A ring, of mass 2.80 kg and radius 14.9 cm, can freely rotate about the axle as shown above.If the ring starts at its highest position and is given a small nudge, it will begin to rotate around the axle. What is the speed of the lowest point of the ring when the ring has reached its lowest position?

To solve this problem, we can use the principle of conservation of energy. This principle states that the total energy of an isolated system remains constant if no external forces do work on it. In this case, the ring's initial potential energy is converted into kinetic energy as it falls.

Step 1: Calculate the initial potential energy of the ring. The potential energy (PE) is given by the formula PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. In this case, the height is the radius of the ring, so h = 14.9 cm = 0.149 m. The mass m = 2.80 kg, and g = 9.8 m/s². So, PE = 2.80 kg * 9.8 m/s² * 0.149 m = 4.08 Joules.

Step 2: At the lowest point of the ring, all the potential energy has been converted into kinetic energy. The kinetic energy (KE) is given by the formula KE = 1/2 mv², where m is the mass and v is the velocity. We can set the kinetic energy equal to the potential energy and solve for v: 4.08 Joules = 1/2 * 2.80 kg * v².

Step 3: Solve the equation for v. This gives v = sqrt((2*4.08 Joules) / 2.80 kg) = sqrt(2.92) = 1.71 m/s.

So, the speed of the lowest point of the ring when the ring has reached its lowest position is 1.71 m/s.