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

Answer:

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

$I_{1}=\frac{ML^{2}}{12}$

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

$I_{2}=2\times m\times \frac{L^{2}}{4}$

$I_{2}=\frac{mL^{2}}{2}$

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

I = I1 + I2

$I=\frac{ML^{2}}{12}+\frac{mL^{2}}{2}$

$I=\frac{(M +6m)L^{2}}{12}$