If vec{L} and vec{P} represent the angular mo | vedclass

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(A) Given the position vector $\vec{r} = a(\hat{i} \cos \omega t + \hat{j} \sin \omega t)$.To find the force,we first calculate the acceleration $\vec

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Opposite to the direction of $\vec{r}$

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(A) Given the position vector $\vec{r} = a(\hat{i} \cos \omega t + \hat{j} \sin \omega t)$.To find the force,we first calculate the acceleration $\vec{a} = \frac{d^2\vec{r}}{dt^2}$.Velocity $\vec{v} = \frac{d\vec{r}}{dt} = a\omega(-\hat{i} \sin \omega t + \hat{j} \cos \omega t)$.Acceleration $\vec{a} = \frac{d\vec{v}}{dt} = -a\omega^2(\hat{i} \cos \omega t + \hat{j} \sin \omega t) = -\omega^2\vec{r}$.Since $\vec{F} = m\vec{a}$,we have $\vec{F} = -m\omega^2\vec{r}$.This indicates that the force $\vec{F}$ is directed opposite to the position vector $\vec{r}$.

If $\vec{L}$ and $\vec{P}$ represent the angular momentum and linear momentum respectively of a particle of mass $m$ having position vector $\vec{r} = a(\hat{i} \cos \omega t + \hat{j} \sin \omega t)$,the direction of the force acting on the particle is:

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