Question

Angular momentum of a particle of linear momentum \(\vec{p}\) and at a distance \(\vec{r}\) from the origin is:
1. \(\vec{r} \times \vec{p}\)
2. \(\vec{r}\cdot \vec{p}\)
3. \(\dfrac{\vec{r}\times \vec{p}}{|\vec{r}|}\)
4. \(\dfrac{\vec{r}\times \vec{p}}{|\vec{r}||\vec{p}|}\)

Answer 1

Correct Answer - Option 1 : \(\vec{r} \times \vec{p}\)

Concept:

• Angular momentum:  It is the cross product of its position vectors and its linear momentum.
  • The moment of inertia of the particle is the product of the particle's mass and its perpendicular distance from the axis of rotation.

It is given by:

\( L = M \times v \times r = \overrightarrow p \times \overrightarrow r \)

where, L = Angular momentum, M = mass , v = velocity, r = radius.

Explanation:

• Suppose the particle of m mass rotates about a fixed axis, then the angular momentum is given by a cross product of its position vector and linear momentum

​\( L = M \times v \times r = \overrightarrow p \times \overrightarrow r \)​

• The resultant vector of the cross product of two vectors is also a vector.
• Therefore, angular vector momentum is a vector quantity.
• The value of angular momentum is given as,

\(L = \overrightarrow p \times \overrightarrow r \)