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The angular acceleration of a particle moving along a circle is given by $α = ksin θ,$ where $θ$ is the angle turned by particle and k is constant. The centripetal acceleration of the particle in terms of k, $θ$ and R (radius of circle) is given by :

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$2 kR (1 + \cos θ)$

$2 kR (\sin θ)$

$2 kR (\sin^{2} θ)$

$2 kR (1 - \cos θ)$

answer is D.

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Detailed Solution

$\begin{array}{l} α = ksin θ \\ \int ωdω = \int ksin θdθ \\ \frac{ω^{2}}{2} = - [kcos θ]_{0}^{θ} \Rightarrow \frac{ω^{2}}{2} = - k [\cos θ - 1] \\ ω^{2} = 2 k (1 - \cos θ) \\ a_{c} = ω^{2} R = 2 kR (1 - \cos θ) \end{array}$

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