The angular acceleration when the pencil is 10° from the vertical is 17 rad/s². The time for the pencil to hit the ground is 0.43 s.
Look at the attachment for the complete question. The weight of the pencil is in the middle of it. When the pencil has an angle of 10.0° from the vertical
- The force that applies is the weight force.
F = w = mg
g = the acceleration due to gravity = 9.8 m/s² - The position of weight force is r in the picture.
The weight force make a torque. - Look at the triangle
sin 10° = r ÷ 0.5L
r = 0.5L sin 10° - Moment of inertia for the pencil
I = (1/3) mL²
m = mass of the pencil = 10.0 gr = 0.010 kg
L = the length of the pencil = 15 cm = 0.150 m - According to Newton's second angular law
∑τ = Iα
Fr = Iα
mg 0.5L sin 10° = (1/3) mL²α
3g × 0.5 × sin 10° = Lα
1.5 × 9.8 × 0.174 = 0.150 α
α = 2.553 ÷ 0.150
α = 17 rad/s²
When the pencil hit the ground
- The angle from vertical
θ = 90° = π/2 radians = 3.14 ÷ 2 = 1.57 rad - The pencil moves in non-uniform circular motion
- The pencil's initial stop, ω₀ = 0
θ = ω₀ t + 0.5 αt²
θ = 0.5 αt²
2θ = αt²
2 × 1.57 = 17t²
t² = 3.14 ÷ 17
t² = 0.1847
[tex]t \:=\: \sqrt{0.1847}[/tex]
t = 0.43 s
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