Question

A thin, uniform stick of length 1.9 m and mass 3.1 kg is pinned through one end and is free to rotate. The stick is initially hanging vertically and at rest. You then rotate the stick so that you are holding it horizontally. You release the stick from that horizontal position. What is the magnitude of the angular acceleration of the stick when it has traveled 23.4 degrees (the stick makes an angle of 23.4 degrees with the horizontal)

Answer 1

Answer:

The acceleration is  $\alpha = 7.10 \ rad/s^2$

Explanation:

From the question we are told that

  The length of the stick is  $d = 1.9 \ m$

  The mass of the stick is  $m = 3.1 \ kg$

  The angular displacement is  $\theta = 23.4^o$

Generally the torque of this uniform stick after this displacement is mathematically represented as

    $\tau = \frac{1}{2} * d * [m*g]* sin (90 - \theta)$

     $\tau = \frac{1}{2} * 1.9 * [3.1*9.8]* sin (90 - 23.4)$

    $\tau = 26.49 \ kg\cdot m^2 \cdot s^{-2}$

Generally the moment of inertia of the uniform stick is mathematically represented as  

         $I = \frac{1}{3} * m * d^2$

=>      $I = \frac{1}{3} * 3.1 * 1.9 ^2$

=>      $I = 3.73 \ kg \cdot m^2$

Generally the angular acceleration is mathematically represented as

       $\alpha = \frac{\tau}{I}$

=>    $\alpha = \frac{26.49}{3.73}$

=>    $\alpha = 7.10 \ rad/s^2$