A student skates up a ramp that makes an angle $30^\circ$ with the horizontal. He/she starts at the bottom of the ramp with speed $v_0$ 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)

Step 1: Analyze the geometry and effective gravity on the ramp.

The ramp is inclined at an angle of $30^\circ$ to the horizontal. The component of gravity acting along the plane of the ramp is directed downwards along the slope.

$$g_{\text{eff}} = g \sin 30^\circ$$

The highest point of the semicircular path is $y$, which is at a height $h$ from the ground. The points $x$ and $z$ lie on the horizontal diameter of the semicircle in the ramp plane. The vertical height of points $x$ and $z$ from the ground is given by:

$$h_x = h_z = h - R \sin 30^\circ$$

Step 2: Apply conservation of energy.

Since energy loss is negligible, conserve total mechanical energy between the bottom of the ramp (where speed is $v_0$) and the highest point $y$ (where speed is $v_y$).

$$\frac{1}{2}mv_0^2 = mgh + \frac{1}{2}mv_y^2$$ $$v_0^2 = 2gh + v_y^2$$

Step 3: Analyze the dynamics at the highest point $y$.

At the highest point $y$, the centripetal force required for the turn is provided solely by the component of the student's weight along the ramp.

$$\frac{mv_y^2}{R} = mg \sin 30^\circ$$ $$v_y^2 = gR \sin 30^\circ$$ $$v_y^2 = \frac{1}{2}gR$$

Step 4: Calculate the value of $v_0^2 - 2gh$.

Substitute the expression for $v_y^2$ into the energy conservation equation.

$$v_0^2 = 2gh + \frac{1}{2}gR$$ $$v_0^2 - 2gh = \frac{1}{2}gR$$

Step 5: Determine the centripetal force at points $x$ and $z$.

The centripetal force at any point on the semicircular path of radius $R$ is given by:

$$F_c = \frac{mv^2}{R}$$

Since $x$ and $z$ are at the lowest vertical level of the semicircular path, the potential energy is minimum at these points, which means the speed $v$ is maximum.

$$v_x^2 = v_y^2 + 2g(h - h_x)$$ $$v_x^2 = v_y^2 + 2gR \sin 30^\circ$$ $$v_x^2 = \frac{1}{2}gR + gR = \frac{3}{2}gR$$

Since the speed is maximum at $x$ and $z$, the centripetal force required for the turn is also maximum at these points.

Answer:

• $v_0^2 - 2gh = \frac{1}{2}gR$
• The centripetal force required for the turn is maximum at $x$ and $z$.