Answer:
The average angular acceleration of the Earth is; α = 6.152 X 10⁻²⁰ rad/s²
Explanation:
We are given;
The period of 365 revolutions of Earth in 2006, T₁ = 365 days, 0.840 sec
Converting to seconds, we have;
T₁ = (365 × 24 × 60 × 60) + 0.84
T₁ = (3.1536 x 10⁷) + 0.840
T₁ = 31536000.84 s
Now, the period of 365 rotation of Earth in 2006 is; T₀ = 365 days
Converting to seconds, we have;
T₀ = 31536000 s
Hence, time period of one rotation in the year 2006 is;
Tₐ = 31536000.84/365
Tₐ = 86400.0023 s
The time period of rotation is given by the formula;
Tₐ = 2π/ωₐ
Making ωₐ the subject;
ωₐ = 2π/Tₐ
Plugging in the relevant values;
ωₐ = 2π/ 365.046306
ωₐ = 7.272205023 x 10⁻⁵ rad/s
Therefore, the time period of one rotation in the year 1906 is;
Tₓ = 31536000/365
Tₓ = 86400 s
Time period of rotation,
Tₓ = 2π /ωₓ
ωₓ = 2π / T
Plugging in the relevant values;
ωₓ = 2π/86400
ωₓ = 7.272205217 x 10⁻⁵ rad/s
The average angular acceleration is given by;
α = (ωₓ - ωₐ) / T₁
α = ((7.272205217 × 10⁻⁵) - (7.272205023 × 10⁻⁵)) / 31536000.84
α = 6.152 X 10⁻²⁰ rad/s²
Thus, the average angular acceleration of the Earth is; α = 6.152 X 10⁻²⁰ rad/s²
