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
[tex]I = \frac{ML^2}{12} + \frac{m_1L^2}{4} + \frac{m_2L^2}{4}[/tex]
now we have
[tex]M = 0.120 kg[/tex]
[tex]m_1 = 0.02 kg[/tex]
[tex]m_3 = 0.08 kg[/tex]
now we have
[tex]I = \frac{0.120(0.90)^2}{12} + \frac{0.02(0.90)^2}{4} + \frac{0.08(0.90)^2}{4}[/tex]
so we have
[tex]I = 8.1 \times 10^[-3} + 4.05 \times 10^[-3} + 0.0162[/tex]
[tex]I = 0.02835[/tex]
now by energy conservation we can say work done by gravity must be equal to change in kinetic energy
so we have
[tex]\frac{1}{2}I\omega^2 = m_1g \frac{L}{2} - m_2 g\frac{L}{2}[/tex]
[tex]\frac{1}{2}(0.02835)\omega^2 = (0.08 - 0.02)(9.81)(0.45)[/tex]
[tex]\omega = 4.32 rad/s[/tex]
Now speed of 0.08 kg mass when it reaches to bottom point is given as
[tex]v = \omega \frac{L}{2}[/tex]
[tex]v = 4.32 (0.45)[/tex]
[tex]v = 1.94 m/s[/tex]