2.20 ĵ) m.

Answer 1

The angular momentum of the particle about the origin is
`\vec l = -19.305\,k\,\left[kg\cdot (m)/(s) \right]`.

Step-by-step explanation:

Vectorially speaking, the angular momentum is given by the following cross product:

`\vec l = \vec r * m\vec v`

This cross product can be solved with the help of determinants and its properties, that is:

`\vec l = \left|\begin{array}{ccc}i&j&k\\r_(x)&r_(y)&0\\m\cdot v_(x)&m\cdot v_(y)&0\end{array}\right|`

`\vec l = m\left|\begin{array}{ccc}i&j&k\\r_(x)&r_(y)&0\\v_(x)& v_(y)&0\end{array}\right|`

The 3 x 3 determinant is solved by the Sarrus Law:

`\vec l = m \cdot (r_(x)\cdot v_(y) - r_(y)\cdot v_(x))k`

If
`m = 1.30\,kg`,
`\vec r = 1.50\,i + 2.20\,j\,[m]` and
`\vec v = 4.50\,i-3.30\,j\,\left[(m)/(s) \right]`, the angular momentum of the particle about the origin is:

`\vec l = (1.30\,kg)\cdot \left[\left(1.50\,m\right)\cdot\left(-3.30\,(m)/(s) \right)-\left(2.20\,m\right)\cdot\left(4.50\,(m)/(s) \right)\right]k`

`\vec l = -19.305\,k\,\left[kg\cdot (m)/(s) \right]`

The angular momentum of the particle about the origin is
`\vec l = -19.305\,k\,\left[kg\cdot (m)/(s) \right]`.