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

Question

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?

in progress 0
Trúc Chi 2 weeks 2021-09-05T13:46:50+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-09-05T13:48:17+00:00

    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

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )