Q.(a) Derive the expression for Centripetal Acceleration. OR

Differentiating the circular position vector twice gives a = -omega^2 r(t), i.e., an acceleration of magnitude omega^2 r = v^2/r always pointing toward the centre -- this is centripetal acceleration. (This question offered an internal choice; part (a), the derivation of centripetal acceleration, is answered here.)

Set up: Consider a particle moving with constant speed v along a circle of radius r, with constant angular speed omega (so v = omega r). Place the centre of the circle at the origin. At time t, the particle's angular position is theta = omega t (measuring from the x-axis), so its position vector is:

r(t) = r cos(omega t) i + r sin(omega t) j

Step 1 -- Velocity (first derivative):

v(t) = dr/dt = -r omega sin(omega t) i + r omega cos(omega t) j

Magnitude: |v| = r omega sqrt(sin^2 + cos^2) = r omega -- consistent with v = omega r, confirming the speed is constant.

Note also that v(t) . r(t) = -r^2 omega sin cos + r^2 omega sin cos = 0, so velocity is always perpendicular to the position vector (tangential), as expected for circular motion.

Step 2 -- Acceleration (second derivative):

a(t) = dv/dt = -r omega^2 cos(omega t) i - r omega^2 sin(omega t) j

= -omega^2 [r cos(omega t) i + r sin(omega t) j]

= -omega^2 r(t)

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