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.

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?

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RobertKer 6 months 2021-07-30T12:14:31+00:00 1 Answers 3 views 0

Answers ( )

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    2021-07-30T12:15:40+00:00

    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

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