A frictionless inclined plane terminates smoothly into a frictionless vertical circular loop mounted on a plank that can slide on a frictionless horizontal floor. This assembly has a total mass of 0.8 kg. A small ball P of mass 0.2 kg is released on the inclined plane from a height that is triple of the radius of the circular loop. Find force of normal reaction between the ball (in N) and the circular loop, when the ball is at its topmost position in the circular loop. $(T a k e g = 10 m / s^{2})$

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Detailed Solution

$C O E; Δ U + Δ K = 0 \Rightarrow m g (2 r - 3 r) + \frac{1}{2} μ v_{r}^{2} = 0 \Rightarrow m g r = \frac{1}{2} \times \frac{4 m}{5} \times 25 V^{2} \Rightarrow V = \sqrt{\frac{g r}{10}}$

Notice: we have used $Δ K = Δ \tilde{K} = \frac{1}{2} μ v_{r}^{2}$
Because $| {\bar{p}}_{i} | = | {\bar{p}}_{f} | = 0$
Question Image

Motion of ball is circular wrt C.

$m g + N = m \times \frac{25 v^{2}}{r} \Rightarrow N = m \times \frac{25}{r} \times \frac{g r}{10} - m g = \frac{3}{2} m g = \frac{3}{2} \times 0.2 \times 10 = 3 N$

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