If $\vec{L}$ and $\vec{P}$ represent the angular momentum and linear momentum respectively of a particle of mass ' $m$ ' having position vector as $\vec{r}=a(\hat{i} \cos \omega t+\hat{j} \sin \omega t)$. The direction of force is

<p>Let's analyze the situation step by step.</p> <p><p>The particle’s position vector is given by</p> <p>$\vec{r} = a(\hat{i}\cos\omega t + \hat{j}\sin\omega t),$</p> <p>which represents uniform circular motion in the xy-plane.</p></p> <p><p>Its velocity is the time derivative of the position:</p> <p>$$\vec{v} = \frac{d\vec{r}}{dt} = -a\omega\sin\omega t\,\hat{i} + a\omega\cos\omega t\,\hat{j}.$$</p></p> <p><p>The acceleration, obtained by differentiating the velocity, is:</p> <p>$$\vec{a} = \frac{d\vec{v}}{dt} = -a\omega^2\cos\omega t\,\hat{i} - a\omega^2\sin\omega t\,\hat{j}.$$</p> <p>Notice that this can be rewritten as:</p> <p>$\vec{a} = -\omega^2 \vec{r}.$</p> <p>This indicates that the acceleration (and hence the force, since $\vec{F}=m\vec{a}$) is directed opposite to the position vector $\vec{r}$—that is, radially inward.</p></p> <p><p>To check against the given options:</p></p> <p><p>Option A (Opposite to $\vec{L}$): The angular momentum $\vec{L}=m\vec{r}\times\vec{v}$ is perpendicular to the plane of motion (along $\pm\hat{k}$) and does not indicate a direction in the plane.</p></p> <p><p>Option B (Opposite to $\vec{L}\times\vec{P}$): For a particle in circular motion, when you compute $\vec{L}\times\vec{P}$ (with $\vec{P} = m\vec{v}$), you’ll find that it is proportional to $-\vec{r}.$ Thus, the direction opposite to $\vec{L}\times\vec{P}$ would be the same as $\vec{r}$, which is the outward radial direction—not matching the inward (centripetal) force.</p></p> <p><p>Option D (Opposite to $\vec{P}$): The linear momentum is tangential; the force (centripetal) is not tangential but directed radially inward.</p></p> <p><p>Option C (Opposite to $\vec{r}$): As we found, the acceleration (and hence force) is given by $\vec{F}= m\vec{a} = -m\omega^2 \vec{r},$ which is clearly opposite to $\vec{r}.$</p></p> <p>Therefore, the force is directed opposite to $\vec{r},$ making Option C the correct answer.</p>