(C) At the top of the hill, the forces acting on the rider are the gravitational force $(mg)$ acting downwards and the normal reaction $(N)$ acting up

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(C) At the top of the hill, the forces acting on the rider are the gravitational force $(mg)$ acting downwards and the normal reaction $(N)$ acting upwards.The centripetal force required for circular motion is provided by the net force towards the center: $mg - N = \frac{mv^2}{R}$.For the condition of "weightlessness", the normal reaction $N$ must be zero.Therefore, $mg = \frac{mv^2}{R}$.Solving for velocity $v$: $v = \sqrt{Rg}$.Given $R = 20\, m$ and taking $g = 10\, m/s^2$, we have:$v = \sqrt{20 \times 10} = \sqrt{200} \approx 14.14\, m/s$.Thus, the speed is between $14\, m/s$ and $15\, m/s$.

Class 11 Physics 3-2.Motion in Plane Dynamics of circular Motion (Centrifugal force) and Pendulum and Motion on Curved path

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$A$ roller coaster is designed such that riders experience "weightlessness" as they go round the top of a hill whose radius of curvature is $20\, m$. The speed of the car at the top of the hill is between

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A
$16\, m/s$ and $17\, m/s$
B
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D
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3-2.Motion in Plane

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Dynamics of circular Motion (Centrifugal force) and Pendulum and Motion on Curved path

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