To solve this problem, we can use conservation of mechanical energy.
The mechanical energy of the pendulum at point A (initial point) is given by:
E_A = mgh_A + (1/2)mv_A^2
where:
m = mass of the pendulum (1.6 kg)
g = acceleration due to gravity (9.8 m/s^2)
h_A = height of the pendulum at point A (1.5 m in this case)
v_A = velocity of the pendulum at point A
Since the pendulum is released from rest at point A, the initial velocity v_A is zero. Therefore, the mechanical energy at point A simplifies to:
E_A = mgh_A
At point D (final point), the mechanical energy is given by:
E_D = mgh_D + (1/2)mv_D^2
where:
h_D = height of the pendulum at point D (half the height of point A, so h_A/2 = 1.5/2 = 0.75 m)
v_D = velocity of the pendulum at point D (which we want to find)
Conservation of mechanical energy states that the mechanical energy at point A is equal to the mechanical energy at point D:
E_A = E_D
mgh_A = mgh_D + (1/2)mv_D^2
mgh_A - mgh_D = (1/2)mv_D^2
mg(h_A - h_D) = (1/2)mv_D^2
Simplifying further:
g(h_A - h_D) = (1/2)v_D^2
v_D^2 = 2g(h_A - h_D)
Substituting the known values:
v_D^2 = 2(9.8 m/s^2)(1.5 m - 0.75 m)
v_D^2 = 2(9.8 m/s^2)(0.75 m)
v_D^2 = 15.18 m^2/s^2
Taking the square root of both sides to find the speed:
v_D = √(15.18 m^2/s^2)
v_D ≈ 3.89 m/s
Therefore, the speed of the pendulum at point D, half the height up to point E, is approximately 3.89 m/s.
A 1.5 m long frictionless pendulum of mass 1.6 kg is released from point A at an angle 0 of 10 degrees. What is the speed of the pendulum at point D, half the height up to point E?
1 answer