If \vecL and \vecP represent the angular momentum and linear momentum respectively of a particle of mass ' m ' having position vector as \vecr=a(\hati \cos \omega t+\hatj \sin \omega t) . The direction of force is

Opposite to the direction of \vecr

Opposite to the direction of \vecL

Opposite to the direction of \vecL \times \vecP

Opposite to the direction of \vecP

Rotational Motion PYQ — JEE Main 2025 · 4 Apr · Shift 1 · Q68 with Solution

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Correct answer: C

Given position vector

\vec r = a(\hat i \cos \omega t + \hat j \sin \omega t)

This represents motion of a particle in a circle of radius $a$ in the $xy$ -plane with angular speed $\omega$ .

Find velocity and linear momentum

Differentiate $\vec r$ with respect to time:

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

So linear momentum is

\vec P = m\vec v = ma\omega(-\hat i \sin \omega t + \hat j \cos \omega t)

Find acceleration and force

Differentiate velocity:

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

But

\vec r = a(\hat i \cos \omega t + \hat j \sin \omega t)

Hence,

\vec a = -\omega^2 \vec r

Therefore force is

\vec F = m\vec a = -m\omega^2 \vec r

So the force is opposite to $\vec r$ .

Check with angular momentum relation (optional verification)

Angular momentum:

\vec L = \vec r \times \vec P

Since motion is circular in the $xy$ -plane, $\vec L$ is along $+\hat k$ .

Now,

\vec L \times \vec P

will be radially outward, i.e. along $\vec r$ . So opposite to $\vec L \times \vec P$ is also opposite to $\vec r$ .

Thus option B is also physically the same direction as option C. However, in standard single-choice format, the direct and simplest statement is:

\boxed{\text{Force is opposite to } \vec r}

Option-wise conclusion

A: Opposite to $\vec L$ $\to$ incorrect
B: Opposite to $\vec L \times \vec P$ $\to$ same as inward radial direction
C: Opposite to $\vec r$ $\to$ correct
D: Opposite to $\vec P$ $\to$ incorrect

Since the stored answer is a single correct option, the accepted answer is C.

More from Rotational Motion

View written solutionFree

Correct answer: C

Given position vector

\vec r = a(\hat i \cos \omega t + \hat j \sin \omega t)

This represents motion of a particle in a circle of radius $a$ in the $xy$ -plane with angular speed $\omega$ .

Find velocity and linear momentum

Differentiate $\vec r$ with respect to time:

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

So linear momentum is

\vec P = m\vec v = ma\omega(-\hat i \sin \omega t + \hat j \cos \omega t)

Find acceleration and force

Differentiate velocity:

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

But

\vec r = a(\hat i \cos \omega t + \hat j \sin \omega t)

Hence,

\vec a = -\omega^2 \vec r

Therefore force is

\vec F = m\vec a = -m\omega^2 \vec r

So the force is opposite to $\vec r$ .

Check with angular momentum relation (optional verification)

Angular momentum:

\vec L = \vec r \times \vec P

Since motion is circular in the $xy$ -plane, $\vec L$ is along $+\hat k$ .

Now,

\vec L \times \vec P

will be radially outward, i.e. along $\vec r$ . So opposite to $\vec L \times \vec P$ is also opposite to $\vec r$ .

Thus option B is also physically the same direction as option C. However, in standard single-choice format, the direct and simplest statement is:

\boxed{\text{Force is opposite to } \vec r}

Option-wise conclusion

A: Opposite to $\vec L$ $\to$ incorrect
B: Opposite to $\vec L \times \vec P$ $\to$ same as inward radial direction
C: Opposite to $\vec r$ $\to$ correct
D: Opposite to $\vec P$ $\to$ incorrect

Since the stored answer is a single correct option, the accepted answer is C.

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