A bob of a pendulum of length 0.5 ~m has a speed of 6 ~ms ⁻¹ at its lowest point. Find the speed of the bob when the string of the pendulum makes 60^ with the vertical, (take .g=10 ~ms ⁻² )

Step-by-step solution

Given, length of pendulum, l=0.5 ~m Speed at lowest point, v₁=6 ~m / s Angle made by string with vertical, =60^ In O B C, O C O B = = 60^ l-h l = 60^ h=l (1- 60^ )h=0.5 (1- 1 2 )=0.25 ~m Using law of conservation of mechanical energy, we get aligned & K_i+U_i=K_f+U_f & 1 2 m v₁^2+0= 1 2 m v₂^2+m g h aligned [Taking initial potential energy at ground is equal to zero.] aligned 1 2 m v₂^2 & = 1 2 m v₁^2-m g h v₂^2 & =v₁^2-2 g h aligned Substituting the values, we get v₂= (6)^2-2 10 0.25 = 31 ~m / s Hence, the final v