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Direction of acceleration or force in uniform circular motion from a position or momentum vector

This type of question, asked 4 times: 25 Jun 2022, 29 Jan 2023, 5 Apr 2024, 4 Apr 2025. All 4 questions below, oldest first.

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  1. Question 1

    25 Jun 2022

    97
    For a particle in uniform circular motion, the acceleration a→\overrightarrow{a} at any point P(R,θ)P(R,\theta) on the circular path of radius RR is (when θ\theta is measured from the positive xx-axis and vv is uniform speed):

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  2. Question 2

    29 Jan 2023

    108
    An object moves at a constant speed along a circular path in a horizontal plane with center at the origin. When the object is at x=+2 mx = + 2\text{ }m, its velocity is −4ȷ^ m/s- 4\widehat{\jmath}\text{ }m/s. The object's velocity (v) and acceleration (a)(a) at x=−2 mx = - 2\text{ }m will be

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  3. Question 3

    5 Apr 2024

    97
    A particle moves in x−yx - y plane under the influence of a force F→\overrightarrow{F} such that its linear momentum is P→(t)=i^cos⁡(kt)−j^sin⁡(kt)\overrightarrow{P}(t) =\widehat{i}\cos(kt) -\widehat{j}\sin(kt). If kk is constant, the angle between F→\overrightarrow{F} and P→\overrightarrow{P} will be:

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  4. Question 4

    4 Apr 2025

    11
    If L→\overrightarrow{L} and P→\overrightarrow{P} represent the angular momentum and linear momentum respectively of a particle of mass ' m ' having position vector r→=a(i^cos⁡ωt+j^sin⁡ωt)\overrightarrow{r}= a(\widehat{i}\cos\omega t +\widehat{j}\sin\omega t). The direction of force is.

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