Question 106: Centre of Mass and Rotational Motion

Question

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

Accepted Answer

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

Answer

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

Answer

Opposite to the direction of $\vec{L} 	imes \vec{P}$

Answer

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

Answer

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

Centre of Mass and Rotational Motion

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

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