A slender rod is 90.0 cm long and has mass 0.120 kg. A small 0.0200 kg sphere is welded to one end of the rod, and a small 0.0800 kg sphere

Question

A slender rod is 90.0 cm long and has mass 0.120 kg. A small 0.0200 kg sphere is welded to one end of the rod, and a small 0.0800 kg sphere is welded to the other end. The rod, pivoting about a stationary, frictionless axis at its center, is held horizontal and released from rest.

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Yến Oanh 4 years 2021-08-07T02:59:44+00:00 1 Answers 1002 views 0

Answers ( )

  1. ThanhThu
    1
    2021-08-07T03:00:51+00:00
    Reply

    Answer:

    Speed of 0.08 kg mass when it will reach to the bottom position is 1.94 m/s

    Explanation:

    When rod is released from rest then due to unbalanced torque about the hinge the system will rotate

    Now moment of inertia of the system is given as

    I = \frac{ML^2}{12} + \frac{m_1L^2}{4} + \frac{m_2L^2}{4}

    now we have

    M = 0.120 kg

    m_1 = 0.02 kg

    m_3 = 0.08 kg

    now we have

    I = \frac{0.120(0.90)^2}{12} + \frac{0.02(0.90)^2}{4} + \frac{0.08(0.90)^2}{4}

    so we have

    I = 8.1 \times 10^[-3} + 4.05 \times 10^[-3} + 0.0162

    I = 0.02835

    now by energy conservation we can say work done by gravity must be equal to change in kinetic energy

    so we have

    \frac{1}{2}I\omega^2 = m_1g \frac{L}{2} - m_2 g\frac{L}{2}

    \frac{1}{2}(0.02835)\omega^2 = (0.08 - 0.02)(9.81)(0.45)

    \omega = 4.32 rad/s

    Now speed of 0.08 kg mass when it reaches to bottom point is given as

    v = \omega \frac{L}{2}

    v = 4.32 (0.45)

    v = 1.94 m/s

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