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MCQEasyJEE 2025Uniform Circular Motion

JEE Physics 2025 Question with Solution

If L→\overrightarrow{\mathrm{L}} and P→\overrightarrow{\mathrm{P}} represent the angular momentum and linear momentum respectively of a particle of mass 'mm' having position vector r→=a(i^cos⁡ωt+j^sin⁡ωt)\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 r→\overrightarrow{\mathrm{r}}

  • B

    Opposite to the direction of L→\overrightarrow{\mathrm{L}}

  • C

    Opposite to the direction of P→\overrightarrow{\mathrm{P}}

  • D

    Opposite to the direction of L→×P→\overrightarrow{\mathrm{L}} \times \overrightarrow{\mathrm{P}}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The position vector is

r→=a(i^cos⁡ωt+j^sin⁡ωt)\overrightarrow{\mathrm{r}} = \mathrm{a}(\hat{\mathrm{i}} \cos \omega \mathrm{t} + \hat{\mathrm{j}} \sin \omega \mathrm{t})

Find: The direction of the force acting on the particle.

From the given position vector, differentiate with respect to time to obtain acceleration through uniform circular motion relations:

a→=−ω2r→\overrightarrow{\mathrm{a}} = -\omega^2 \overrightarrow{\mathrm{r}}

Therefore, force is

F→=ma→=−mω2r→\overrightarrow{\mathrm{F}} = m\overrightarrow{\mathrm{a}} = -m\omega^2 \overrightarrow{\mathrm{r}}

The negative sign shows that the force is directed opposite to r→\overrightarrow{\mathrm{r}}.

Therefore, the correct option is A.

Differentiation-Based Derivation

Given:

r⃗=a(i^cos⁡ωt+j^sin⁡ωt)\vec{r} = a(\hat{i} \cos \omega t + \hat{j} \sin \omega t)

Find: The direction of F⃗\vec{F}.

First differentiate r⃗\vec{r} to find velocity:

v⃗=dr⃗dt=a(−ωi^sin⁡ωt+ωj^cos⁡ωt)\vec{v} = \frac{d\vec{r}}{dt} = a(-\omega \hat{i} \sin \omega t + \omega \hat{j} \cos \omega t)

Hence linear momentum is

P⃗=mv⃗=maω(−i^sin⁡ωt+j^cos⁡ωt)\vec{P} = m\vec{v} = ma\omega(-\hat{i} \sin \omega t + \hat{j} \cos \omega t)

Now differentiate velocity to find acceleration:

a⃗=dv⃗dt=−aω2(i^cos⁡ωt+j^sin⁡ωt)\vec{a} = \frac{d\vec{v}}{dt} = -a\omega^2(\hat{i} \cos \omega t + \hat{j} \sin \omega t)

Thus force is

F⃗=ma⃗=−maω2(i^cos⁡ωt+j^sin⁡ωt)\vec{F} = m\vec{a} = -ma\omega^2(\hat{i} \cos \omega t + \hat{j} \sin \omega t)

But

(i^cos⁡ωt+j^sin⁡ωt)(\hat{i} \cos \omega t + \hat{j} \sin \omega t)

is the same direction as r⃗\vec{r}. Hence F⃗\vec{F} is opposite to r⃗\vec{r}.

Therefore, the direction of force is opposite to the direction of r⃗\vec{r}, so the correct option is A.

Common mistakes

  • Confusing the force direction with the direction of linear momentum P→\overrightarrow{\mathrm{P}}. In circular motion, P→\overrightarrow{\mathrm{P}} is tangential, whereas the force is centripetal. Differentiate the position vector or use centripetal acceleration to identify the inward direction.

  • Assuming that force is along r→\overrightarrow{\mathrm{r}} instead of opposite to it. The acceleration is −ω2r→-\omega^2\overrightarrow{\mathrm{r}}, and the negative sign is essential because it indicates direction toward the center.

  • Using angular momentum L→\overrightarrow{\mathrm{L}} to infer the force direction directly. L→\overrightarrow{\mathrm{L}} is perpendicular to the plane of motion, while the force lies in the plane. Work from a→=d2r→dt2\overrightarrow{\mathrm{a}} = \frac{d^2\overrightarrow{\mathrm{r}}}{dt^2} instead.

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