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Physics Chp 7. 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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Presentation on theme: "Physics Chp 7. 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."— Presentation transcript:

1 Physics Chp 7

2 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

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

4 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

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

6 How fast is the earth turning in angular terms?

7 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

8 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

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

10 Tangential acceleration – changing direction a T = rα

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

12 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?

13 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

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

15 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

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

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

18 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

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

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


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