4.1 Motion in a Circle p. 120 Gravity and Motion The force that keeps a body firmly attached to the ground is the same force that keeps the Moon in orbit around the Earth and keeps the Earth going around the Sun and is called gravitational force. All satellites and planets orbit larger bodies in elliptical orbits. These orbits are very close to being circles, and therefore circular orbits will be studied in detail in this chapter. Uniform Circular Motion Any object that is travelling in a circle at a constant speed is said to be in uniform circular motion. Acceleration is the change in velocity which is a vector and has both speed and direction. An object undergoing uniform circular motion will be accelerating, even though it has a constant speed, due to its changing direction.

4.1 Motion in a Circle p. 121 Identifying Centripetal Acceleration vovo vfvf vovo vfvf -Δv-Δv An object moving in uniform circular motion. The velocity of the object at two different positions of the object at two different times are shown above. Using just the velocity vectors, the change in velocity shows the direction of the acceleration. a = ΔvΔv ΔtΔt As the change in time approaches zero, Δt  0, the direction of the change in velocity becomes ever closer to the center of the circular path. The direction of the acceleration of an object moving in uniform circular motion is towards the center of the circle and is called centripetal acceleration

4.1 Motion in a Circle p. 122 vovo v1v1 Direction of Centripetal Acceleration R0R0 R1R1 ѳ vovo v1v1 -Δv-Δv R0R0 ΔRΔR R1R1 ѳ Radius Vectors: R 1 + (-R 0 ) = ΔR ѳ Velocity Vectors: v 1 + (-v 0 ) = Δv Similar Triangles R 0 + ΔR = R 1 (-v 0 ) + Δv = v 1 Examine two triangles formed when a body moves through uniform circular motion. ΔvΔv ΔRΔR v R = At very small time intervals R 1 = R 0 = R and v 1 = v o = v and:

4.1 Motion in a Circle p. 122 ΔvΔv ΔRΔR v R = Defining Centripetal Acceleration From previous slide: Therefore: ΔvΔv v R = ΔRΔR And:a = ΔvΔv ΔtΔt v R = ΔRΔR ΔtΔt R0R0 ΔRΔR R1R1 ѳ ΔsΔs As Δt, time interval, becomes smaller, Δt  0, R 0 = R 1 and ΔR = Δs (arc length). a = v R * v * Δt ΔtΔt v R * v* v v = ΔsΔs ΔtΔt ΔRΔR ΔtΔt = ΔR = v * Δt = a c = v2v2 R Finally: Centripetal Acceleration Equation (Magnitude of acceleration)

4.1 Motion in a Circle p. 123 Another Way to Calculate Centripetal Acceleration Distance around a circle is called circumference and is equal to C = 2πR. R v = d ΔtΔt = 2πR2πR ΔtΔt a c = v2v2 R And: = (2πR) 2 Δt2Δt2 a c = = 4π2R24π2R2 Δt2Δt2 R * 4π2R4π2R T2T2 Centripetal Acceleration (where Δt = T, the period)

4.1 Motion in a Circle p. 124 Centripetal Force Remember Newton’s 2nd Law:F net = m * a Therefore centripetal Force (center-seeking force) must be equal to: F c = m v2v2 R F c = m * 4π2R4π2R T2T2 or

4.1 Motion in a Circle In this section, you should understand how to solve the following key questions. Page#124Quick Check #1 – 3 Page #125 Practice Problems Centripetal Force #1– 3 Page #129– Review Questions #1 – 12