A heavy particle slides under gravity down the inside of a smooth vertical tube held in vertical plane. It starts from the highest point with velocity $\sqrt{2 a g}$ , where $a$ is the radius of the circle. The angular position $θ$ (as shown in figure) at which the vertical acceleration of the particle is maximum, is given by $\cos^{- 1} (\frac{p}{q}) .$ Find the value of $p + q$ .

Difficult

Solution

$\begin{array}{l} At position θ, \\ v^{2} = v_{0}^{2} + 2 g h \\ where, h = a (1 - \cos θ) \\ ∴ v^{2} = {(\sqrt{2 a g})}^{2} + 2 a g (1 - \cos θ) \\ \Rightarrow v^{2} = 2 a g (2 - \cos θ) \\ A l s o, N + m g \cos θ = \frac{m v^{2}}{a} \\ \Rightarrow N + m g \cos θ = 2 m g (2 - \cos θ) \\ \Rightarrow N = m g (4 - 3 \cos θ) \\ Net vertical force, F = N \cos θ + m g \\ \Rightarrow F = m g (4 \cos θ - 3 \cos^{2} θ + 1) \\ This force (or acceleration) will be maximum when \\ \frac{d F}{d θ} = 0 \\ \Rightarrow - 4 \sin θ + 6 \sin θ \cos θ = 0 \\ \Rightarrow \sin θ (- 4 + 6 \cos θ = 0) \\ \Rightarrow \cos θ = \frac{2}{3} \\ \Rightarrow θ = \cos^{- 1} (\frac{2}{3}) \\ ∴ p + q = 2 + 3 = 5 \end{array}$

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