A body of $m \mathrm{~kg}$ slides from rest along the curve of vertical circle from point $A$ to $B$ in friction less path. The velocity of the body at $B$ is:
$\text { (given, } R=14 \mathrm{~m}, g=10 \mathrm{~m} / \mathrm{s}^2 \text { and } \sqrt{2}=1.4 \text { ) }$

Question

A body of $m \mathrm{~kg}$ slides from rest along the curve of vertical circle from point $A$ to $B$ in friction less path. The velocity of the body at $B$ is:

$\text { (given, } R=14 \mathrm{~m}, g=10 \mathrm{~m} / \mathrm{s}^2 \text { and } \sqrt{2}=1.4 \text { ) }$

Aicharya · Accepted Answer

Apply W.E.T. from A to B $\begin{aligned} & \Rightarrow \mathrm{W}_{\mathrm{mg}}=\mathrm{K}_{\mathrm{B}}-\mathrm{K}_{\mathrm{A}} \\ & \Rightarrow \mathrm{mg} \times\left(\frac{\mathrm{R}}{\sqrt{2}}+\mathrm{R}\right)=\frac{1}{2} \mathrm{mv}_{\mathrm{B}}^2-0\left\{\mathrm{v}_{\mathrm{A}}=0 \text { rest }\right\} \\ & \Rightarrow \mathrm{mgR} \frac{(\sqrt{2}+1)}{\sqrt{2}}=\frac{1}{2} \mathrm{mv}_{\mathrm{B}}^2 \\ & \Rightarrow \sqrt{\mathrm{gR} \frac{2(\sqrt{2}+1)}{\sqrt{2}}}=\mathrm{v}_{\mathrm{B}} \\ & \Rightarrow \sqrt{\frac{10 \times 14 \times 2(2.4)}{1.4}}=\mathrm{v}_{\mathrm{B}} \\ & \Rightarrow 21.9=\mathrm{v}_{\mathrm{B}} \end{aligned}$

A body of $m \mathrm{~kg}$ slides from rest along the curve of vertical circle from point $A$ to $B$ in…

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