Answer:
The acceleration is [tex]\alpha = 7.10 \ rad/s^2[/tex]
Explanation:
From the question we are told that
The length of the stick is [tex]d = 1.9 \ m[/tex]
The mass of the stick is [tex]m = 3.1 \ kg[/tex]
The angular displacement is [tex]\theta = 23.4^o[/tex]
Generally the torque of this uniform stick after this displacement is mathematically represented as
[tex]\tau = \frac{1}{2} * d * [m*g]* sin (90 - \theta)[/tex]
[tex]\tau = \frac{1}{2} * 1.9 * [3.1*9.8]* sin (90 - 23.4)[/tex]
[tex]\tau = 26.49 \ kg\cdot m^2 \cdot s^{-2}[/tex]
Generally the moment of inertia of the uniform stick is mathematically represented as
[tex]I = \frac{1}{3} * m * d^2[/tex]
=> [tex]I = \frac{1}{3} * 3.1 * 1.9 ^2[/tex]
=> [tex]I = 3.73 \ kg \cdot m^2[/tex]
Generally the angular acceleration is mathematically represented as
[tex]\alpha = \frac{\tau}{I}[/tex]
=> [tex]\alpha = \frac{26.49}{3.73}[/tex]
=> [tex]\alpha = 7.10 \ rad/s^2[/tex]
