Answer:
α = 17 rad / s²
, t = 0.4299 s
Explanation:
Let's use Newton's second angular law or torque to find angular acceleration
τ = I α
W r = I α
The weight is applied in the middle of the pencil,
sin 10 = r / (L/2)
r = L/2 sin 10
The pencil can be approximated by a bar that rotates on one end, in this case its moment of inertia is
I = 1/3 M L²
Let's calculate
mg L / 2 sin 10 = (1/3 m L²) α
α
f = 3/2 g / L sin 10
α = 3/2 9.8 / 0.150 sin 10
α = 17 rad / s²
If the pencil has a constant acceleration we can use the angular kinematics relationships, as part of the rest wo = 0
θ = w₀ t + ½ α t²
t = RA (2θ / α
)
The angle from the vertical to the ground is
θ = π / 2
t = √ (2 π / (2 α
))
t = √ (π / α
)
t = √ (π / 17)
t = 0.4299 s