Question

The angular momentum of a particle with respect to the origin will not zero, if

• A. the directional lone of linear momentum passes through the origin
• B. the particle is the origin
• C. the angle between the position vector and linear momentum is 180°
• D. the linear momentum vanishes
• E. the angle between the position vector and linear momentum is 90°

Answer 1

Answer: (E)

The angular momentum \(L\) of a particle with respect to the origin is given by the cross product of the position vector \(\vec{r}\) and the linear momentum \(\vec{p}\): \[ \vec{L} = \vec{r} \times \vec{p} \] The magnitude of angular momentum is: \[ L = r p \sin(\theta) \] Where \(r\) is the magnitude of the position vector, \(p\) is the magnitude of the linear momentum, and \(\theta\) is the angle between \(\vec{r}\) and \(\vec{p}\). For the angular momentum to be non-zero, \(\sin(\theta)\) must not be zero, which means \(\theta\) must not be 0° or 180°. The maximum value of \(L\) occurs when \(\theta = 90^\circ\), which means the angle between the position vector and the linear momentum is 90°.

The correct option is (E) : the angle between the position vector and linear momentum is 90°

Answer 2

Angular momentum of a particle with respect to the origin is given by:
$ \vec{L} = \vec{r} \times \vec{p} $
where $\vec{r}$ is the position vector and $\vec{p}$ is the linear momentum.

The magnitude of angular momentum is:
$ L = r p \sin\theta $
where $\theta$ is the angle between $\vec{r}$ and $\vec{p}$.

For angular momentum to be non-zero, $\sin\theta$ must be non-zero.
This happens when the angle between $\vec{r}$ and $\vec{p}$ is 90°, because:
$ \sin(90^\circ) = 1 $

Hence, the angular momentum is maximum and certainly not zero when the angle is 90°.

Answer: the angle between the position vector and linear momentum is 90°