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

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 = (ML^2)/(12)+(m_1L^2)/(4)+(mL^2)/(4)`

`I = (0.39* 0.8^2)/(12)+(0.02* 0.8^2)/(4)+(0.05* 0.8^2)/(4)`

`I =0.032\ kg.m^2`

using conservation of energy

`(1)/(2)I\omega^2 = (m_1-m_2)g(L)/(2)`

`\omega=\sqrt{((m_1-m_2)gL)/(I)}`

`\omega=\sqrt{((0.05-0.02)* 9.8 * 0.8)/(0.032)}`

`\omega=2.71 \rad/s`

we know,

v = r ω

`v = (L)/(2) * 2.71`

`v = (0.8)/(2) * 2.71`

v = 1.084 m/s