Question

A student skates up a ramp that makes an angle $30^{\circ}$ with the horizontal. He/she starts (as shown in the figure) 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)

• A. $v _{0}^{2}-2 gh =\frac{1}{2} gR$
• B. $v _{0}^{2}-2 gh =\frac{\sqrt{3}}{2} gR$
• C. the centripetal force required at points $x$ and $z$ is zero
• D. the centripetal force required is maximum at points $x$ and $z$

Answer 1

Answer: (A), (D)

Let $v$ be the speed at $y$
Energy conservation $\Rightarrow \frac{1}{2} m v_{0}^{2}=m g h+\frac{1}{2} m v^{2}$ .... (1)
FBD at $y \Rightarrow m g \sin 30^{\circ}=\frac{m v^{2}}{R}$ .... (2)
From (1) and (2)
$\Rightarrow v _{0}^{2}-2 gh =\frac{ gR }{2}$, (A) option is correct.
(D) is true as on the circular path he/she will have maximum speed at $x$ and $z$.