Problem 104
θ = 4.0t – 3.0t
2
+ t
3
a)
At t = 2.0
ω = 4.0 rad/s
b)
At t = 4.0s
ω = 28 rad/s
e)
α = 18 rad/s
2
The angular position of a point on the rim of a rotating wheel is given by θ = 4.0 t – 3.0
t
2
+
t
3
, where θ is in
radians and
t
is in seconds. What are the angular velocities at (a)
t
= 2.0 s and (b)
t
= 4.0 s? (c) What is the
average angular acceleration for the time interval that begins at
t
= 2.0 s and ends at
t
= 4.0 s? What are the
instantaneous angular accelerations at (d) the beginning and (e) the end of this time interval?
ω = dθ/dt =4.0 – 6.0t + 3.0t
2
α = dω/dt = 6.0 + 6.0t
c)
α
av
= ∆ω/∆t = (24 rad/s)/2.0 s
α
av
= 12 rad/s
2
d)
α = 6.0 rad/s
2
ω = 4.0 – 6.0(2.0) + 3.0(2.0)
2
ω = 4.0 – 6.0(4.0) + 3.0(4.0)
2
α =
6.0 + 6.0(2.0)
At t = 2.0
At t = 4.0s
α =
6.0 + 6.0(4.0)
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View Full DocumentIn the time it takes the light to travel 2L, the
wheel turns through an angle of
∆θ = 2π/500
∆θ = 1.26 x 10
2
rad
The time required to travel 2L is
∆t = 2L/c
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 Spring '08
 Dr.Schrieber
 Physics, General Relativity, Light, Physical quantities

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