Physics of Amusement Park Rides The Carousel Ferris Wheel Loop-the-Loop The Rotor

The Carousel: An Example of Uniform Circular Motion Turning about axis O w/ Constant Angular Speed  Axis O

Draw the Forces on the Rider  What is the nature of the net force on the rider ? Where is this pointing ? Central force pointing to O What provides the net force on him ? The normal support force of the back of the seat What kind of net force along the y-axis ? Is there motion along y ? y: stationary uniform motion What is the sensation he feels ? Pressed against the chair (or alternatively, chair pressing against his back) How can this be enhanced ? Increase  O y N’ = mg N = mv2/R x mg

The Ferris Wheel “Weightless” Feeling at the top ‘Heavy’ feeling at the bottom

Draw the Forces on the Rider constant  2 What is the nature of the net force on the rider ? Where is this pointing ? a central force mg At 2: mg - N = mv2/R or N = mg - mv2/R ‘feels lighter’ 1 At 1: N - mg = mv2/R or N = mg + mv2/R ‘feels heavier’ 3 N What is the sensation he feels At locations 2 and 4 ? 4 mg What happens if the rotational speed is increased beyond (gR)1/2? He flies off upon reaching position 2

The Roller Coaster SLOW FAST Weightless Sensation Heavy Sensation

How is Energy conserved in the roller coaster ride ? Potential Energy = mgh Kinetic Energy= 0 A. C Vo = 0 h R B. KE = (1/2) mv2 PE = 0 To clear the top of the loop, h  2R. In fact, ignoring friction, minimum h = 2.5 R

Draw the Forces on the Rider constant  What is the nature of the net force on the rider ? Where is this pointing ? A central force towards O 2 v At 2: N + mg = mv2/R or N = mv2/R - mg Radius R N mg 1 O At 1: N - mg = mv2/R or N = mg + mv2/R N 3 What is the sensation he feels at locations 2 and 4 ? Heavier at 4, lighter at 2 4 mg What happens if the speed v is decreased below (gR)1/2? Rider falls out of car at 2, if not wearing harness.

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.

Draw the forces on the rider ‘pinned’ to the rotor’s wall: Friction f = N N= Fc = mv2/R rotor wall Floor pulled out mg The central force Fc is provided by the normal or support force N from the rotor’s walls. For a minimum rotor speed, the normal force is large enough that the friction f is enough to overcome the weight mg, keeping the man pinned to the wall. The critical minimum speed is solved from mg = N = mv2/R or v = (gR/)1/2