A simple pendulum of length 2.5 m is displaced such that its string makes an angle of 37^ with the downward vertical. From this position, the pendulum bob is given an initial speed of 6 m s ⁻¹ along its circular path towards the lowest point. The speed of the bob when it reaches the lowest point of its path is: (Take g = 10 m s ⁻² and 37^ = 0.8 )

Step-by-step solution

Let the lowest point be the reference level for potential energy ( U = 0 ). The initial vertical height h of the bob above the lowest point is: h = L(1 - 37^ ) h = 2.5 (1 - 0.8) = 2.5 0.2 = 0.5 m Applying conservation of mechanical energy between the initial position and the lowest point: 1 2 m u^2 + m g h = 1 2 m v^2 v^2 = u^2 + 2 g h Substituting the given values ( u = 6 m s ⁻¹ , g = 10 m s ⁻² , h = 0.5 m ): v^2 = ( 6 )^2 + 2 10 0.5 v^2 = 6 + 10 = 16 v = 4 m s ⁻¹ Answer: 4 m s ⁻¹