Question

A uniform bar has two small balls glued to its ends. The bar has length L and mass M, while the balls each have mass m and can be treated as point masses. What is the moment of inertia of this bar about an axis perpendicular to the bar through its center?

Answer 1

Step-by-step explanation:

Length of bar = L

mass of bar = M

mass of each ball = m

Moment of inertia of the bar about its centre perpendicular to its plane is

`I_(1)=(ML^(2))/(12)`

Moment of inertia of the two small balls about the centre of the bar perpendicular to its plane is

`I_(2)=2* m* (L^(2))/(4)`

`I_(2)=(mL^(2))/(2)`

Total moment of inertia of the system about the centre of the bar perpendicular to its plane is

I = I1 + I2

`I=(ML^(2))/(12)+(mL^(2))/(2)`

`I=((M +6m)L^(2))/(12)`