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 conservation of mechanical energy. At the bottom of the swing, all the initial potential energy is converted into kinetic energy. At the highest point of the swing, all the kinetic energy is converted into potential energy.
a) To find the height of the child at the highest point of their swinging motion, we can equate the kinetic energy at the bottom to the potential energy at the highest point:
1/2 * m * v^2 = m * g * h
where m is the mass of the child (40 kg), v is the velocity at the bottom (5 m/s), g is the acceleration due to gravity (9.8 m/s^2), and h is the height at the highest point.
Simplifying the equation:
1/2 * 40 kg * (5 m/s)^2 = 40 kg * 9.8 m/s^2 * h
100 J = 392 kg * m^2/s^2 * h
h = 0.255 m
Therefore, the child is approximately 0.255 m above the ground at the highest point of their swinging motion.
b) To find the angle between the chains and their natural resting position at the highest point of the swing, we can use trigonometry. At the highest point, the child is at the same height as the pivot point of the swing, forming a right triangle.
Using the Pythagorean theorem:
(2.1 m)^2 = 0.255 m^2 + d^2
4.41 m^2 = 0.065 m^2 + d^2
d^2 = 4.345 m^2
d = 2.085 m
Now, we can find the angle using the tangent function:
tan θ = 0.255 m / 2.085 m
θ ≈ tan^(-1)(0.122)
θ ≈ 6.99 degrees
Therefore, at the highest point of the swinging motion, the angle between the chains and their natural resting position is approximately 6.99 degrees.