A body takes the following path and moves with constant speed. If $a_{A}$ and $a_{B}$ are the magnitude of its radial acceleration at A and B, then.

a_{A} = a_{B}

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a_{A} < a_{B}

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a_{A} > a_{B}

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Solution

The correct option is C $a_{A} > a_{B}$
The radius of curvature is the radius of the osculating circle at that point of the curve.

We can clearly see that $R_{A} < R_{B}$
$∴ \frac{v^{2}}{R_{A}} > \frac{v^{2}}{R_{B}}$ (as v is the speed and is constant)
$\Rightarrow a_{A} > a_{B}$

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A body takes the following path and moves with constant speed. If $a_{A}$ and $a_{B}$ are the magnitude of its radial acceleration at A and B.

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