Fig 1 shows a pendulum of length L = 1.0 m. Its ball has speed of vo=2.0 m/s when the cord makes an angle of 30 degrees with the verti

Fig 1 shows a pendulum of length L = 1.0 m. Its ball has speed of vo=2.0
m/s when the cord makes an angle of 30 degrees with the vertical. What
is the speed (V) of the ball when it passes the lowest position?

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  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 = [tex]\sqrt{ v_1^2 +2gL(1-cos30 )}[/tex]

    let’s calculate

               v = [tex]\sqrt{ 2^2 + 2 \ 9.8 \ 1.0 (1- cos 30)}[/tex]

               v = 2.57 m / s

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