Question

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

• A. Opposite to the direction of $\overrightarrow{\mathrm{r}}$
• B. Opposite to the direction of $\overrightarrow{\mathrm{L}}$
• C. Opposite to the direction of $\overrightarrow{\mathrm{P}}$
• D. Opposite to the direction of $\overrightarrow{\mathrm{L}} \times \overrightarrow{\mathrm{P}}$

Hint:

The force is proportional to the acceleration, which is opposite to the position vector in this case.

Answer 1

The Correct Option is A

1. Position vector: \[ \overrightarrow{\mathrm{r}} = \mathrm{a}(\hat{\mathrm{i}} \cos \omega \mathrm{t} + \hat{\mathrm{j}} \sin \omega \mathrm{t}) \]
2. Acceleration: \[ \overrightarrow{\mathrm{a}} = -\omega^2 \overrightarrow{\mathrm{r}} \]
3. Force: \[ \overrightarrow{\mathrm{F}} = m \overrightarrow{\mathrm{a}} = -m \omega^2 \overrightarrow{\mathrm{r}} \] Therefore, the direction of force is opposite to the direction of $\overrightarrow{\mathrm{r}}$.