Answer to Question #83303 in Classical Mechanics for Bianca
1) The radius of the earth at the equator is 6.4 x 10^6 m. Assume that the earth is spherical.
a) determine the instantaneous velocity of a point at the equator
b) determine the centripetal acceleration of a point at the equator
b) 3.4 x 10^-2 ms-2
a) Since the earth rotation is uniform, the instantaneous velocity is given by the length l of a path of a point at the equator divided by the time t during which this path is passed. Taking the circumference of the equator as a path, we have l = 2πr, where r is the radius of the earth. This path is passed in 24 hours, which constitutes 86400 seconds. The velocity is v = 2πr/t ≈ 465 m s–1.
Answer: 465 m s–1.
b) The centripetal acceleration is given by the formula a = v2/r. Hence, a = (465 m s–1)2/(6.4 × 106 m) ≈ 3.4 × 10–2 m s–2.
Answer: 3.4 × 10–2 m s–2.
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