Question

A slender rod is 80.0 cm long and has mass 0.390 kg . A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.0500-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.
What is the linear speed of the 0.0500-kg sphere as its passes through its lowest point?

Answer 1

Answer

given,

length of slender rod =80 cm = 0.8 m

mass of rod = 0.39 Kg

mass of small sphere = 0.0200 kg

mass of another sphere weld = 0.0500 Kg

calculating the moment of inertia of the system

$I = \dfrac{ML^2}{12}+\dfrac{m_1L^2}{4}+\dfrac{mL^2}{4}$

$I = \dfrac{0.39\times 0.8^2}{12}+\dfrac{0.02\times 0.8^2}{4}+\dfrac{0.05\times 0.8^2}{4}$

$I =0.032\ kg.m^2$

using conservation of energy

$\dfrac{1}{2}I\omega^2 = (m_1-m_2)g\dfrac{L}{2}$

$\omega=\sqrt{\dfrac{(m_1-m_2)gL}{I}}$

$\omega=\sqrt{\dfrac{(0.05-0.02)\times 9.8 \times 0.8}{0.032}}$

$\omega=2.71 \rad/s$

we know,

v = r ω

$v = \dfrac{L}{2} \times 2.71$

$v = \dfrac{0.8}{2} \times 2.71$

v = 1.084 m/s