Transcript Physics Chp 7

Physics Chp 7
• Angular displacement θ or ∆θ
• ∆θ = ∆s/r
s is the arc length r is the radius
• s = 2πr
• 2π = 360o or π = 180o
• Counterclockwise is the traditional + direction
while clockwise is the traditional –
• This is based on the right-hand rule (cover
later)
• 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
• Angular acceleration α
• α = ∆ω/t in rad/s2
• How fast is the earth turning in angular terms?
r = 6.38x106 m T = 24 hrs
• s = 2 π r = 2 π (6.38x106m) =
• ∆θ = s/r = 2 π r /r = 2 π
• ω = ∆θ/t
t = T = 24hrs or 86400s
• ω = 2 π /86400s =
rad/s
Angular motion equations
• ωf = ωo + αt
• θ = ½(ωf + ωo)t
• θ = ωot + 1/2αt2
• ω f 2 = ωo2 + 2αθ
• Tangential Speed – linear on the edge of the
circle
• vT = rω
• Tangential acceleration – changing direction
• aT = rα
• Centripetal acceleration – moving in towards
the center always
• ac = vT 2/r
• ac = (rω)2/r
• ac = rω2
• 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?
• D = 25 cm r = 12.5 cm ∆x = 15 m t = 2.6 s
• vT = 15m/2.6s = 5.8 m/s
• vT = rω ω =vT/r ω = 5.8m/s / 0.125m =
• ω = 46 rad/s
• What is it’s angular acceleration if it stops
after hitting the pins and coming to rest in a
total of 0.75 m.
• ωo = 46 rad/s ωf = 0 ∆x = 0.75m
• θ = s/r = 0.75m / 0.125m = 6 rad
• ω f 2 = ωo2 + 2αθ
• 02 = (46 rad/s)2 + 2α (6 rad)
• α = 180 rad/s2
• Centripetal Force
• Fc = mac
•
•
•
•
You can substitute into ac either
ac = vT 2/r
or
ac = rω2
• Newton’s Law of Gravitation
• Fg = Gm1m2/r2
• G = 6.673 x 10-11 Nm2/kg2
• The larger the masses or the closer they are to
each other the greater the force of attraction.
• The value for “g” can be determined for
different locations compared to the center of
the earth using Newton’s Law
• Fg = Gm1m2/r2 = m2g m1 = mass of the earth
and r is how far from the center of the earth
to the center of the m2
• g = Gm1/r2
• So at the top of Mt. McKinley or Denali how
large is the value of g?
• r = radius of earth + elevation of Denali
• r = 6.37 x 106 m + 6200 m = 6.3762 x 106 m
• m1 is the mass of the earth
• g = Gm1/r2
• g = (6.673 x 10-11 Nm2/kg2)(5.98x1024kg) / (6.3762 x 106m)2
• g = 9.?? m/s2