Presentation on theme: "Physics of Amusement Park Rides"— Presentation transcript:
1 Physics of Amusement Park Rides The CarouselFerris WheelLoop-the-LoopThe Rotor
2 The Carousel: An Example of Uniform Circular Motion Turning about axis Ow/ Constant Angular Speed Axis O
3 Draw the Forces on the Rider What is the nature of the net force onthe rider ? Where is this pointing ?Central force pointing to OWhat provides the net force on him ?The normal support forceof the back of the seatWhat kind of net force along the y-axis ?Is there motion along y ?y: stationary uniform motionWhat is the sensation he feels ?Pressed against the chair(or alternatively, chair pressing against his back)How can this be enhanced ?Increase OyN’ = mgN = mv2/Rxmg
4 The Ferris Wheel “Weightless” Feeling at the top ‘Heavy’ feeling at the bottom
5 Draw the Forces on the Rider constant 2 What is the nature of the net force onthe rider ? Where is this pointing ?a central forcemgAt 2: mg - N = mv2/Ror N = mg - mv2/R‘feels lighter’1At 1: N - mg = mv2/Ror N = mg + mv2/R‘feels heavier’3NWhat is the sensation he feelsAt locations 2 and 4 ?4mgWhat happens if the rotational speed is increased beyond (gR)1/2?He flies off upon reaching position 2
6 The Roller CoasterSLOWFASTWeightless SensationHeavy Sensation
7 How is Energy conserved in the roller coaster ride ? Potential Energy = mghKinetic Energy= 0A.CVo = 0hRB. KE = (1/2) mv2PE = 0To clear the top of the loop, h 2R.In fact, ignoring friction, minimum h = 2.5 R
8 Draw the Forces on the Rider constant What is the nature of the net force onthe rider ? Where is this pointing ?A central force towards O2vAt 2: N + mg = mv2/Ror N = mv2/R - mgRadius RNmg1OAt 1: N - mg = mv2/Ror N = mg + mv2/RN3What is the sensation he feelsat locations 2 and 4 ?Heavier at 4, lighter at 24mgWhat happens if the speed v is decreased below (gR)1/2?Rider falls out of car at 2, if not wearing harness.
9 The Rotor Radius R A large cylinder spins. You are thrown and pinned against the wall.The floor then slides out.Yet you do not fall.
10 Draw the forces on the rider ‘pinned’ to the rotor’s wall: Frictionf = NN= Fc = mv2/Rrotor wallFloor pulled outmgThe central force Fc is provided by the normal orsupport force N from the rotor’s walls.For a minimum rotor speed, the normal force is large enoughthat the friction f is enough to overcome the weight mg, keepingthe man pinned to the wall.The critical minimum speed is solved from mg = N = mv2/Ror v = (gR/)1/2