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The angular acceleration of the toppling pole shown in figure is given by $α = k \sin θ,$ where $θ$ is the angle between the axis of the pole and the vertical, and k is a constant. The pole starts from rest at $θ = 0 .$ Choose the correct options

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The centripetal acceleration of the upper end of the pole is $2 k / (1 - \cos θ)$

The tangential acceleration of the upper end of the pole is $/ k \sin θ$

The centripetal acceleration of the upper end of the pole is $/ k \sin θ$

The tangential acceleration of the upper end of the pole is $2 k / (1 - \cos θ)$

answer is A, B.

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Detailed Solution

$α = k \sin θ$

$\frac{ω d ω}{d θ} = k \sin θ$

$\frac{ω^{2}}{2} = k (\cos θ)$

$ω^{2} = k (\cos θ - 1)$

$a_{t} = l α, a_{c} = l ω^{2}$

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