Question

An object with a mass of 9 kg is moving at the inside of a circular loop of radius 9 m which is frictionless, like a roller cart. If the amount of normal force acting on the mass at the top of the loop is 2.2 times the weight of the mass, how fast is the object moving at the bottom of the loop in m/s

Answer 1

At the top of the loop, the centripetal force acts upward, the weight and the normal force acts downward Thus, the equation of motion of object at the top of the loop - 3.2W p3.2mg 3.2x9x9.81 - 16.8m/s Apply principle of conservation of energy Energy at the top of the loop - Energy at the bottom of the loop +4gr botom top (16.8) +(4x9.81x9) 25.2 m/s Thus, the velocity of the object at the bottom of the loop is 25.2 m/s