- #1

- 28

- 0

## Homework Statement

Determine the location of y of the center of mass G of the assembly and then calculate the moment of inertia about an axis perpendicular to the page and passing through G. The block has a mass of 3 kg and the mass of the semicylinder is 5 kg.

## Homework Equations

-parallel axis theorem

-moment of inertia plate

-moment of inertia semicylinder

- y= (Σydm)/(Σdm)

## The Attempt at a Solution

The block has center at y=350, dm= 3, ydm= 1050

The semicylinder has center at y=115,117, dm=5 and ydm= 575,585

So y(center)= (1050+575,585)/(8)= 203,198 mm

(answer in back of book is 203 mm, so this seems correct)

now I(block) = (1/12)*m*(a^2+b^2)+ md^2= (1/12)*3*(0.3^2+0.4^2) + 3*(0.203198-0.350)^2

=> 0.1271524816 kg*m^2 (as you can see I converted mm in m, which explains the 0.203198 instead of 203.198 for example)

now I(semicylinder)= (1/2)*m*R^2 + md^2= (1/2)*5*(0.2)^2+ 5* (0.203198-0.115117)^2

=> 0.13879 kg*m^2 (I heard, but also verified myself that the moment of inertia for semicylinder is same as for full cylinder, hence 1/2*m*R^2.

The answer should be 0.230 kg* m^2

but i get 0.266 kg* m^2

So what did I do wrong? Is the moment of inertia for semicylinder not correct afterall?