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5.Work, Energy, Power and Collision
hard
A particle of mass $m$ , attached with a string of length $l$ is moving in a vertical circle. If the particle is just looping the loop without slackening of the string and $v_A, v_B, v_D$ are speeds at positions $A, B, D$ shown in figure, then
A
$v_B > v_D > v_A$
B
tension in the string at $D$ is $3\ mg$
C
$v_D = \sqrt {3gl}$
D
All of the above
Solution
At $\mathrm{A}, \quad v_{A}=\sqrt{g l}$
At $\mathrm{B}, \quad v_{B}=\sqrt{5 g l}$
and at $D,$ $\quad v_{D}=\sqrt{3 g l}$
Thus, $v_{B}>v_{D}>v_{A}$
$\mathrm{Also}, T=3 m g(1+\cos \theta)$
$\mathrm{So},$ at $\mathrm{D}, \theta=90^{\circ}$
$\therefore T=3 m g(1+0)=3 m g$
Standard 11
Physics
