A pendulum bob has a speed 4 ~m / s at its lowest position. The pendulum is 1 m long. When the length of the string makes an angle of 60^ with the vertical, the speed of the bob at that position is (acceleration due to gravity, g=10 ~m / s ^2, 60^ =0 5 )

Step-by-step solution

The speed v₂ of the pendulum bob when the string makes an angle of 60^ with the vertical can be found using conservation of mechanical energy, taking the lowest position as the reference for potential energy. At the lowest position, the speed is v₁ = 4 , m/s , so the total energy is 1 2 mv₁^2 = 8m . At = 60^ , the bob is at a height h = L(1 - ) = 0.5 , m above the reference level, giving a potential energy of mgh = 5m . By conservation of energy, 1 2 mv₁^2 = mgh + 1 2 mv₂^2 . Cancelling m and multiplying by 2 yield