Question

Fig 1 shows a pendulum of length L = 1.0 m. Its ball has speed of vo=2.0

Answer 1

Answer:

v = 2.57 m / s

Explanation:

For this exercise let's use conservation of energy

starting point. When it is at an angle of 30º

          Em₀ = K + U = ½ m v₁² + m g y₁

final point. Lowest position

          Em_f = K = ½ m v²

as there is no friction, the energy is conserved

          Em₀ = Em_f

          ½ m v₁² + m g y₁ = ½ m v²

Let's find the height(y₁), which is the length of the thread minus the projection (L ') of the 30º angle

         cos 30 = L ’/ L

         L ’= L cos 30

         y₁ = L -L '

          y₁ = L- L cos 30

we substitute

          ½ m v₁² + m g L (1- cos 30) = ½ m v²

           v = $\sqrt{ v_1^2 +2gL(1-cos30 )}$

let's calculate

           v = $\sqrt{ 2^2 + 2 \ 9.8 \ 1.0 (1- cos 30)}$

           v = 2.57 m / s