Presentation on theme: "Physics of Amusement Park Rides"— Presentation transcript:
1Physics of Amusement Park Rides The CarouselFerris WheelLoop-the-LoopThe Rotor
2The Carousel: An Example of Uniform Circular Motion Turning about axis Ow/ Constant Angular Speed Axis O
3Draw 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
4The Ferris Wheel “Weightless” Feeling at the top ‘Heavy’ feeling at the bottom
5Draw 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
7How 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
8Draw 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.
9The 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.
10Draw 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