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.

Answer 1

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

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

$v = 4.32 (0.45)$

$v = 1.94 m/s$