A small body slides down a smooth uneven surf | vedclass

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Question

(A) Let the mass of the body be $m$. At point $A$,the normal force $N$ provides the centripetal force required for circular motion. Since the weight $

Vedclass · Accepted Answer

$H=\frac{3 R}{2}$

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(A) Let the mass of the body be $m$. At point $A$,the normal force $N$ provides the centripetal force required for circular motion. Since the weight $mg$ acts vertically downwards and the normal force acts horizontally towards the center $O$,the net force on the body at $A$ is the vector sum of $N$ and $mg$.The magnitude of the net force is $F_A = \sqrt{N^2 + (mg)^2}$.Given $F_A = \sqrt{2} mg$,we have $\sqrt{N^2 + (mg)^2} = \sqrt{2} mg$.Squaring both sides,$N^2 + m^2g^2 = 2m^2g^2$,which gives $N^2 = m^2g^2$,so $N = mg$.The centripetal force equation at $A$ is $N = \frac{mv_A^2}{R}$,so $mg = \frac{mv_A^2}{R}$,which implies $v_A^2 = Rg$.By the law of conservation of energy,the total energy at the starting point $P$ (height $H$) equals the total energy at point $A$ (height $R$):$mgH = \frac{1}{2}mv_A^2 + mgR$.Substituting $v_A^2 = Rg$:$mgH = \frac{1}{2}m(Rg) + mgR = \frac{3}{2}mgR$.Therefore,$H = \frac{3}{2}R$.

$A$ small body slides down a smooth uneven surface from a height $H$,which eventually emerges into a circular loop of radius $R (< H)$. What should be the value of $H$,so that the force on the body at $A$ is $\sqrt{2}$ times its weight?

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