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For a particle in uniform circular motion of radius R at uniform speed v, the acceleration vector at a point P located at angle theta (measured from the positive x-axis) on the circle is:

1. -(v²/R) sin(theta) i + (v²/R) cos(theta) j
2. -(v²/R) cos(theta) i + (v²/R) sin(theta) j
3. -(v²/R) cos(theta) i - (v²/R) sin(theta) j
4. -(v²/R) i + (v²/R) j

Correct answer: -(v²/R) cos(theta) i - (v²/R) sin(theta) j

Solution

The point P has position vector R(cos theta i + sin theta j). The centripetal acceleration points toward the centre, i.e. opposite to the radial unit vector, with magnitude v²/R: a = -(v²/R)(cos theta i + sin theta j).

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