LCPA βͺ PHYSICS (MS.APPEL) 5-1 NOTES: CIRCULAR MOTION Name: Page 1 of 2 Date: 10-30-17 Period: Circular Motion Notes Linear vs. Circular Motion Quantity Linear Motion Speed + Direction Direction of the object’s motion Circular Motion Speed stays the same Direction is tangent to the circle πΆβππππ ππ π£ππππππ‘π¦ Speed does not change; Direction changes Direction—Toward the center of the circle Velocity π‘πππ Acceleration Force Due to a push/pull Direction of the push/pull Push or pull πΉ = π∗π In the direction of the acceleration Looking at Circular Motion Centripetal force – “center seeking” Centrifugal force – “center fleeing” ο Fictitious! (Fake!) Direction is toward the center of circle LCPA βͺ PHYSICS (MS.APPEL) 5-1 NOTES: CIRCULAR MOTION Name: Page 2 of 2 Date: 10-30-17 Period: Circular Motion Problems Important Quantities ο· Frequency (f): πππππ’ππππ¦ = Example: 5 revolutions in 40 seconds. π= ο· ππ’ππππ ππ πππ£πππ’π‘ππππ πππ£ → [ ] π‘πππ π ππ Velocity (v): π£ = πππ π‘ππππ π‘πππ → [ π π ππ 5 πππ£ πππ£ = 0.125 40 sec π ππ ] π£ = 2πππ ο· Centripetal Acceleration (ac): ππ = ο· π£2 π Centripetal Force (Fc): ππ£ 2 πΉπ = ππ = → [π] π Example: Captain Chip, the pilot of a 60,500 kg jet plane, is told that he must remain in a holding pattern over the airport until it is his turn to land. If Captain Chip flies his plane in a circle whose radius is 50 km once every 30 min, what centripetal force must the air exert against the wings to keep the plane moving in a circle? m = 60,500 kg # rev = 1 Time = 30 minutes = 180 seconds f =? v=? r = 50 km = 50,000 m Fc= ? 1 πππ£ πππ£ = 0.0006 180 π ππ π πππ£ π π£ = 2π(50,000 π) (0.0006 ) = 175 π π π= π 2 (175 π ) πΉπ = π ∗ π = (60,500 ππ) ( ) = 36, 900 π 50,000 π
