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 $\overrightarrow{\mathrm{r}}$.

Answer 2

We are given that the position vector of a particle is:

\[ \vec{r} = a(\hat{i} \cos \omega t + \hat{j} \sin \omega t) \]

We need to determine the direction of the force acting on the particle.

Concept Used:

The particle moves in a circular path of radius \( a \). In uniform circular motion, the acceleration (and hence the force) is always directed towards the center of the circle, i.e., opposite to the radius vector \( \vec{r} \).

The relevant relations are:

\[ \vec{P} = m \vec{v}, \quad \vec{L} = \vec{r} \times \vec{P} \] \[ \vec{F} = m \vec{a} = \frac{d\vec{P}}{dt} \]

Step-by-Step Solution:

Step 1: Differentiate \( \vec{r} \) with respect to time to find velocity.

\[ \vec{v} = \frac{d\vec{r}}{dt} = a(-\omega \hat{i} \sin \omega t + \omega \hat{j} \cos \omega t) \]

Step 2: Compute linear momentum.

\[ \vec{P} = m \vec{v} = m a \omega (-\hat{i} \sin \omega t + \hat{j} \cos \omega t) \]

Step 3: Find acceleration by differentiating velocity.

\[ \vec{a} = \frac{d\vec{v}}{dt} = -a \omega^2 (\hat{i} \cos \omega t + \hat{j} \sin \omega t) \]

Step 4: Compute the force acting on the particle.

\[ \vec{F} = m \vec{a} = -m a \omega^2 (\hat{i} \cos \omega t + \hat{j} \sin \omega t) \]

Step 5: Observe the direction of the force.

The term \( (\hat{i} \cos \omega t + \hat{j} \sin \omega t) \) is the same as the direction of \( \vec{r} \). The negative sign indicates that the force is directed opposite to \( \vec{r} \).

Final Computation & Result:

Therefore, the direction of the force is:

\[ \boxed{\text{Opposite to the direction of } \vec{r}} \]

Final Answer: Opposite to the direction of \(\vec{r}\).