Question

A uniform, 4.5 kg, square, solid wooden gate 2.0 mm on each side hangs vertically from a frictionless pivot at the center of its upper edge. A 1.2 kg raven flying horizontally at 5.0 m/s flies into this door at its center and bounces back at 1.5 m/s in the opposite direction. What is the angular speed of the gate just after it is struck by the unfortunate raven?

Answer 1

Answer:

The angular velocity is  $w = 1.43\ rad/sec$

Explanation:

From the question we are told that

   The  mass of wooden gate  is $m_g = 4.5 kg$

    The  length of side is  L = 2 m

    The mass of the raven is  $m_r = 1.2 kg$

     The initial speed of the raven is $u_r = 5.0m/s$

     The final speed of the raven is   $v_r = 1.5 m/s$

From the law of  conservation of angular momentum we express this question mathematically as

       Total initial angular momentum  of both the Raven and  the Gate =  The Final angular momentum of both the Raven and the Gate  

The initial angular momentum of the Raven is $m_r * u_r * \frac{L}{2}$

Note: the length is half because the Raven hit the gate at the mid point

The initial angular momentum of the Gate is  zero

Note: This above is the generally formula for angular momentum of  square objects

  The final angular velocity  of the Raven is  $m_r * v_r * \frac{L}{2}$

   The  final angular velocity of the Gate  is   $\frac{1}{3} m_g L^2 w$

Substituting this formula

  $m_r * u_r * \frac{L}{2} = \frac{1}{3} m_g L^2 w + m_r * v_r * \frac{L}{2}$

  $\frac{1}{3} m_g L^2 w = m_r * v_r * \frac{L}{2} - m_r * u_r * \frac{L}{2}$

  $\frac{1}{3} m_g L^2 w = m_r * \frac{L}{2} * [u_r - v_r]$

Where $w$ is the angular velocity

     Substituting value  

   $\frac{1}{3} (4.5)(2)^2 w = 1.2 * \frac{2}{2} * [5 - 1.5]$

     $6w = 4.2$

       $w = \frac{6}{4.2}$

            $w = 1.43\ rad/sec$