Respuesta :
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
The time for the pencil to hit the ground if it falls from standing perfectly vertical and maintains the angular acceleration is 0.4299s.
How to calculate the time taken?
Newton's second angular law will be used to find the angular acceleration. This will be:
τ = I α
Wr = I α
Here, the weight is applied in the middle of the pencil, therefore,
sin 10 = r / (L/2)
Make r the subject of the formula
r = L/2 sin 10
Then, the moment of inertia will be:
I = 1/3 M L²
mg L / 2 sin 10 = (1/3 m L²) α
αf = 3/2 g / L sin 10
α = [3/2 × 9.8] / [0.150 sin10]
α = 17 rad/s²
The angle from the vertical to the ground will be denoted as:
θ = π / 2
t = √ (2 π / (2 α ))
t = √(π / α )
t = √(π / 17)
t = √3.142 / 17
t = [tex]\sqrt{0.1848}[/tex]
t = 0.4299s
Therefore, the time for the pencil to hit the ground is 0.4299s.
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