- #1

- 68

- 10

- Homework Statement
- A bar of length 10cm and diameter 2cm is positioned vertically, it has another bar of length 5cm and diameter 2cm attached at a right angle to its middle. The density is 800 kg m^-3. Find the position of the centre of mass. Then find the moments of inertia about 3 perpendicular axis through its centre of mass.

- Relevant Equations
- Xcm = m1x1 + m2x2 / m1 + m2

I = mr^2/2 (for a bar)

Using the equation above I get Xcm = 0.022 m. I set the origin be at the left of the vertical rod parallel to its centre of mass as in the diagram. But I’m not sure if the equation is correct for 3d.

for the moments of inertia I am using

I = Icm + md^2

= (mr^2)/2 + md^2

where d is the distance from the centre of mass to the axis. Does this equation seem correct?

for the moments of inertia I am using

I = Icm + md^2

= (mr^2)/2 + md^2

where d is the distance from the centre of mass to the axis. Does this equation seem correct?