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 find the relationship between the position vector \( \vec{r} \), linear momentum \( \vec{p} \), and angular momentum \( \vec{L} \) of a particle, we can follow these steps:

### Step 1: Define Angular Momentum
The angular momentum \( \vec{L} \) of a particle with respect to the origin is defined as:
\[ \vec{L} = \vec{r} \times \vec{p} \]
where \( \vec{r} \) is the position vector of the particle and \( \vec{p} \) is the linear momentum of the particle.

### Step 2: Express Linear Momentum
The linear momentum \( \vec{p} \) of a particle is given by:
\[ \vec{p} = m \vec{v} \]
where \( m \) is the mass of the particle and \( \vec{v} \) is its velocity.

### Step 3: Differentiate Angular Momentum
To find the relationship involving time, we differentiate the angular momentum \( \vec{L} \) with respect to time \( t \):
\[ \frac{d\vec{L}}{dt} = \frac{d}{dt}(\vec{r} \times \vec{p}) \]
Using the product rule for differentiation, we have:
\[ \frac{d\vec{L}}{dt} = \frac{d\vec{r}}{dt} \times \vec{p} + \vec{r} \times \frac{d\vec{p}}{dt} \]

### Step 4: Substitute Velocity
Since \( \frac{d\vec{r}}{dt} \) is the velocity \( \vec{v} \), we can rewrite the equation as:
\[ \frac{d\vec{L}}{dt} = \vec{v} \times \vec{p} + \vec{r} \times \frac{d\vec{p}}{dt} \]

### Step 5: Analyze the Terms
Now, substituting \( \vec{p} = m\vec{v} \) into the equation:
\[ \vec{v} \times \vec{p} = \vec{v} \times (m\vec{v}) \]
Since the cross product of a vector with itself is zero:
\[ \vec{v} \times \vec{v} = 0 \]
Thus, the equation simplifies to:
\[ \frac{d\vec{L}}{dt} = 0 + \vec{r} \times \frac{d\vec{p}}{dt} \]
This leads to:
\[ \frac{d\vec{L}}{dt} = \vec{r} \times \frac{d\vec{p}}{dt} \]

### Step 6: Final Relation
We can rearrange this to express the relationship between \( \vec{L} \), \( \vec{r} \), and \( \vec{p} \):
\[ \frac{d\vec{L}}{dt} - \vec{r} \times \frac{d\vec{p}}{dt} = 0 \]

### Conclusion
This final equation shows the relationship between the angular momentum \( \vec{L} \), position vector \( \vec{r} \), and the change in linear momentum \( \frac{d\vec{p}}{dt} \).