Determine the mass moment of inertia of the assembly

  • #1
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Homework Statement


media%2Fdca%2Fdca53f80-5326-4c2b-98d0-3142a2aa5bd8%2FphpUyg13w.png

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?
 

Answers and Replies

  • #2
TSny
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EDIT: When getting ##I## for the semi-cylinder, does ##(1/2)MR^2## give you the rotational inertia about the CM of the semi-cylinder?
 
  • #3
BvU
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What does the parallel axis theorem say about which moments of inertia ?
 
  • #4
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EDIT: When getting ##I## for the semi-cylinder, does ##(1/2)MR^2## give you the rotational inertia about the CM of the semi-cylinder?

Ow, so the moment of inertia for the semi cylinder with (1/2)MR^2 is not around the the center of the semi cylinder but around the center of the full cylinder it would make? So in this case it would be around the z-axis which passes through y=200 mm?
 
  • #5
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So 0.5*5*(0.2)^2+(0.203-0.2)^2*5= I(cylinder)

Together with the I(plate) gives 0.227. So why is it still off?
 
  • #6
BvU
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What does the parallel axis theorem say about which moments of inertia ?
Meaning you want to make a different kind of calculation. Answer to #4 is yes.
 
  • #7
TSny
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So 0.5*5*(0.2)^2+(0.203-0.2)^2*5= I(cylinder)

Together with the I(plate) gives 0.227. So why is it still off?
The parallel axis theorem ##I = I_{cm} + md^2## involves the rotational inertia about the center of mass of the object, ##I_{cm}##. Your calculation uses ##(1/2)MR^2## instead of ##I_{cm}##.
 

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