A bullet with a mass of 4.5 g is moving with a speed of 300 m/s (with respect to the ground) when it collides with a rod with a mass of 3 kg

Question

A bullet with a mass of 4.5 g is moving with a speed of 300 m/s (with respect to the ground) when it collides with a rod with a mass of 3 kg and a length of L = 0.25 m. The rod is initially at rest, in a vertical position, and pivots about an axis going through its center of mass which is located exactly halfway along the rod. The bullet imbeds itself in the rod at a distance L/4 from the pivot point. As a result, the bullet-rod system starts rotating together. What is the magnitude of the angular velocity (in rad/s) of the rotation (with respect to the ground) immediately after the collision? You may treat the bullet as a point particle.

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Eirian 4 years 2021-07-27T02:56:25+00:00 1 Answers 11 views 0

Answers ( )

  1. hongcuc2
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    2021-07-27T02:57:59+00:00
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    Answer:

    ω = 5.41 rad/s

    Explanation:

    Since the rod is rotating around its axis, angular momentum will play part in this question.

    The conservation of angular momentum implies that

    L_1 = L_2

    So, the initial angular momentum is

    L_1 = m_b v_b \frac{L}{4} = (4.5\times 10^{-3}~{\rm kg})(300~{\rm m/s})(\frac{0.25}{4}~{\rm m}) = 0.0844~{\rm kg.m^2/s}

    The final angular momentum includes the rod and the bullet together. So,

    L_2 = I\omega\\I = I_{rod} + I_{bullet} = \frac{1}{12}m_r L^2 + m_b(\frac{L}{4})^2 = \frac{1}{12}3(0.25)^2 + (4.5\times 10^{-3})(\frac{0.25}{4})^2 = 0.0156~{\rm kg.m^2}\\L_2 = L_1 = I\omega\\0.0844 = 0.0156\omega\\\omega = 5.41~{\rm rad/s}

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