A thin rod (length = 2.97 m) is oriented vertically, with its bottom end attached to the floor by means of a frictionless hinge.
1 answer:
Answer:
a) w = 2.57 rad / s , b) α = 3.3 rad / s²
Explanation:
a) Let's use the conservation of mechanical energy, we will write it in two points the highest and when touching the ground
Initial. Higher
Em₀ = U = m g h
Final. Touching the ground
= K = ½ I w²
How energy is conserved
Em₀ =
mg h = ½ I w2
The moment of specific object inertia
I = m L²
We replace
m g h = ½ (mL²) w²
w² = 2g h / L²
The height of the object is the length of the bar
h = L
w = √ 2g / L
w = √ (2 9.8 / 2.97)
w = 2.57 rad / s
b) the angular acceleration can be found from Newton's second rotational law
τ = I α
W L = I α
mg L = (m L²) α
α = g / L
α = 9.8 / 2.97
α = 3.3 rad / s²
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