A 40kg child is swinging on a swing. At the bottom of the swinging motion, the child is 0.6m above the ground and travelling at a speed of 5m/s. The chains the swing is attached to are 2.1m long.
a) How high above the ground is the child at the highest point of their swinging motion?
b) At the highest point of the swinging motion, what is the angle between the chains and their natural resting position?
To solve this problem, we can use the principles of conservation of mechanical energy.
a) At the bottom of the swinging motion, the child is at ground level and traveling at a speed of 5 m/s. We can assume that at the highest point of the swinging motion, the child momentarily stops before swinging back in the other direction.
At the bottom of the swing, the child's mechanical energy is given by the equation:
E = 1/2 * m * v^2 + m * g * h
where m is the mass of the child (40 kg), v is the velocity of the child (5 m/s), g is the acceleration due to gravity (9.8 m/s^2), and h is the height above the ground (0.6 m).
E = 1/2 * 40 * (5)^2 + 40 * 9.8 * 0.6
E = 500 + 235.2
E = 735.2 J
At the highest point of the swinging motion, the child's mechanical energy is entirely potential energy, given by the equation:
E = m * g * h'
where h' is the height above the ground at the highest point.
735.2 J = 40 * 9.8 * h'
h' = 735.2 / (40 * 9.8)
h' ≈ 1.88 m
Therefore, the child is approximately 1.88 m above the ground at the highest point of their swinging motion.
b) At the highest point of the swinging motion, the chains are at equilibrium (no net force acting on them) and their tension is equal to the weight of the child.
The angle between the chains and their natural resting position can be found using trigonometry.
cos(θ) = adjacent/hypotenuse
cos(θ) = 0.6 m / 2.1 m
θ = arccos(0.6 / 2.1)
θ ≈ 71.86°
Therefore, at the highest point of the swinging motion, the angle between the chains and their natural resting position is approximately 71.86°.