A smooth inclined plane ends in a vertical circular loop, as shown in the figure. A small body is re

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A smooth inclined plane ends in a vertical circular loop, as shown in the figure. A small body is released from height $h$ as shown. If the body exerts a force of three times its weight on the plane at the highest point of circle then the height $h = \alpha R$. The value of $\alpha$ is

Accepted Answer

Sol. [Key Concept: At the top the plane pushes with $3mg$, so $mg + 3mg = \frac{{mv}^{2}}{R}$ gives $v^{2} = 4gR$; energy conservation $mg(h - 2R) = \frac{1}{2}\ {mv}^{2}$ then gives $h = 4R$.]at highest point$${mg + N = \frac{{mv}^{2}}{R} }{4mg = \frac{{mv}^{2}}{R} }{v = \sqrt{4gR} }$$Now energy conservation$$mg(h - 2R) = \frac{1}{2}	ext{ }m 	imes 4gR $$( $h = 4R$ )

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