Answer:
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Part A
Calculate the angular velocity of the merry-go-round as a function of time.
Express your answer in terms of the variables β, γ, and t.
Part B
What is the initial value of the angular velocity?
Part C
Calculate the instantaneous value of the angular velocity ωz at t= 5.35 s .
Part D
Calculate the average angular velocity ωav−z for the time interval t=0 to t= 5.35 s .
Part A:
Angular velocity is the rate of change of the angle, so we can calculate it by taking the derivative of θ(t) with respect to time:
ω(t) = dθ(t)/dt = γ + 3βt^2
Therefore, the angular velocity of the merry-go-round as a function of time is given by:
ω(t) = γ + 3βt^2
Part B:
The initial value of the angular velocity is the angular velocity at t=0. Plugging in t=0 into the equation above, we get:
ω(0) = γ + 3β(0)^2 = γ
So the initial value of the angular velocity is simply γ.
Part C:
To find the instantaneous value of the angular velocity at t=5.35 s, we can plug in t=5.35 into the equation we found in part A:
ω(5.35) = γ + 3β(5.35)^2 = 0.422 + 3(1.00×10^−2)(5.35)^2 ≈ 1.05 rad/s
Therefore, the instantaneous value of the angular velocity at t=5.35 s is approximately 1.05 rad/s.
Part D:
The average angular velocity ωav−z for the time interval t=0 to t=5.35 s is given by the change in the angle divided by the time interval:
ωav−z = [θ(5.35) - θ(0)]/(5.35 - 0)
Plugging in the values we are given, we get:
ωav−z = [γ(5.35) + β(5.35)^3/3 - γ(0) - β(0)^3/3]/5.35
ωav−z = [γ(5.35) + β(5.35)^3/3]/5.35
ωav−z ≈ 0.667 γ + 0.101 β
Therefore, the average angular velocity ωav−z for the time interval t=0 to t=5.35 s is approximately 0.667 times the value of γ plus 0.101 times the value of β.