Atall, cylindrical chimney falls over when its base is ruptured. treat the chimney as a thin rod of length 53.2 m. at the instant it makes an angle of 34.1° with the vertical as it falls, what are (a) the radial acceleration of the top, and (b) the tangential acceleration of the top. (hint: use energy considerations, not a torque.) (c) at what angle θ is the tangential acceleration equal to g? assume free-fall acceleration to be equal to 9.81 m/s2
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Ответ:
ω^2 = 3g / h sin θ
Substituting the given values:
ω^2 = 3 (9.81) / 53.2 sin 34.1
ω^2 = 0.987 /s
The formula for radial acceleration is:
a = rω^2
So,
a = 53.2 (0.987) = 52.494 /s^2
The linear velocity is:
v^2 = ar
v^2 = 52.949 (53.2) = 2816.887
The tangential acceleration is:
a = r v^2
a = 53.2 (2816.887)
a = 149858.378 m/s^2
If the tangential acceleration is equal to g:
g = r^2 3g / sin θ
Solving for θ
θ = 67°
Ответ: