Work, Power & Energy

The minimum speed V at position A, which is required by the bob to reach the top end at position B, is $\sqrt{5gL}$. From conservation of energy, we get $\frac{1}{2} m v^{2}$ = $\frac{1}{2} m (v')^{2}$ + mgl (1 - cos θ) T - mg cos θ = $\frac{m v^{2}}{l}$ Therefore, $(v')^{2}$ = $v^{2}$ + 2gl cos θ - 2gl $\left(\frac{v}{2}\right)^{2}$ = $v^{2}$ + 2gl (cosθ - 1) $\frac{3 v^{2}}{4}$ = 2gl (1- cos θ) $\frac{15 g l}{4}$ = 2gl (1- cos θ) ⇒ cos θ = 1 - $\frac{15}{8}$ = $-\frac{7}{8}$ = - 0.87 Since we have cos $\left(\frac{3\pi}{4}\right)$ = $\frac{-1}{\sqrt{2}}$ = - 0.71, we get $\frac{3\pi}{4}$ < θ < π.