Motion of Skater on Inclined Ramp and Semicircular Turn

To determine the correct options, we analyze the skater's motion on the inclined plane using Newton's laws of motion and the principle of conservation of mechanical energy.

1. Analysis at the Highest Point $y$

The inclined ramp makes an angle $\theta = 30^\circ$ with the horizontal plane. The component of gravitational acceleration acting along the plane of the incline (directed down the slope) is given by: $g_{\parallel} = g \sin 30^\circ = \frac{1}{2}g$

At point $y$, which is the peak of the semicircular path of radius $R$, the vector pointing towards the center of curvature is directed straight down the incline plane.

It is given that the centripetal force required for the circular turn at the highest point $y$ is supplied solely by the component of the skater's weight along the incline. Therefore: $\frac{m v_y^2}{R} = m g \sin 30^\circ = \frac{1}{2}mg$

Solving for $v_y^2$: $v_y^2 = \frac{1}{2}gR$

2. Conservation of Mechanical Energy

Since energy dissipation is negligible, the total mechanical energy is conserved between the starting point at the bottom of the ramp (height $0$, speed $v_0$) and point $y$ (height $h$, speed $v_y$): $\frac{1}{2}mv_0^2 = mgh + \frac{1}{2}mv_y^2$

Dividing by $\frac{1}{2}m$: $v_0^2 = 2gh + v_y^2$

Substituting $v_y^2 = \frac{1}{2}gR$ into the equation: $v_0^2 - 2gh = \frac{1}{2}gR$

This proves that Option (A) is correct and Option (B) is incorrect.

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3. Evaluation of Centripetal Force at Points $x$ and $z$

The required centripetal force at any point on the semicircular path of radius $R$ is given by: $F_c = \frac{mv^2}{R}$

The height of points $x$ and $z$ on the inclined ramp is less than the peak height $h$ at point $y$: $h_x = h_z = h - R \sin 30^\circ = h - \frac{R}{2}$

By conservation of energy, the speed $v$ at any point at height $y_{\text{height}}$ is: $v^2 = v_0^2 - 2g y_{\text{height}}$

Since points $x$ and $z$ are at the lowest elevation along the semicircular arc $xyz$, the skater's speed $v$ reaches its maximum value at points $x$ and $z$: $v_x^2 = v_z^2 = v_0^2 - 2g\left(h - \frac{R}{2}\right) = (v_0^2 - 2gh) + gR = \frac{1}{2}gR + gR = \frac{3}{2}gR$

Since the speed $v$ is maximum at points $x$ and $z$, the required centripetal force $F_c = \frac{mv^2}{R}$ is also maximum at points $x$ and $z$.

This proves that Option (D) is correct and Option (C) is incorrect.

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Conclusion

• Correct Options: (A) and (D)