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(D) In a vertical circular motion,the minimum velocity required at the highest point $A$ to complete the circle is $v_A = \sqrt{gr}$.Using the princip

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$1:\sqrt{3}:\sqrt{5}$

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(D) In a vertical circular motion,the minimum velocity required at the highest point $A$ to complete the circle is $v_A = \sqrt{gr}$.Using the principle of conservation of energy between point $A$ and point $B$ (horizontal position),we have:$\frac{1}{2}mv_A^2 + mg(2r) = \frac{1}{2}mv_B^2 + mgr$$\frac{1}{2}mgr + 2mgr = \frac{1}{2}mv_B^2 + mgr$$\frac{3}{2}mgr = \frac{1}{2}mv_B^2 \implies v_B = \sqrt{3gr}$.Using the principle of conservation of energy between point $A$ and point $C$ (lowest point),we have:$\frac{1}{2}mv_A^2 + mg(2r) = \frac{1}{2}mv_C^2 + 0$$\frac{1}{2}mgr + 2mgr = \frac{1}{2}mv_C^2$$\frac{5}{2}mgr = \frac{1}{2}mv_C^2 \implies v_C = \sqrt{5gr}$.Thus,the ratio of velocities is $v_A : v_B : v_C = \sqrt{gr} : \sqrt{3gr} : \sqrt{5gr} = 1 : \sqrt{3} : \sqrt{5}$.

If a particle completes a vertical circular motion,what is the ratio of its velocities at points $A$,$B$,and $C$?

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