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Physics Chp 7

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Angular displacement θ or ∆θ ∆θ = ∆s/r s is the arc length r is the radius s = 2πr 2π = 360 o or π = 180 o

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Counterclockwise is the traditional + direction while clockwise is the traditional – This is based on the right-hand rule (cover later)

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Moving in an angle in an amount of time causes a velocity and if that changes it also makes for an acceleration. Angular Velocity ω in rad/s ω = ∆θ/t

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Angular acceleration α α = ∆ω/t in rad/s 2

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How fast is the earth turning in angular terms?

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r = 6.38x10 6 m T = 24 hrs s = 2 π r = 2 π (6.38x10 6 m) = ∆θ = s/r = 2 π r /r = 2 π ω = ∆θ/t t = T = 24hrs or 86400s ω = 2 π /86400s = rad/s

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Angular motion equations ω f = ω o + αt θ = ½(ω f + ω o )t θ = ω o t + 1/2αt 2 ω f 2 = ω o 2 + 2αθ

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Tangential Speed – linear on the edge of the circle v T = rω

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Tangential acceleration – changing direction a T = rα

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Centripetal acceleration – moving in towards the center always a c = v T 2 /r a c = (rω) 2 /r a c = rω 2

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If a 25 cm diameter bowling ball leaves your hand and travels the 15 m before hitting the pins in 2.6 s, what is it’s angular velocity assuming it maintains a constant velocity?

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D = 25 cm r = 12.5 cm ∆x = 15 m t = 2.6 s v T = 15m/2.6s = 5.8 m/s v T = rω ω =v T /r ω = 5.8m/s / 0.125m = ω = 46 rad/s

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What is it’s angular acceleration if it stops after hitting the pins and coming to rest in a total of 0.75 m.

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ω o = 46 rad/s ω f = 0 ∆x = 0.75m θ = s/r = 0.75m / 0.125m = 6 rad ω f 2 = ω o 2 + 2αθ 0 2 = (46 rad/s) 2 + 2α (6 rad) α = 180 rad/s 2

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Centripetal Force F c = ma c You can substitute into ac either a c = v T 2 /r or a c = rω 2

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Newton’s Law of Gravitation F g = Gm 1 m 2 /r 2 G = 6.673 x 10 -11 Nm 2 /kg 2 The larger the masses or the closer they are to each other the greater the force of attraction.

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The value for “g” can be determined for different locations compared to the center of the earth using Newton’s Law F g = Gm 1 m 2 /r 2 = m 2 g m 1 = mass of the earth and r is how far from the center of the earth to the center of the m 2

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g = Gm 1 /r 2 So at the top of Mt. McKinley or Denali how large is the value of g?

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r = radius of earth + elevation of Denali r = 6.37 x 10 6 m + 6200 m = 6.3762 x 10 6 m m 1 is the mass of the earth g = Gm 1 /r 2 g = (6.673 x 10 -11 Nm 2 /kg 2 )(5.98x10 24 kg) / (6.3762 x 10 6 m) 2 g = 9.?? m/s 2

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