Question

3. A roller coaster sits at the top of a hill and is preparing to enter a loop-the-loop as shown below. The hill has a height of 86.0 meters, the loop has a radius of 22.0 meters and the roller coaster has a mass of 525 kg. Assume h=0 at the bottom of the hill. a) What will be the velocity of the roller coaster when it reaches point C? A b) What will be the velocity of the roller coaster when it reaches point A? RAB

Answer 1

525kg af 1/32 ner we can Ans Height of the hills h=86m. Mass of the roller coasteram= Let the velocity of the roller coaster posiad cov. When the roller coaster is at the top of the hill its lotal energy is potential energy = mgh. Where ins the go ravitational acceleration = 9.8 m But When it is at point č its gg=1/2 mu2 So, from energy conservalim say 12 m vzamgh o /2x9:8886 m/s. 41 m/s Now let the velocity of roller coaster at point A :V From energy conserralien we total energy at ca total energy kinetic > ate the total energy energy & potential energy (m gx 2R) 27 £ mul2+ mg x2R = 12 m 2 + 4 gr u vien v2-49R /412-489.8 x 22 m/s v'e 28.6 m/s. R: Radias of the loop 222m varaga can sag at A. 012 2 v2

a> velocity, at ca 41 m/s by Velocity at A= /28.6 m/s