What will be the minimum speed of the roller- | vedclass

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(B) At the highest point of the vertical circular path, the forces acting on the passenger are the gravitational force $mg$ (acting downwards) and the

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$10$

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(B) At the highest point of the vertical circular path, the forces acting on the passenger are the gravitational force $mg$ (acting downwards) and the normal reaction $N$ (acting downwards).The net centripetal force required for circular motion is provided by the sum of these forces:$N + mg = \frac{mv^2}{R}$For the passenger not to fall out, the minimum condition is that the normal reaction $N$ becomes zero at the top point.Setting $N = 0$, we get:$mg = \frac{mv^2}{R}$$v^2 = gR$$v = \sqrt{gR}$Given $g = 10 \ m \ s^{-2}$ and $R = 10 \ m$, we substitute these values:$v = \sqrt{10 	imes 10} = \sqrt{100} = 10 \ m \ s^{-1}$.Thus, the minimum speed required is $10 \ m \ s^{-1}$.

What will be the minimum speed of the roller-coaster so that the passenger at the top, when becomes upside down, does not fall out (in $m \ s^{-1}$)? Consider the acceleration due to gravity as $10 \ m \ s^{-2}$, and the radius of curvature of the roller coaster is $10 \ m$.

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