A student skates up a ramp that makes an angle 30 ° with the...

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

{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 student skates up a ramp that makes an angle 30 ° with the horizontal. He/she starts (as shown in the figure) at the

Options

A ${v}_{0}^{2} - 2gh = \dfrac{1}{2}gR$
B ${v}_{0}^{2} - 2gh = \dfrac{\sqrt{3}}{2}gR$
Cthe centripetal force required at points $x$ and $z$ is zero
Dthe centripetal force required is maximum at points $x$ and $z$

Correct answer

A. ${v}_{0}^{2} - 2gh = \dfrac{1}{2}gR$

Step-by-step solution

Speed at $x$ and $z$ is equal and also maximum in circular track.

From energy conversation,

$\dfrac{1}{2}m{v}_{0}^{2} = mgh + \dfrac{1}{2}m{v}^{2}$ .......(1)

at $y\text{,} mg \sin 30 ° = \dfrac{m{v}^{2}}{R}$

$⇒ {v}^{2} = \dfrac{Rg}{2}$

$∴$ From equation (1)

${v}_{0}^{2} - 2gh = \dfrac{Rg}{2}$

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