A student skates up a ramp that makes an angle 30° with the horizontal. He/she starts (as shown in the figure) at the bottom of the ramp with speed 𝑣₀ and wants to turn around over a semicircular path xyz of radius 𝑅 during which he/she reaches a maximum height ℎ (at point y) from the ground as shown in the figure. Assume that the energy loss is negligible and the force required for this turn at the highest point is provided by his/her weight only. Then (g is the acceleration due to gravity)

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$v_{0}^{2} - 2 gh = \frac{1}{2} gR$

the centripetal force required at points x and z is zero

$v_{0}^{2} - 2 gh = \frac{\sqrt{3}}{2} gR$

the centripetal force required is maximum at points x and z

answer is A, D.

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Detailed Solution

By the energy conservation (ME) between bottom point and point Y

$\begin{array}{l} \frac{1}{2} {mv}_{0}^{2} = mgh + \frac{1}{2} {mv}_{1}^{2} \\ ∴ v_{1}^{2} = v_{0}^{2} - 2 gh \dots (i) \end{array}$

Now at point Y the centripetal force provided by the component of mg

$\begin{array}{l} ∴ mgsin 30^{\circ} = \frac{{mv}_{1}^{2}}{R} \\ ∴ v_{1}^{2} = \frac{gR}{2} \\ ∴ from (i) \\ \frac{g R}{2} = v_{0}^{2} - 2 g h \end{array}$

As gravitation force is not providing centripetal force at x and z that's why maximum force is applied on x and z for circular motion.

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