The angular acceleration of the pencil when it makes an angle of 10.0 degrees with the vertical is 2.553 rad/s² of every meter length of the pencil.
If the pencil is in the position 10.0 degrees with the vertical, the weight of the pencil in the middle of the pencil will go down. Look at the triangle, x is the distance to calculate the torque.
- [tex]\frac{x}{\frac{1}{2} L} = sin \: 10^o[/tex]
[tex]x = \frac{1}{2} L \: sin \: 10^o[/tex] - Moment of inertia for a cylindrical rod-shaped object with a shaft at the end [tex]\frac{1}{3} mL^2[/tex].
Newton's second law of rotation
∑τ = Iα
wx = Iα
[tex]w \frac{1}{2} L \: sin \: 10^o = \frac{1}{3} m L^2 \alpha[/tex]
[tex]mg \frac{1}{2} \: sin \: 10^o = \frac{1}{3} mL \alpha[/tex]
[tex]g \frac{1}{2} \: sin \: 10^o = \frac{1}{3} L \alpha[/tex]
[tex]\alpha = \frac{g \frac{1}{2} \: sin \: 10^o}{\frac{1}{3} L}[/tex]
[tex]\alpha = \frac{3 g \: sin \: 10^o}{2L}[/tex]
[tex]\alpha = \frac{3 \times 9.8 \times 0.1736}{2L}[/tex]
[tex]\alpha = \frac{2.553}{L}[/tex]
Because the length of pencil unknown
α = 2.553 rad/s² of every meter length of the pencil.
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