Acceleration in Circular Path – Rankers Physics
Topic: Circular Motion
Subtopic: Kinematics of Circular Motion

Acceleration in Circular Path

A particle moves in a circular path so that its distance travel varies with time \(t\) as \(s = 3t^2 + 6t\). Then its acceleration at \(t = 1\text{ sec.}\) is (radius of path is \(12\text{ m}\)) -
\(6\sqrt{5}\text{ m/s}^2\)
\(6\text{ m/s}^2\)
\(12\text{ m/s}^2\)
\(12\sqrt{3}\text{ m/s}^2\)

Solution:

Speed is \(v = \frac{ds}{dt} = 6t + 6\). At \(t = 1\text{ s}\), \(v = 12\text{ m/s}\). Tangential acceleration is \(a_t = \frac{dv}{dt} = 6\text{ m/s}^2\). Centripetal acceleration is \(a_c = \frac{v^2}{R} = \frac{12^2}{12} = 12\text{ m/s}^2\). Total acceleration is \(a = \sqrt{a_t^2 + a_c^2} = \sqrt{6^2 + 12^2} = 6\sqrt{5}\text{ m/s}^2\).

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