Consider a particle of mass m having linear momentum `vecp` at position `vecr` relative to the origin O. Let `vecL` be the angular momentum of the particle with respect to the origin. Which of the following equations correctly relate(s) `vecr`, `vecp` and `vecL`?

Options

A. `(dvecL)/(dt) + vecr xx (dvecp)/(dt)=0`
B. `(dvecL)/(dt) + (dvecr)/(dt) xx vecp=0`
C. `(dvecL)/(dt) - (dvecr)/(dt) xx vecp=0`
D. `(dvecL)/(dt)- vecr xx (dvecp)/(dt)=0`

Explanation

To solve the problem, we need to establish the relationship between the position vector \(\vec{r}\), linear momentum vector \(\vec{p}\), and angular momentum vector \(\vec{L}\) of a particle of mass \(m\).

### Step-by-Step Solution:

1. **Understanding Angular Momentum**:
The angular momentum \(\vec{L}\) of a particle with respect to a point (in this case, the origin O) is defined as:
\[ \vec{L} = \vec{r} \times \vec{p} \]
where \(\vec{r}\) is the position vector of the particle relative to the origin O, and \(\vec{p}\) is the linear momentum of the particle.

2. **Differentiating Angular Momentum**:
To find a relationship involving time, we differentiate both sides of the angular momentum equation with respect to time \(t\):
\[ \frac{d\vec{L}}{dt} = \frac{d}{dt}(\vec{r} \times \vec{p}) \]

3. **Applying the Product Rule**:
Using the product rule for differentiation of the cross product, we have:
\[ \frac{d\vec{L}}{dt} = \frac{d\vec{r}}{dt} \times \vec{p} + \vec{r} \times \frac{d\vec{p}}{dt} \]

4. **Substituting Linear Momentum**:
Since the linear momentum \(\vec{p}\) is defined as \(\vec{p} = m\vec{v}\) (where \(\vec{v} = \frac{d\vec{r}}{dt}\)), we can express \(\frac{d\vec{p}}{dt}\) as:
\[ \frac{d\vec{p}}{dt} = m\frac{d\vec{v}}{dt} = m\vec{a} \]
where \(\vec{a}\) is the acceleration of the particle.

5. **Analyzing the Terms**:
The first term \(\frac{d\vec{r}}{dt} \times \vec{p}\) becomes \(\vec{v} \times \vec{p}\). However, since \(\vec{p} = m\vec{v}\), we find that:
\[ \vec{v} \times \vec{p} = \vec{v} \times (m\vec{v}) = 0 \]
because the cross product of any vector with itself is zero.

6. **Final Equation**:
Thus, we can simplify our earlier equation:
\[ \frac{d\vec{L}}{dt} = 0 + \vec{r} \times \frac{d\vec{p}}{dt} \]
or,
\[ \frac{d\vec{L}}{dt} = \vec{r} \times \frac{d\vec{p}}{dt} \]

7. **Setting Up the Final Relation**:
Rearranging gives us:
\[ \frac{d\vec{L}}{dt} - \vec{r} \times \frac{d\vec{p}}{dt} = 0 \]

### Conclusion:
The correct relationship that relates \(\vec{r}\), \(\vec{p}\), and \(\vec{L}\) is:
\[ \frac{d\vec{L}}{dt} - \vec{r} \times \frac{d\vec{p}}{dt} = 0 \]
Thus, the answer is option D.